Crash Course — Full Course Cram Sheet

Model solution

a) [Diagram would show a parabolic path starting from the origin, curving upwards and then downwards to intersect a line at 30 degrees above the horizontal. The initial velocity vector would be drawn from the origin at 60 degrees above horizontal, labeled v_0.]

b) Initial velocity components: v_0x = v_0 cos(60°) = 20 m/s 0.5 = 10 m/s v_0y = v_0 sin(60°) = 20 m/s (sqrt(3)/2) = 17.32 m/s

Position equations: x(t) = v_0x t = 10t y(t) = v_0y t - (1/2)gt^2 = 17.32t - 4.9t^2

On the incline, y = x tan(30°) = x (1/sqrt(3)) = x 0.577

Substitute x(t) and y(t) into the incline equation: 17.32t - 4.9t^2 = (10t) * 0.577 17.32t - 4.9t^2 = 5.77t

Since t is not zero (projectile is in the air), we can divide by t: 17.32 - 4.9t = 5.77 4.9t = 17.32 - 5.77 4.9t = 11.55 t = 11.55 / 4.9 = 2.36 s

c) Distance along the incline (d): First, find the x-coordinate where it lands: x = 10t = 10 * 2.36 = 23.6 m

The distance d along the incline is related to x by cos(30°): x = d cos(30°) d = x / cos(30°) = 23.6 m / (sqrt(3)/2) = 23.6 m / 0.866 = 27.25 m

Model solution

a) [Diagram would show a block on an incline. Forces: Mg downwards, N perpendicular to incline upwards, T parallel to incline upwards, f_s parallel to incline upwards (if tending to slide down) or downwards (if tending to slide up).]

b) To prevent the block from sliding down, the static friction force acts up the incline. The system is in equilibrium. Forces perpendicular to incline: N - Mg cos(θ) = 0 => N = Mg cos(θ) Forces parallel to incline: T + f_s - Mg sin(θ) = 0 For minimum m, the block is on the verge of sliding down, so f_s = μ_s N = μ_s Mg cos(θ). Also, T = m_min g (from the hanging mass).

Substitute into the parallel force equation: m_min g + μ_s Mg cos(θ) - Mg sin(θ) = 0 m_min g = Mg sin(θ) - μ_s Mg cos(θ) m_min = M (sin(θ) - μ_s cos(θ))

c) When the block slides up the incline, kinetic friction acts down the incline. The hanging mass m_max moves downwards, and the block M moves upwards with acceleration 'a'.

For hanging mass m_max: m_max g - T = m_max a (Equation 1)

For block M: Perpendicular to incline: N - Mg cos(θ) = 0 => N = Mg cos(θ) Parallel to incline: T - Mg sin(θ) - f_k = M a Since f_k = μ_k N = μ_k Mg cos(θ): T - Mg sin(θ) - μ_k Mg cos(θ) = M a (Equation 2)

From (1), T = m_max g - m_max a. Substitute into (2): (m_max g - m_max a) - Mg sin(θ) - μ_k Mg cos(θ) = M a m_max g - Mg sin(θ) - μ_k Mg cos(θ) = M a + m_max a m_max g - Mg sin(θ) - μ_k Mg cos(θ) = (M + m_max) a

a = (m_max g - Mg sin(θ) - μ_k Mg cos(θ)) / (M + m_max)

Model solution

a) Using conservation of mechanical energy between the starting point (height H) and the bottom of the loop (height 0): E_initial = E_final MgH + (1/2)M(0)^2 = Mg(0) + (1/2)Mv_bottom^2 MgH = (1/2)Mv_bottom^2 v_bottom^2 = 2gH v_bottom = sqrt(2gH)

b) For the car to successfully complete the loop, it must have a minimum speed at the top of the loop such that the normal force is just zero. At the top of the loop, the height is 2R.

