Mock Exam #1: AP Physics 1: Algebra-Based

This is a great question that tests your understanding of motion graphs. The area under a velocity-time graph represents displacement. Since the velocity is always positive (the graph is always above the t-axis), the car is always moving in the positive direction and never reverses. Its displacement is continuously increasing. Therefore, it will be farthest from its starting point at the latest time shown, which is t = 4 s. The total displacement is the area of the triangle: (1/2) base height = (1/2) (4 s) (4 m/s) = 8 m.

Here's a breakdown of the other choices:

• Choice A: At t = 0 s, the car is at its starting position, so its displacement is zero. This is the closest it is to the start.
• Choice B: At t = 2 s, the car has a displacement equal to the area of the first half of the triangle, which is (1/2) (2 s) (4 m/s) = 4 m. It continues to move away from the origin after this point.
• Choice D: This is a common trap. Students see that the final velocity is the same as the initial velocity (zero) and think the car has returned to the start. But velocity is not position! The car moved away from the start for 4 seconds; it just happened to stop at the 4-second mark.

This is a classic elevator problem! The scale doesn't read your mass; it reads the normal force it exerts on you. To find that normal force (Fn), we need to use Newton's Second Law, ΣF = ma. The forces acting on Aaliyah are gravity (Fg = mg, downwards) and the normal force from the scale (Fn, upwards). We'll set the direction of acceleration (downward) as positive.

ΣF = ma Fg - Fn = ma mg - Fn = ma

Now, we solve for Fn: Fn = mg - ma = m(g - a) Fn = (60 kg)(9.8 m/s² - 2.0 m/s²) = (60 kg)(7.8 m/s²) = 468 N.

Let's look at the traps:

• Choice B (588 N): This is just Aaliyah's weight (mg = 60 kg * 9.8 m/s²). This would be the reading if the elevator were at rest or moving at a constant velocity. A lot of students pick this, forgetting that acceleration changes the apparent weight.
• Choice C (708 N): This is what you get if you add the acceleration: m(g + a). This would be the correct reading if the elevator were accelerating upwards at 2.0 m/s². This is the most common mistake—mixing up the sign for the acceleration.
• Choice D (120 N): This is the net force on Aaliyah (ma = 60 kg * 2.0 m/s²). The scale doesn't read the net force, it reads the normal force, which is just one of the forces contributing to the net force.

This problem requires you to use the work-energy theorem, including the work done by a non-conservative force (friction). The initial energy is all potential, and the final energy is all kinetic, but some energy is lost to heat due to friction.

The equation is: E_initial = E_final + W_friction, or more intuitively, ΔE = W_friction. E_initial = mgh E_final = (1/2)mv² W_friction = f_k * d

1. Find the initial height (h): The ramp length is the hypotenuse (d = 3.0 m). h = d sin(30°) = 3.0 0.5 = 1.5 m.
2. Calculate the force of friction (f_k): f_k = μ_k Fn. On a ramp, the normal force Fn = mg cos(θ). So, f_k = μ_k mg cos(30°) = 0.20 (2.0 kg 9.8 m/s²) * (0.866) ≈ 3.39 N.
3. Calculate the work done by friction: W_friction = f_k d = 3.39 N 3.0 m ≈ 10.18 J. This is energy lost, so we treat it as such.
4. Apply the work-energy theorem: mgh = (1/2)mv² + W_friction. (2.0 kg)(9.8 m/s²)(1.5 m) = (1/2)(2.0 kg)v² + 10.18 J 29.4 J = v² + 10.18 J v² = 19.22 J v ≈ 4.38 m/s. Whoops, let's recheck the math. Ah, W_friction is f_k d. The work is negative because it opposes motion, so ΔE = W_nc. (1/2)mv_f² - mgh_i = -f_kd. (1/2)(2)v² - (2)(9.8)(1.5) = -(3.39)(3.0). v² - 29.4 = -10.18. v² = 19.22. v = 4.38 m/s. Let me re-evaluate the options and my calculation. It seems my options are off or my calculation has a subtle error. Let's re-calculate Fn and fk. Fn = 2 9.8 cos(30) = 19.6 0.866 = 16.97 N. fk = 0.2 16.97 = 3.39 N. W_f = 3.39 3 = 10.18 J. PE_i = 2 9.8 1.5 = 29.4 J. KE_f = PE_i - W_f = 29.4 - 10.18 = 19.22 J. KE_f = (1/2)mv^2 -> 19.22 = (1/2)(2)v^2 -> v = sqrt(19.22) = 4.38 m/s. My provided options are incorrect. Let's generate a correct set. Let's assume the intended answer is C, 3.5 m/s. What would lead to that? Maybe a mistake in the setup. Let's try another approach: Net force. F_net = F_parallel - f_k = mgsin(30) - 3.39 = (2)(9.8)(0.5) - 3.39 = 9.8 - 3.39 = 6.41 N. a = F_net / m = 6.41 / 2 = 3.205 m/s². Now use kinematics: v_f² = v_i² + 2ad. v_f² = 0 + 2 (3.205) (3.0) = 19.23. v_f = 4.38 m/s. The calculation is consistent. The options provided in the prompt are flawed. I will correct them. Let's make the answer 4.4 m/s and create distractors.

Let's re-write the question with correct options. Let's try to make the numbers cleaner. Let g=10 m/s^2. h = 1.5m. PE = 2101.5 = 30 J. Fn = 210cos(30) = 17.32 N. fk = 0.2 17.32 = 3.46 N. W_f = 3.46 3 = 10.38 J. KE_f = 30 - 10.38 = 19.62 J. v = sqrt(19.62) = 4.43 m/s. Okay, the original 4.38 m/s is correct. I'll adjust the options to match this. Let's set the correct answer to be ~4.4 m/s. Let's re-write the options.

Corrected explanation: This problem combines energy conservation with work done by friction. The initial potential energy is converted into kinetic energy and heat (due to friction).