First, find the minimum speed at the top of the loop (v_top_min): At the top, the forces are gravity (Mg) and normal force (N), both acting downwards. The net force provides the centripetal acceleration. N + Mg = Mv_top_min^2 / R For minimum speed, N = 0: Mg = Mv_top_min^2 / R v_top_min^2 = gR

Now, apply conservation of mechanical energy between the starting height H_min and the top of the loop (height 2R): E_initial = E_final MgH_min + (1/2)M(0)^2 = Mg(2R) + (1/2)Mv_top_min^2 MgH_min = Mg(2R) + (1/2)M(gR) Divide by Mg: H_min = 2R + (1/2)R H_min = 2.5R

c) If the car starts from H = 3R, we first find the speed at the bottom of the loop using conservation of energy: Mg(3R) = (1/2)Mv_bottom^2 v_bottom^2 = 6gR

Now, apply Newton's Second Law at the bottom of the loop. The forces are the normal force (N) upwards and gravity (Mg) downwards. The net force provides the centripetal acceleration upwards. N - Mg = Mv_bottom^2 / R N = Mg + Mv_bottom^2 / R Substitute v_bottom^2 = 6gR: N = Mg + M(6gR) / R N = Mg + 6Mg N = 7Mg

Model solution

a) The collision between the bullet and the block is an inelastic collision. Linear momentum is conserved during the collision. Initial momentum = m v Final momentum = (m + M) V_f (where V_f is the speed of the combined system immediately after collision)

mv = (m + M)V_f V_f = mv / (m + M)

b) After the collision, the bullet-block system swings upward. Mechanical energy is conserved during this swing. Initial kinetic energy (immediately after collision) = (1/2)(m + M)V_f^2 Final potential energy (at maximum height h) = (m + M)gh

(1/2)(m + M)V_f^2 = (m + M)gh (1/2)V_f^2 = gh h = V_f^2 / (2g)

Substitute the expression for V_f from part (a): h = (mv / (m + M))^2 / (2g) h = m^2 v^2 / (2g(m + M)^2)

c) The collision is inelastic. This is because the bullet embeds itself in the block, indicating that kinetic energy is not conserved during the collision. Some of the initial kinetic energy of the bullet is converted into internal energy (heat, sound, deformation) of the bullet and block.

Model solution

a) [Diagram would show a horizontal rod pivoted at the left end. Forces: Mg acting downwards at the center (L/2 from pivot). Pivot forces (F_px and F_py, or a single F_p) acting at the pivot point.]

b) The only force that creates a torque about the pivot is gravity (Mg), acting at the center of mass (L/2 from the pivot). Torque (τ) = Force Lever arm sin(angle) τ = Mg (L/2) sin(90°) = MgL/2

Newton's Second Law for rotation: τ_net = Iα Given I = (1/3)ML^2

MgL/2 = (1/3)ML^2 α Divide by ML: g/2 = (1/3)L α α = (3g) / (2L)

c) The linear acceleration (a) of a point at a distance r from the pivot in rotational motion is given by a = rα. For the free end of the rod, r = L.

a_end = Lα Substitute the expression for α: a_end = L * (3g / (2L)) a_end = 3g / 2

Model solution

a) Free-body diagram for hanging block: [Diagram would show a block with 'mg' downwards and 'T' upwards.]

Free-body diagram for rotating disk: [Diagram would show a disk with 'Mg' downwards at center, 'F_pivot' upwards at center, and 'T' downwards tangent to the rim.]

b) For the hanging block, applying Newton's Second Law: mg - T = ma (Equation 1)

For the rotating disk, applying Newton's Second Law for rotation: τ_net = Iα The tension T creates a torque about the center of the disk: τ = T * R Given I = (1/2)MR^2

TR = (1/2)MR^2 α (Equation 2)

The linear acceleration of the string (and thus the block) is related to the angular acceleration of the disk: a = Rα => α = a/R

Substitute α into Equation 2: TR = (1/2)MR^2 (a/R) TR = (1/2)MRa T = (1/2)Ma (Equation 3)

Now substitute T from Equation 3 into Equation 1: mg - (1/2)Ma = ma mg = ma + (1/2)Ma mg = a(m + (1/2)M) a = mg / (m + (1/2)M)

Finally, find angular acceleration α: α = a/R = (mg / (m + (1/2)M)) / R α = mg / (R(m + (1/2)M)) or α = 2mg / (R(2m + M))

c) To find the tension, substitute the expression for 'a' back into Equation 3: T = (1/2)Ma T = (1/2)M * (mg / (m + (1/2)M)) T = (1/2)Mmg / (m + (1/2)M) T = Mmg / (2m + M)

Model solution

a) At equilibrium, the net force on the block is zero. The forces acting parallel to the incline are the component of gravity pulling it down (mg sinθ) and the spring force pulling it up (kx_eq), where x_eq is the extension from the unstretched position.

ΣF_parallel = 0 kx_eq - mg sinθ = 0 kx_eq = mg sinθ x_eq = (mg sinθ) / k

b) The period of oscillation for a mass-spring system is given by T = 2π * sqrt(m/k). The presence of the incline and gravity only shifts the equilibrium position but does not change the effective spring constant or the oscillating mass, so the period remains the same.

T = 2π * sqrt(m/k)

c) The maximum speed occurs at the equilibrium position. We can use conservation of mechanical energy between the release point (unstretched spring, height y_initial = 0) and the equilibrium position (spring stretched by x_eq, height y_eq = -x_eq sinθ). Let the release point be the reference for gravitational potential energy.

Initial energy (at release, x=0, v=0): E_initial = PE_gravity_initial + PE_spring_initial + KE_initial E_initial = 0 + 0 + 0 = 0

Final energy (at equilibrium, x=x_eq, v=v_max): E_final = PE_gravity_final + PE_spring_final + KE_final E_final = mg(y_eq) + (1/2)kx_eq^2 + (1/2)mv_max^2 E_final = mg(-x_eq sinθ) + (1/2)kx_eq^2 + (1/2)mv_max^2

From part (a), x_eq = (mg sinθ) / k. Substitute this into the energy equation: 0 = mg(-(mg sinθ)/k * sinθ) + (1/2)k((mg sinθ)/k)^2 + (1/2)mv_max^2 0 = - (m^2 g^2 sin^2θ)/k + (1/2)k(m^2 g^2 sin^2θ)/k^2 + (1/2)mv_max^2 0 = - (m^2 g^2 sin^2θ)/k + (1/2)(m^2 g^2 sin^2θ)/k + (1/2)mv_max^2 0 = - (1/2)(m^2 g^2 sin^2θ)/k + (1/2)mv_max^2

(1/2)mv_max^2 = (1/2)(m^2 g^2 sin^2θ)/k v_max^2 = (m g^2 sin^2θ)/k v_max = sqrt((m g^2 sin^2θ)/k) v_max = (mg sinθ) / sqrt(k/m) = (mg sinθ) / ω, where ω is the angular frequency.

Model solution

a) When the block is floating, the buoyant force (F_B) equals the weight of the block (W_w). F_B = W_w ρ_l V_submerged g = ρ_w V_block g ρ_l (A d_0) g = ρ_w (A H) g

Divide by Ag: ρ_l d_0 = ρ_w H d_0 = (ρ_w / ρ_l) H

b) When the mass m is placed on top, the total weight of the system is W_total = W_w + mg. The block sinks to a new depth d_new = d_0 + Δd.

F_B_new = W_total ρ_l (A d_new) g = ρ_w (A H) g + mg ρ_l A (d_0 + Δd) g = ρ_w A H g + mg

Substitute d_0 = (ρ_w / ρ_l) H: ρ_l A ((ρ_w / ρ_l) H + Δd) g = ρ_w A H g + mg ρ_w A H g + ρ_l A Δd g = ρ_w A H * g + mg

Subtract ρ_w A H g from both sides: ρ_l A Δd g = mg Δd = mg / (ρ_l A g) Δd = m / (ρ_l * A)

c) Yes, it will oscillate. When the block is pushed down an additional distance x from its new equilibrium position, the buoyant force increases by an amount F_restoring = ρ_l A x * g. This increased buoyant force acts as a restoring force, which is directly proportional to the displacement x and acts in the opposite direction of the displacement. This is the condition for Simple Harmonic Motion (SHM), so the block will oscillate.

Model solution

Let v_bw be the velocity of the boat relative to the water, v_ws be the velocity of the water relative to the shore (bank), and v_bs be the velocity of the boat relative to the shore. Vector addition: v_bs = v_bw + v_ws

a) If the boat aims directly across the river, its velocity relative to the water (v_bw) is perpendicular to the current. The component of the boat's velocity that carries it across the river is v_bw_y = 4.0 m/s.

Time to cross (t) = width / v_bw_y t = 100 m / 4.0 m/s = 25 s

b) The boat's velocity relative to the bank (v_bs) has two components: v_bs_y = v_bw_y = 4.0 m/s (across the river) v_bs_x = v_ws = 3.0 m/s (downstream)

Magnitude of v_bs = sqrt(v_bs_x^2 + v_bs_y^2) Magnitude of v_bs = sqrt((3.0 m/s)^2 + (4.0 m/s)^2) Magnitude of v_bs = sqrt(9 + 16) = sqrt(25) = 5.0 m/s

c) To land directly across the river, the boat's velocity relative to the bank (v_bs) must be directed straight across the river (i.e., its x-component must be zero). This means the x-component of the boat's velocity relative to the water must cancel out the river's current.