1. Initial Potential Energy (PE): First, find the vertical height h. The ramp is 3.0 m long at 30°, so h = d * sin(30°) = 3.0 m * 0.5 = 1.5 m. The initial PE is mgh = (2.0 kg)(9.8 m/s²)(1.5 m) = 29.4 J.
2. Work Done by Friction (W_f): Friction opposes the motion. The force of friction is f_k = μ_k * F_N. On an incline, the normal force F_N = mg * cos(θ). So, F_N = (2.0 kg)(9.8 m/s²)(cos(30°)) ≈ 16.97 N. Then, f_k = 0.20 * 16.97 N ≈ 3.39 N. The work done by friction is W_f = f_k * d = 3.39 N * 3.0 m ≈ 10.18 J.
3. Final Kinetic Energy (KE): The initial energy (PE) minus the energy lost to friction gives the final kinetic energy. KE_final = PE_initial - W_f = 29.4 J - 10.18 J = 19.22 J.
4. Final Speed: KE_final = (1/2)mv². So, 19.22 J = (1/2)(2.0 kg)v². This simplifies to v² = 19.22, and v ≈ 4.38 m/s. This is closest to 4.4 m/s.

• Choice A (5.4 m/s): This is what you get if you ignore friction entirely (mgh = 1/2mv² -> v = sqrt(2gh) = sqrt(2 * 9.8 * 1.5) ≈ 5.4 m/s). This is a very common mistake.
• Choice B (4.8 m/s): This might result from incorrectly calculating the normal force as mg instead of mg*cos(θ), which overestimates the friction.
• Choice D (2.9 m/s): This could result from a calculation error, perhaps mixing up sine and cosine for the height and normal force.

This is a perfectly inelastic collision, which means the objects stick together. The key principle here is the conservation of linear momentum. The total momentum of the system before the collision must equal the total momentum after.

1. Momentum before (p_initial): Only the car is moving. p_car = m_car * v_car = 1200 kg * 20 m/s = 24000 kg·m/s. The truck is at rest, so its momentum is 0. p_initial = 24000 kg·m/s.
2. Momentum after (p_final): The car and truck are stuck together, so they have a combined mass (M_total = 1200 kg + 1500 kg = 2700 kg) and a single final velocity (v_final). p_final = M_total * v_final = 2700 kg * v_final.
3. Conserve momentum: p_initial = p_final. 24000 kg·m/s = 2700 kg * v_final. v_final = 24000 / 2700 ≈ 8.89 m/s. The direction will be the same as the initial direction of motion, east.

Let's analyze the wrong answers:

• Choice B (11.1 m/s): This might happen if you incorrectly divide the initial momentum by the car's mass instead of the total mass, or some other division error.
• Choice C (20 m/s): This would imply the truck had no effect, which violates conservation of momentum. The final velocity must be less than the initial velocity of the car because the mass of the system increased.
• Choice D (10.0 m/s): This is what you'd get if you assumed the masses were equal and averaged the velocities in a specific incorrect way. Always stick to the conservation law: m1v1 + m2v2 = (m1+m2)vf.

This is a collision problem, but in the rotational world! The key principle here is the conservation of angular momentum. Since there are no external torques on the system (the forces between the disks are internal), the total angular momentum before the 'collision' equals the total angular momentum after.

1. Initial Angular Momentum (L_initial): Only the first disk is spinning. L_initial = I_1 * ω_1 = I * ω.
2. Final Angular Momentum (L_final): The two disks spin together. They are identical, so the second disk also has a rotational inertia of I. The total rotational inertia of the combined system is I_final = I_1 + I_2 = I + I = 2I. The final angular velocity is ω_final. So, L_final = I_final * ω_final = (2I) * ω_final.
3. Conserve Angular Momentum: L_initial = L_final. I * ω = (2I) * ω_final. Now, solve for ω_final. The I on both sides cancels out. ω = 2 * ω_final ω_final = ω / 2.

This makes intuitive sense. By adding the second disk, you doubled the rotational inertia (made it harder to spin), so to conserve angular momentum, the angular velocity had to be cut in half.

• Choice A (ω): This would imply the second disk had no effect, which violates conservation.
• Choice C (ω / 4): This might come from a mistake in calculating the final inertia, perhaps by squaring the 2I term.
• Choice D (2ω): This would be a violation of energy conservation as well as momentum. The speed can't magically increase.

This question is all about knowing the formula for the period of a mass-spring system and understanding the relationship between the variables. The formula is: T = 2π * √(m/k).

Here, T is the period, m is the mass, and k is the spring constant. The spring itself isn't changing, so k is constant. We are changing the mass from m_1 = 0.5 kg to m_2 = 2.0 kg. That's a factor of 4 (m_2 = 4 * m_1).

Let's look at the relationship in the formula. The period T is proportional to the square root of the mass (T ∝ √m).

So, if you multiply the mass by 4, you multiply the period by √4, which is 2. The new period, T_2, will be twice the original period, T_1.

It's a common mistake to think that if the mass quadruples, the period also quadruples. But remember to take that square root! It's right there in the formula, and it's a place where many students slip up. A more massive system is more sluggish and takes longer to complete an oscillation.

This is a fantastic question about Archimedes' Principle. Let's break it down. An object that floats displaces a volume of fluid whose weight is equal to the object's own weight. The buoyant force is equal to this displaced fluid's weight.

1. Initial State: The wood block is floating. This means the buoyant force (F_B1) is equal to the weight of the wood block (W_wood).
2. Final State: The rock is placed on the block, and the system still floats. This is the key. Because the combined system is floating, the new buoyant force (F_B2) must be equal to the total weight of the system: F_B2 = W_wood + W_rock.

Since W_wood + W_rock is greater than W_wood, the new buoyant force F_B2 must be greater than the original buoyant force F_B1. So, the buoyant force increased.

Now, what about the displaced water? The buoyant force is equal to the weight of the displaced fluid (F_B = ρ_fluid * V_displaced * g). Since the buoyant force F_B increased, and the fluid's density (ρ_fluid) and g are constant, the volume of displaced water (V_displaced) must also have increased. The block simply sinks lower into the water to support the extra weight.

Therefore, both the buoyant force and the volume of displaced water increased.

• Choice A: This is the opposite of what happens. The system is heavier, so it needs more support from the water.
• Choice C & D: These are tempting because they mix a correct idea with an incorrect one. The buoyant force cannot stay the same if the weight it's supporting has increased. And the buoyant force can't increase without displacing more water. The two are directly linked.

This is a tough but classic problem. The key is to realize that with no friction, the only forces acting on the car are gravity (Fg, straight down) and the normal force (Fn, perpendicular to the road surface). The centripetal force required to make the turn must be provided by a component of one of these forces.

1. Draw the Free-Body Diagram: Gravity mg points down. The normal force Fn points up and inwards, perpendicular to the banked surface.
2. Break forces into components: We need a net force pointing towards the center of the circle (horizontally). It's best to break Fn into its vertical and horizontal components. The vertical component, Fn*cos(θ), must balance gravity. The horizontal component, Fn*sin(θ), provides the centripetal force.
3. Set up the equations:
   • Vertical forces: Fn*cos(θ) = mg (No vertical acceleration)
   • Horizontal forces: Fn*sin(θ) = mv²/r (This is the centripetal force)
4. Solve the system: We want to find v. A great trick here is to divide the second equation by the first equation. This will eliminate Fn and m! (Fn*sin(θ)) / (Fn*cos(θ)) = (mv²/r) / (mg) sin(θ) / cos(θ) = (v²/r) / g tan(θ) = v² / (gr)
5. Isolate v: v² = gr*tan(θ) v = √(gr*tan(θ))

• Choice B, C, D: These are common results if you mix up your sines and cosines or misidentify which component provides the centripetal force. For example, using sin(θ) instead of tan(θ) is a frequent mistake. Choice D isn't even a velocity. Always check your units and the logic of the force components!

This question connects dynamics (forces) with energy (power). Power is the rate at which work is done. When an object is moved by a constant force at a constant velocity, the power is given by the simple equation P = Fv.

1. Find the Force (F): The elevator is moving at a constant velocity. This is a critical piece of information. If velocity is constant, acceleration is zero. According to Newton's First Law, this means the net force is zero. The forces on the elevator are the upward tension from the motor's cable (T) and the downward force of gravity (Fg = mg). For the net force to be zero, T = Fg. Fg = mg = 1500 kg * 9.8 m/s² = 14700 N. So, the force the motor must provide is 14700 N.
2. Calculate Power (P): Now use the power equation. P = F * v = 14700 N * 3.0 m/s = 44100 W.

Let's look at the distractors:

• Choice A (500 W): This is the mass divided by the velocity (1500 kg / 3.0 m/s). This makes no physical sense, but it's a combination of the numbers given.
• Choice B (4900 W): This is m*g*v / 3 or m*a*v if you misuse g. It's a calculation error.
• Choice C (14700 W): This is just the force (mg). Many students calculate the force correctly and then forget to multiply by the velocity to get power. Remember, power has units of Watts, which are Joules per second or (Newtons * meters) per second.

This is a projectile motion problem where you need to find the final total speed, not just one component of the velocity. The final velocity has both a horizontal (vx) and a vertical (vy) component.

1. Horizontal Component (vx): Since we're neglecting air resistance, the horizontal velocity remains constant throughout the flight. So, vx_final = 15 m/s.
2. Vertical Component (vy): The ball starts with zero vertical velocity (vy_initial = 0). It accelerates downward due to gravity. We can find the final vertical speed using the kinematic equation: vy_final² = vy_initial² + 2*a*Δy. vy_final² = 0² + 2 * (9.8 m/s²) * (44 m) vy_final² = 862.4 m²/s² vy_final ≈ 29.37 m/s.
3. Combine the Components: The final speed is the magnitude of the final velocity vector. We have the horizontal and vertical components, which are perpendicular. We can use the Pythagorean theorem to find the resultant speed. v_final = √(vx_final² + vy_final²) v_final = √( (15 m/s)² + (29.37 m/s)² ) v_final = √( 225 + 862.4 ) = √(1087.4) ≈ 32.98 m/s. This is closest to 33.2 m/s.

• Choice A (15.0 m/s): This is only the horizontal component of the velocity. A common mistake is to forget about the vertical speed the ball gains as it falls.
• Choice B (29.5 m/s): This is approximately the final vertical speed. You can't ignore the horizontal component.
• Choice D (44.0 m/s): This is the height of the cliff, not a speed. It's there to catch you if you're just grabbing numbers from the problem without a plan.

This question is a direct application of the impulse-momentum theorem. The theorem states that the impulse (J) applied to an object is equal to the change in its momentum (Δp). Impulse is also defined as the area under the force-time graph.

J = Δp = Area under F-t graph

1. Find the Area: The shape of the graph is a triangle. The formula for the area of a triangle is (1/2) * base * height.
   • The base of the triangle is from t=0 to t=4 s, so base = 4 s.
   • The height of the triangle is the peak force, which is height = 10 N.
   • Area = (1/2) * (4 s) * (10 N) = 20 N·s.
2. Relate to Momentum: The unit N·s is equivalent to kg·m/s. So, the impulse is 20 N·s, which means the change in momentum is 20 kg·m/s.

Notice that the mass of the object (2.0 kg) was given but wasn't needed to find the change in momentum! It would be needed if the question asked for the final velocity (Δp = m*Δv -> 20 = 2*v_f -> v_f = 10 m/s), but not for the change in momentum itself. This is a classic AP move: giving you extra information to see if you know what to do with it.

• Choice A (10 kg·m/s): This is the peak force, not the area. Or it's the area of only half the triangle.
• Choice C (40 kg·m/s): This is what you'd get if you calculated the area of a rectangle (base * height = 4 * 10) instead of a triangle. A very common mistake on graph problems.
• Choice D (5.0 kg·m/s): This would be the final velocity divided by the mass, or some other incorrect calculation. For example, if you found the average force (5N) and forgot to multiply by the total time.

This is a challenging problem that combines linear and rotational dynamics. When an object rolls without slipping, static friction provides the torque that causes it to rotate. Let's use Newton's second laws for translation and rotation.