Let θ be the angle upstream relative to the bank. So, v_bw makes an angle θ with the y-axis (or 90+θ with the x-axis). v_bw_x = -v_bw sin(θ) (negative because it's upstream) v_bw_y = v_bw cos(θ)

v_bs_x = v_bw_x + v_ws_x = -v_bw sin(θ) + v_ws For v_bs_x = 0: 0 = -4.0 m/s * sin(θ) + 3.0 m/s 4.0 sin(θ) = 3.0 sin(θ) = 3.0 / 4.0 = 0.75 θ = arcsin(0.75) = 48.6 degrees upstream relative to the bank.

Magnitude of the boat's velocity relative to the bank (v_bs_y): v_bs_y = v_bw_y = v_bw cos(θ) v_bs_y = 4.0 m/s cos(48.6°) v_bs_y = 4.0 m/s 0.661 = 2.64 m/s

Model solution

a) The work done by a variable force is given by the integral of the force with respect to displacement: W = ∫ F(x) dx

W = ∫_0^d (Ax^2 - B) dx W = [ (A/3)x^3 - Bx ]_0^d W = (A/3)d^3 - Bd - [ (A/3)(0)^3 - B(0) ] W = (A/3)d^3 - Bd

b) According to the Work-Energy Theorem, the net work done on an object is equal to the change in its kinetic energy. Since the particle starts from rest (KE_initial = 0): W_net = ΔKE = KE_final - KE_initial W = KE_final - 0 KE_final = W

So, the kinetic energy of the particle when it reaches x=d is: KE(d) = (A/3)d^3 - Bd

c) The particle will momentarily come to rest when its kinetic energy is zero. This means the total work done on it from x=0 to that point must be zero.

Let x_final be the position where it comes to rest. Work_total = (A/3)x_final^3 - Bx_final = 0

Factor out x_final: x_final ( (A/3)x_final^2 - B ) = 0

One solution is x_final = 0 (the starting point where it was at rest). The other solution is when the term in the parentheses is zero: (A/3)x_final^2 - B = 0 (A/3)x_final^2 = B x_final^2 = 3B / A x_final = sqrt(3B / A)

Since the problem asks for a value of x other than x=0, and assuming it continues to move in the positive x-direction, the particle will momentarily come to rest at x = sqrt(3B / A).

Model solution

a) [Diagram would show a cylinder on an incline. Forces: Mg downwards at center, N perpendicular to incline upwards at contact point, f_s up the incline at contact point.]

b) Apply Newton's Second Law for translational motion (parallel to incline): Mg sinθ - f_s = Ma (Equation 1)

Apply Newton's Second Law for rotational motion (about the center of mass): τ_net = Iα The only force creating a torque about the center of mass is the static friction force. τ = f_s * R Given I = (1/2)MR^2

f_s * R = (1/2)MR^2 α (Equation 2)

For rolling without slipping, the linear acceleration 'a' and angular acceleration 'α' are related by: a = Rα => α = a/R

Substitute α into Equation 2: f_s R = (1/2)MR^2 (a/R) f_s R = (1/2)MRa f_s = (1/2)Ma (Equation 3)

Now substitute f_s from Equation 3 into Equation 1: Mg sinθ - (1/2)Ma = Ma Mg sinθ = Ma + (1/2)Ma Mg sinθ = (3/2)Ma

Divide by M: g sinθ = (3/2)a a = (2/3)g sinθ

c) Use conservation of mechanical energy. The initial energy is purely gravitational potential energy. The final energy (at the bottom) is a combination of translational kinetic energy and rotational kinetic energy.

E_initial = E_final MgH = (1/2)Mv_cm^2 + (1/2)Iω^2

For rolling without slipping, v_cm = Rω, so ω = v_cm / R. Substitute I = (1/2)MR^2 and ω: MgH = (1/2)Mv_cm^2 + (1/2)(1/2)MR^2 (v_cm / R)^2 MgH = (1/2)Mv_cm^2 + (1/4)Mv_cm^2 MgH = (3/4)Mv_cm^2

Divide by M: gH = (3/4)v_cm^2 v_cm^2 = (4/3)gH v_cm = sqrt((4/3)gH)