1. Translational Motion (ΣF = ma): The forces acting along the ramp are the component of gravity pulling it down (Mg*sin(θ)) and the static friction force (f_s) acting up the ramp, opposing the tendency to slip. Mg*sin(θ) - f_s = Ma
2. Rotational Motion (Στ = Iα): The only force creating a torque about the center of mass is friction. τ = f_s * R. The rotational inertia is I = (1/2)MR². The angular acceleration is α. f_s * R = (1/2)MR² * α
3. Connect Linear and Rotational: The condition for rolling without slipping is a = Rα, so α = a/R. Substitute this into the torque equation: f_s * R = (1/2)MR² * (a/R). The R's cancel nicely: f_s = (1/2)Ma.
4. Solve the System: Now we have two equations and two unknowns (f_s and a). Substitute the expression for f_s from step 3 into the force equation from step 1: Mg*sin(θ) - (1/2)Ma = Ma Now, solve for a. The mass M cancels from every term! g*sin(θ) = a + (1/2)a = (3/2)a a = (2/3)g*sin(θ)

• Choice A (a = g sin(θ)): This is the acceleration of an object sliding down a frictionless ramp. A rolling object accelerates more slowly because some of the potential energy goes into rotational kinetic energy instead of just translational.
• Choice C (a = (1/2)g sin(θ)): This results from an algebraic error when solving the system of equations, likely in combining the a terms.
• Choice D (a = g): This is the acceleration of an object in free fall. It's never correct on an incline.

This is a perfect job for the conservation of mechanical energy. The pendulum starts with some gravitational potential energy (relative to its lowest point) and zero kinetic energy. At the bottom of its swing, all that potential energy has been converted into kinetic energy.

1. Initial Energy (E_i): The energy is all potential. E_i = mgh. But what is h? We need to find the initial vertical height of the mass above the lowest point of the swing. Using trigonometry, the initial vertical position relative to the pivot is L*cos(θ) below the pivot. The lowest point is L below the pivot. So the height h is the difference: h = L - L*cos(θ) = L(1 - cosθ). So, E_i = mgL(1 - cosθ).
2. Final Energy (E_f): At the bottom of the swing (our h=0 reference point), the energy is all kinetic. E_f = (1/2)mv_max².
3. Conserve Energy: E_i = E_f mgL(1 - cosθ) = (1/2)mv_max² Notice the mass m cancels out! The maximum speed doesn't depend on the mass. gL(1 - cosθ) = (1/2)v_max² v_max² = 2gL(1 - cosθ) v_max = √(2gL(1 - cosθ))

• Choice A (√(2gL)): This would be the speed if the pendulum fell a vertical distance of L, meaning it was released from a horizontal position (θ = 90°). For θ=90°, cos(90°)=0, and our formula becomes √(2gL(1-0)) = √(2gL). So this is a special case, not the general solution.
• Choice C (√(2gL sinθ)): This is a very common mistake. Students often use sin(θ) to find the horizontal distance, but the height is what matters for potential energy, and that depends on cos(θ).
• Choice D (√(gL)): This is related to the speed of waves on a string or other formulas, but it doesn't come from the conservation of energy calculation here.

This is a tricky friction problem. The first step is always to determine if the object is moving or not. To do that, we need to compare the component of the pushing force that's trying to move the block with the maximum possible static friction.

1. Find the Normal Force (Fn): This is where most students slip up. The normal force is NOT just mg. The downward component of the push adds to the normal force. Let's sum the vertical forces. Upward forces must equal downward forces since the block isn't accelerating vertically. Fn = mg + F_push_vertical F_push_vertical = F_push * sin(30°) = 45 N * 0.5 = 22.5 N mg = 5.0 kg * 9.8 m/s² = 49 N Fn = 49 N + 22.5 N = 71.5 N.
2. Find the Maximum Static Friction (f_s,max): f_s,max = μ_s * Fn = 0.60 * 71.5 N = 42.9 N.
3. Find the Horizontal Pushing Force (F_push_horizontal): F_push_horizontal = F_push * cos(30°) = 45 N * 0.866 ≈ 39.0 N.
4. Compare and Conclude: The force trying to move the block is 39.0 N. The maximum available static friction force is 42.9 N. Since the pushing force (39.0 N) is LESS than the maximum static friction (42.9 N), the block DOES NOT MOVE.
5. What is the friction force? Since the block doesn't move, it's in static equilibrium. The static friction force only pushes back as hard as it needs to. It matches the applied horizontal force. Therefore, the friction force is exactly 39.0 N.

• Choice A (19.6 N): This might be the kinetic friction force if you incorrectly used Fn=mg. μ_k*mg = 0.4*49 = 19.6 N. This is wrong on multiple levels.
• Choice B (29.4 N): This is the static friction force if you incorrectly used Fn=mg. μ_s*mg = 0.6*49 = 29.4 N. You would then incorrectly conclude the block moves, which is the main trap.
• Choice D (45.0 N): This is the magnitude of the pushing force, not the friction force.

This is a classic 'rolling race' problem. It's all about how gravitational potential energy is converted into different kinds of kinetic energy.

Both spheres start with the same potential energy, PE = Mgh. As they roll, this PE is converted into two types of kinetic energy: translational (KE_trans = 1/2 Mv²) and rotational (KE_rot = 1/2 Iω²). Mgh = 1/2 Mv² + 1/2 Iω²

The object that reaches the bottom first will be the one with the greater final translational speed, v.

The key difference is their rotational inertia, I. For a given mass M and radius R:

• Solid sphere: I_solid = (2/5)MR²
• Hollow sphere: I_hollow = (2/3)MR²

Since 2/3 > 2/5, the hollow sphere has a larger rotational inertia. This means it's 'harder' to get the hollow sphere spinning. More of its initial potential energy must be diverted into rotational kinetic energy, leaving less energy available for translational kinetic energy.

The solid sphere, with its smaller rotational inertia, requires less energy to get spinning. Therefore, a larger fraction of its potential energy can be converted into translational kinetic energy, resulting in a higher final speed v and a shorter time to reach the bottom.

• Choice B: While the reasoning about mass distribution is correct, the conclusion is wrong. Having mass farther out increases rotational inertia, which makes it slower.
• Choice C & D: This is the trap for students who think this is like a simple block sliding down a ramp, where mass and size don't matter. For rolling objects, the distribution of that mass (the shape, which determines I) is critical.

This is a fundamental concept in Simple Harmonic Motion (SHM). Let's analyze the forces and energy at different points.

The restoring force from the spring is given by Hooke's Law: F = -kx, where x is the displacement from the equilibrium position.

1. At the equilibrium position (x=0): The spring is neither stretched nor compressed. According to Hooke's Law, the force F = -k(0) = 0. So the net force and acceleration are both zero at this point.
2. Energy: In SHM, total mechanical energy is conserved. It's a constant trade-off between potential energy stored in the spring (PE = 1/2 kx²) and the kinetic energy of the mass (KE = 1/2 mv²).
   • At the amplitudes (maximum x), the mass momentarily stops, so v=0 and KE=0. All the energy is potential.
   • At the equilibrium position (x=0), the potential energy is zero (PE=0). To conserve total energy, all the energy must be kinetic. This means kinetic energy, and therefore speed, is at its maximum.

So, the equilibrium position is the unique point where the net force is zero and the speed is maximum.

• Choice A: At the amplitudes, the displacement x is maximum, so the restoring force F = -kx is also maximum. The speed is momentarily zero.
• Choice B: At any point other than equilibrium or amplitude, there is both a non-zero force and a non-zero speed, but neither is at its maximum or minimum value.
• Choice D: The net force is clearly zero at the equilibrium position, so this is incorrect.

This is a direct application of Pascal's Principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. In a hydraulic lift, this means P_input = P_output.

Since Pressure = Force / Area, we have F_1 / A_1 = F_2 / A_2.

Here's the trap: the problem gives you diameters, not radii or areas. You must calculate the areas first, and remember that Area = πr² = π(d/2)² = (π/4)d². The π/4 term will cancel out, so we can say the ratio of forces is equal to the ratio of the diameters squared.

1. Set up the ratio: F_2 = F_1 * (A_2 / A_1)
2. Use the diameters: A_2 / A_1 = (π(d_2/2)²) / (π(d_1/2)²) = d_2² / d_1². The ratio of the diameters is 24.0 cm / 4.0 cm = 6. The ratio of the areas is 6² = 36.
3. Calculate the output force: F_2 = F_1 * 36 = 100 N * 36 = 3600 N. The lift can support a car with a weight of 3600 N.

• Choice A (600 N): This is what you get if you forget to square the ratio of the diameters. You just do 100 N * (24/4) = 100 N * 6 = 600 N. This is the most common mistake. Remember, pressure depends on area, which is a squared quantity.
• Choice B (1200 N): This might result from a calculation error, perhaps mixing up radii and diameters in an inconsistent way.
• Choice D (21600 N): This is what you might get if you cube the ratio (6³ = 216) instead of squaring it.

This is a fantastic conceptual question that hits on several key physics principles. Let's analyze it carefully.

The system is Priya and Carlos. They start at rest. The push is an internal force.

• Newton's Third Law: The force Priya exerts on Carlos is always equal in magnitude and opposite in direction to the force Carlos exerts on Priya. This is the definition of an action-reaction pair. So, choice A is incorrect.
• Conservation of Momentum: Since they start from rest, the initial momentum of the system is zero. Because the push is an internal force (and ice is frictionless), the total momentum of the system must remain zero. p_Priya + p_Carlos = 0. This means m_Priya * v_Priya = -m_Carlos * v_Carlos. Let's find Carlos's velocity: (50 kg)(3 m/s) = -(75 kg) * v_Carlos 150 = -75 * v_Carlos v_Carlos = -2 m/s. The negative sign means south. So Carlos moves south at 2 m/s. This makes choice D incorrect.
• Kinetic Energy: The initial kinetic energy was zero. The final kinetic energy is (1/2)m_Priya*v_Priya² + (1/2)m_Carlos*v_Carlos². This is clearly a positive number, not zero. The energy came from the chemical potential energy in their muscles. So, choice B is incorrect. Kinetic energy is NOT conserved in this 'explosion'.
• Center of Mass: This is the key. Because the net external force on the system is zero, the velocity of the center of mass cannot change. Since the system started at rest, the center of mass was at rest. It must remain at rest after they push off. Priya and Carlos move away from each other in such a way that their common center of mass doesn't move at all. This makes choice C the correct answer.

Work done by a variable force is found by calculating the area under the Force vs. Position graph. We need to find the total area under the curve from x=0 to x=5.

It's easiest to break the shape into two simpler parts: a rectangle and a triangle.

1. Area of the Rectangle (from x=0 to x=2):
   • The shape is a rectangle with a width of 2 m and a height of 10 N.
   • Area_1 = width * height = 2 m * 10 N = 20 J.
2. Area of the Triangle (from x=2 to x=5):
   • The shape is a triangle with a base that goes from 2 m to 5 m, so base = 5 - 2 = 3 m.
   • The height of the triangle is 10 N.
   • Area_2 = (1/2) * base * height = (1/2) * 3 m * 10 N = 15 J.
3. Total Work:
   • The total work done is the sum of the two areas.
   • Total Work = Area_1 + Area_2 = 20 J + 15 J = 35 J.

• Choice A (20 J): This is only the work done in the first 2 meters. You forgot to include the work done from x=2 to x=5.
• Choice B (25 J): This might be a miscalculation of the triangle's area, perhaps by using the wrong base.
• Choice D (50 J): This is the area of a large rectangle with width 5 and height 10. This ignores the fact that the force was not constant at 10 N for the entire duration.

This question is about the difference between distance and displacement. Distance is the total path length traveled, while displacement is the straight-line separation between the start and end points, including direction.

1. Visualize the motion: The student's path forms a right-angled triangle. The first leg is 3.0 miles east (the base), and the second leg is 4.0 miles north (the height).
2. Find the displacement: The magnitude of the displacement is the length of the hypotenuse of this triangle. We can find it using the Pythagorean theorem: a² + b² = c². c = √(a² + b²) = √((3.0 miles)² + (4.0 miles)²) c = √(9.0 + 16.0) = √25.0 = 5.0 miles.

So the magnitude of the displacement is 5.0 miles.

• Choice A (1.0 mile): This is the difference between the two distances, which doesn't represent displacement.
• Choice C (7.0 miles): This is the total distance traveled (3.0 + 4.0 = 7.0 miles). This is the most common wrong answer because it's easy to confuse distance with displacement. Always remember displacement is the 'as the crow flies' distance from start to finish.
• Choice D (12.0 miles): This is the product of the two distances (3 * 4), which has no physical meaning in this context.

Let's break this down. The key here is the idea of angular impulse. Just like a force applied over time (impulse) changes linear momentum, a torque applied over time (angular impulse) changes angular momentum. The torque on both wheels is the same (τ = F × R), and it's applied for the same time (Δt). This means both wheels receive the same angular impulse (τΔt) and therefore experience the same change in angular momentum. Since they both started from rest, their final angular momentums are equal.

Why are the other choices wrong? Choice A is tempting, but the wheels have different rotational inertias (I_hoop = MR² > I_disk = ½MR²). Since angular momentum L = Iω is the same for both, the wheel with the larger inertia (the hoop) must have a smaller angular velocity. This also rules out Choice D. For Choice B, we can use the formula K_rot = L²/(2I). Since both have the same angular momentum L, the one with the smaller inertia (the disk, A) will have the greater rotational kinetic energy.

The formula for the period of a simple pendulum is T = 2π√(L/g). This equation tells us everything we need to know! The period T is inversely proportional to the square root of the acceleration due to gravity, g. So, if g gets larger, T gets smaller.

In this case, the new gravity g' is 4g. Let's see how that affects the new period T'. T' = 2π√(L/g') = 2π√(L/(4g)) = 2π(√L / √(4g)) = 2π(√L / (2√g)) = (1/2) * [2π√(L/g)]. Since the part in the brackets is the original period T, the new period is T' = T/2.

This is a classic proportional reasoning problem. A common mistake is to forget the square root (leading to T/4) or to flip the relationship (leading to 2T). Always write down the formula and trace how the change in one variable affects the outcome.

This problem tackles a very common misconception about fluid flow. Let's use two key principles. First, the continuity equation (A₁v₁ = A₂v₂). Since the pipe gets narrower (A₂ < A₁), the water must speed up to maintain the same flow rate. So, v₂ > v₁. This eliminates choices C and D.

Now for the pressure. This is where most students slip up. Your intuition might tell you that squeezing water into a smaller pipe increases the pressure. The physics says the opposite! We use Bernoulli's equation for a horizontal pipe: P₁ + ½ρv₁² = P₂ + ½ρv₂². Since we already know v₂ is greater than v₁, the term ½ρv₂² is larger than ½ρv₁². To keep the equation balanced, the pressure P₂ in the narrow section must be less than the pressure P₁ in the wider section. So, v₂ > v₁ and P₂ < P₁. Think of it this way: the fluid speeds up, and the 'energy' for this increase in kinetic energy comes from a decrease in the 'pressure energy'.

This is a great conservation of energy problem. The key is to carefully define your 'before' and 'after' states. 'Before': The block is at rest (K=0) and the spring is compressed by x. All the energy is potential energy in the spring: E_initial = ½kx². 'After': The spring is now only compressed by x/2, and the block is moving with speed v. The energy is a mix of spring potential energy and kinetic energy: E_final = ½k(x/2)² + ½mv².

Since the surface is frictionless, energy is conserved: E_initial = E_final. ½kx² = ½k(x²/4) + ½mv² Let's multiply everything by 2 to clean it up: kx² = k(x²/4) + mv². Now, let's solve for mv²: mv² = kx² - kx²/4 = (3/4)kx². Finally, solve for v: v² = (3/4)kx²/m, which means v = √((3kx²)/(4m)) = x√(3k/(4m)).

Watch out for the traps! Choice A is the speed the block has when the spring is fully uncompressed. Choice B assumes velocity is proportional to distance, which isn't true because the acceleration isn't constant. Choice C comes from incorrectly setting the kinetic energy equal to the remaining potential energy, instead of the change in potential energy.

Impulse is equal to the change in momentum (J = Δp), and momentum is a vector! This is the crucial point. Let's set up a coordinate system. Let the initial velocity be along the x-axis: v_i = <10, 0> m/s. The final velocity is perpendicular, so let's say it's along the y-axis: v_f = <0, 10> m/s.

Now, let's calculate the initial and final momentum vectors (p = mv): p_i = 0.5 kg <10, 0> m/s = <5, 0> kg·m/s. p_f = 0.5 kg <0, 10> m/s = <0, 5> kg·m/s.

The change in momentum is Δp = p_f - p_i = <0, 5> - <5, 0> = <-5, 5> kg·m/s. The question asks for the magnitude of this impulse vector. We use the Pythagorean theorem: |Δp| = √((-5)² + 5²) = √(25 + 25) = √50 ≈ 7.07 N·s.

Common mistakes: Choice A happens if you treat momentum as a scalar (0.5*(10-10)=0). Choice B is just the magnitude of the initial or final momentum, not the change. Choice D is what you get if you incorrectly add the magnitudes (5+5=10). Remember to treat momentum as a vector and subtract tip-to-tail!

This is a fantastic question that tests your fundamental understanding of forces. It looks like a typical elevator problem, but there's a subtle and important twist. Let's analyze the forces on the student: the scale pushes up with a normal force (let's call it F_scale_on_student) and gravity pulls down with a force mg. The net force is F_net = F_scale_on_student - mg = ma. So, the force the scale exerts on the student is F_scale_on_student = mg + ma.

Now, what does a scale measure? A scale does not directly measure your weight (mg). It measures the contact force exerted on it. The question tells you the reading on the scale is N. This means, by definition, the force exerted on the scale is N.

By Newton's Third Law, the force the student exerts on the scale is equal in magnitude and opposite in direction to the force the scale exerts on the student. So the force by the student on the scale has a magnitude of mg + ma. The problem statement has already given this value a name: N. Therefore, the force exerted by the student on the scale is simply N. Don't overthink it!

This is a classic 'race to the bottom' problem. The winner is the object that has the greatest linear acceleration. Let's think about energy. At the top, both objects have the same potential energy (Mgh). At the bottom, this energy is converted into two forms: translational kinetic energy (½Mv²) and rotational kinetic energy (½Iω²).

The object's total energy budget is fixed. The object that 'spends' less energy on rotating will have more energy left over for moving forward (translating). The rotational inertia 'I' is a measure of how hard it is to get something to rotate. The hollow sphere has its mass distributed far from the center, so it has a larger rotational inertia (I_hollow = ⅔MR²) than the solid sphere (I_solid = ⅖MR²).

Because the solid sphere is 'easier' to rotate, a smaller fraction of its potential energy goes into spinning, and a larger fraction goes into moving down the ramp. It will have a greater final velocity and greater average acceleration, so it reaches the bottom first. Choice A perfectly captures this reasoning. Choice C is a common mistake; while the total work done by gravity is the same, how that work is partitioned into different types of kinetic energy determines the speed.

Remember that on a velocity-time graph, the displacement is the area 'under' the curve—or more accurately, the area between the graph and the time axis. Areas above the axis are positive displacement, and areas below are negative displacement.

Because the graph is a straight line from v₀ to -v₀, it must cross the time axis at the midpoint, t=T/2. This creates two triangles. Triangle 1 (from t=0 to t=T/2): It's above the axis. Its area is (1/2) base height = (1/2) (T/2) v₀ = v₀T/4. This is the positive displacement. Triangle 2 (from t=T/2 to t=T): It's below the axis. Its area is (1/2) base height = (1/2) (T/2) (-v₀) = -v₀T/4. This is the negative displacement.

The total displacement is the sum of these two areas: v₀T/4 + (-v₀T/4) = 0. The particle ended up right back where it started!

Be careful: Choice B, (v₀T)/2, represents the total distance traveled (the sum of the absolute values of the areas), not the displacement.

This is a perfectly inelastic collision, which means the objects stick together. In these collisions, momentum is always conserved, but kinetic energy is not.

1. Conserve Momentum: The initial momentum is all in the first cart: p_initial = mv. The final momentum is the combined mass (2m) moving at a new final velocity, v_f: p_final = (2m)v_f. Setting them equal: mv = (2m)v_f. Solving for the final velocity gives v_f = v/2.

2. Calculate Kinetic Energies: The initial kinetic energy is K_initial = ½mv². The final kinetic energy is K_final = ½(total mass)(v_f)² = ½(2m)(v/2)² = m(v²/4) = ¼mv².

3. Find the Ratio: The question asks for K_final / K_initial. (¼mv²) / (½mv²) = (1/4) / (1/2) = 1/2.

So, half of the initial kinetic energy was 'lost' (converted into other forms like heat and sound) during the collision. A common mistake is to assume energy is conserved (Choice A) or to make a mistake with the fractions (Choice C).

This question gets to the heart of the difference between static and kinetic friction. Static friction is the 'stubborn' friction that prevents motion. It can vary, matching your push up to a maximum value of f_s,max = μₛN = μₛMg.

Once your applied force F overcomes this maximum static friction, the crate 'breaks free' and starts to slide. The moment it starts sliding, the type of friction changes. It's no longer static; it's kinetic friction. The kinetic friction force is constant and has a value of f_k = μₖN = μₖMg.

Since we know μₛ > μₖ, the friction force actually drops the instant the crate starts moving. Your applied force F is now greater than the opposing kinetic friction force (F > μₛMg > μₖMg). This means there is a net forward force on the crate, and according to Newton's Second Law (F_net = ma), the crate must accelerate. So, the crate accelerates, and the friction force opposing it is the kinetic friction, μₖMg.

You can solve this with kinematics, but the most direct path is the Work-Energy Theorem. For linear motion, the net work done on an object equals its change in kinetic energy (W = ΔK). The same is true for rotation!

The work done by a constant torque is analogous to the work done by a constant force. Instead of W = Fd, for rotation, we have W_rot = τθ.

The rotational Work-Energy Theorem states that W_rot = ΔK_rot. Since the flywheel starts from rest, its initial rotational kinetic energy is zero (K_rot,initial = 0). Therefore, the final rotational kinetic energy is simply equal to the work done on it. K_rot,final = W_rot = τθ. It's that simple! The other choices come from mixing up formulas. For example, Choice B involves α², which isn't energy, and Choice D is the formula for kinetic energy in terms of angular momentum (K = L²/2I), but L is not equal to τθ.

This is a fantastic question about how energy transforms during simple harmonic motion (SHM). The total mechanical energy of the system is constant: E_total = K + U. At the amplitude (x=A), the mass stops for an instant, so all the energy is potential: E_total = ½kA².

We are looking for the position 'x' where the kinetic energy equals the potential energy (K = U). At this point, the total energy is E_total = K + U = U + U = 2U. So, we can set our two expressions for total energy equal to each other: E_total = ½kA² E_total = 2U = 2(½kx²) = kx²

Setting them equal: ½kA² = kx². We can cancel k from both sides: ½A² = x². Solving for x, we get x = √(A²/2) = A/√2.

This happens at x = ±A/√2. The most common wrong answer is x = ±A/2. If you check that position, you'll find the potential energy is ¼ of the total energy, and the kinetic energy is ¾ of the total energy, so they are not equal.

This is a multi-step problem combining weight, buoyancy, and density. Let's go step-by-step.

1. Find the object's true weight: W = mg = (10 kg)(10 m/s²) = 100 N.

2. Find the buoyant force: The apparent weight is the true weight minus the buoyant force (F_B). W_app = W - F_B 80 N = 100 N - F_B So, the buoyant force is F_B = 20 N.

3. Find the object's volume: The buoyant force is the weight of the fluid displaced. Since the object is fully submerged, the volume of displaced fluid is the volume of the object (V_obj). F_B = ρ_water V_obj g 20 N = (1000 kg/m³) V_obj (10 m/s²) 20 = 10000 * V_obj Solving for the volume: V_obj = 20 / 10000 = 0.002 m³.

4. Find the object's density: Now we have the object's mass (10 kg) and its volume (0.002 m³). ρ_obj = mass / volume = 10 kg / 0.002 m³ = 5000 kg/m³.

Each step is straightforward, but you have to connect them all correctly. A common shortcut is to notice the buoyant force (20N) is 1/5 of the true weight (100N). This means the water's density must be 1/5 of the object's density, so the object's density is 5 times the water's density, or 5000 kg/m³.

This is a classic static equilibrium problem. For the beam to be in equilibrium, the net torque must be zero. The smartest way to solve this is to choose our pivot point at the hinge. Why? Because the hinge exerts forces on the beam, but by picking the hinge as the pivot, the lever arm for those forces is zero, and they create zero torque. This means we can ignore them!

Now, let's sum the torques about the hinge:

1. Torque from gravity: The beam's weight (Mg) acts at its center of mass, which for a uniform beam is at L/2 from the hinge. This creates a clockwise (negative) torque: τ_gravity = -Mg(L/2).
2. Torque from tension: The tension T acts at the end of the beam, a distance L from the hinge. However, only the component of tension perpendicular to the beam creates torque. That component is Tsinθ. This creates a counter-clockwise (positive) torque: τ_tension = +(Tsinθ)L.

Since the net torque is zero: τ_gravity + τ_tension = 0. -Mg(L/2) + (Tsinθ)L = 0 (Tsinθ)L = Mg(L/2) Notice the L cancels on both sides! Tsinθ = Mg/2 Solving for T gives: T = Mg / (2sinθ).

The most common mistakes are using cosθ instead of sinθ, or forgetting that the beam's weight acts at its center (the 1/2 factor).

The key principle here is that an internal explosion cannot change the motion of the system's center of mass. The center of mass must continue along the original parabolic path as if nothing happened.

Let's analyze the momentum right at the moment of the explosion. At the highest point of its trajectory, the firecracker's velocity is purely horizontal, let's call it v_cm. The total mass is M = 2m. The initial momentum of the system is P_initial = (2m)v_cm.

After the explosion, we have two pieces of mass m. One piece falls straight down, meaning its horizontal velocity is zero. The other piece has some unknown horizontal velocity, v₂. The final momentum of the system is P_final = (m)(0) + (m)(v₂).

Since horizontal momentum is conserved during the explosion (gravity is a vertical external force), we have P_initial = P_final. (2m)v_cm = mv₂ Solving for v₂ gives v₂ = 2v_cm.

So, to keep the center of mass moving forward at speed v_cm, the second piece must be shot forward with twice that speed to compensate for the first piece stopping.

This is a classic, and somewhat counter-intuitive, physics problem often called the 'monkey and the hunter'. The key is to realize that gravity affects both the projectile and the target in the exact same way.

Imagine a world with no gravity. If you aim the projectile at the target and fire, it will travel in a straight line and hit the stationary target.

Now, let's turn gravity back on. In any amount of time 't' after the launch, gravity will have pulled the projectile down from its straight-line path by a vertical distance of ½gt². In that same amount of time 't', gravity will have also pulled the falling target down from its starting position by the exact same distance, ½gt².

Since both the projectile and the target have 'fallen' the same vertical distance from their gravity-free positions, the projectile will still be on a collision course. As long as the projectile was aimed at the target's initial position and has enough speed to reach the target's horizontal location, it will hit the falling target.

This is a great example of conservation of angular momentum. Since there is no external torque on the skater (frictionless ice), her angular momentum L = Iω must remain constant.

1. Find the new angular velocity: L_initial = L_final => Iᵢωᵢ = I_fω_f Iᵢωᵢ = (Iᵢ/3)ω_f Solving for ω_f, we find ω_f = 3ωᵢ. She spins three times faster!

2. Calculate the ratio of kinetic energies: K_initial = ½Iᵢωᵢ² K_final = ½I_fω_f² = ½(Iᵢ/3)(3ωᵢ)² = ½(Iᵢ/3)(9ωᵢ²) = 3(½Iᵢωᵢ²) = 3K_initial. So, the ratio K_f / K_i = 3.

Wait, where did this extra energy come from? The skater did positive work by using her muscles to pull her arms in. This internal work increased the rotational kinetic energy of the system. A common mistake is to assume kinetic energy is conserved (Choice B). It is not! Only angular momentum is conserved here.

This question is a test of precision. Power is the rate at which work is done, and for a specific force, the power it delivers is given by P = (Force) × (velocity).

The question specifically asks for the power delivered by the engine. The force provided by the engine is F. The car is moving at speed v. Therefore, the power delivered by the engine is simply P_engine = Fv.

It's easy to get distracted by the other information. The net force on the car is F_net = F - f. The power that goes into increasing the car's kinetic energy (the 'net power') is P_net = (F - f)v. The power 'lost' to drag is P_drag = fv. Notice that P_engine = P_net + P_drag. The question, however, only asks for the power from the engine, which is Fv.

For any object in a stable circular orbit, the gravitational force is providing the necessary centripetal force to keep it moving in a circle. All we have to do is set these two expressions equal to each other and solve for the speed, v.

1. Gravitational Force: F_g = GMm/R²
2. Centripetal Force: F_c = mv²/R

Set them equal: GMm/R² = mv²/R

Now we solve for v. Notice that the mass of the satellite, m, appears on both sides, so it cancels out! This means the orbital speed doesn't depend on how heavy the satellite is. We can also cancel one factor of R.

GM/R = v²

Taking the square root of both sides gives the orbital speed: v = √(GM/R).

Choice B is a common mistake of using the satellite's mass instead of the planet's. Choice D is the expression for the gravitational acceleration 'g' at that radius, not the speed.

This is a challenging problem that combines linear and rotational dynamics. We need two equations from Newton's Second Law.

1. Linear Motion (F_net = Ma): The applied force F pulls the cylinder forward, but the ground exerts a static friction force, f, backward to make it roll. So, the net force is F - f = Ma.

2. Rotational Motion (τ_net = Iα): The only force creating a torque about the center of mass is friction. τ = fR. The moment of inertia for a solid cylinder is I = ½MR². So, fR = (½MR²)α.

3. The 'Rolling without Slipping' Link: The key that connects these two equations is the condition a = αR, which means α = a/R.

Let's substitute this into the torque equation: fR = (½MR²)(a/R). This simplifies to f = ½Ma. This is powerful—it tells us exactly what the friction force must be.

Now, substitute this expression for f back into our linear equation: F - (½Ma) = Ma F = Ma + ½Ma F = (3/2)Ma

Finally, solve for the acceleration, a: a = F / ((3/2)M) = 2F/(3M).

The most common mistake is to ignore friction (Choice A), which would mean the object slides, not rolls.