AP Physics C Mechanics: 4.3 Conservation of Linear Momentum, Collisions – Exam Style questions with Answer- FRQ

Question

A pendulum of length L consists of block 1 of mass 3M attached to the end of a string. Block 1 is released from rest with the string horizontal, as shown above. At the bottom of its swing, block 1 collides with block 2 of mass M, which is initially at rest at the edge of a table of height 2L. Block 1 never touches the table. As a result of the collision, block 2 is launched horizontally from the table, landing on the floor a distance 4L from the base of the table. After the collision, block 1 continues forward and swings up. At its highest point, the string makes an angle θMAX to the vertical. Air resistance and friction are negligible. Express all algebraic answers in terms of M, L, and physical constants, as appropriate.
(a) Determine the speed of block 1 at the bottom of its swing just before it makes contact with block 2.
(b) On the dot below, which represents block 1, draw and label the forces (not components) that act on block 1 just before it makes contact with block 2. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot. Forces with greater magnitude should be represented by longer vectors.

(c) Derive an expression for the tension FT in the string when the string is vertical just before block 1 makes contact with block 2. If you need to draw anything other than what you have shown in part (b) to assist in your solution, use the space below. Do NOT add anything to the figure in part (b). For parts (d)–(g), the value for the length of the pendulum is L = 75 cm.
(d) Calculate the time between the instant block 2 leaves the table and the instant it first contacts the floor.
(e) Calculate the speed of block 2 as it leaves the table.
(f) Calculate the speed of block 1 just after it collides with block 2.
(g) Calculate the angle θMAX that the string makes with the vertical, as shown in the original figure, when block 1 is at its highest point after the collision.

Answer/Explanation

Ans:

(a)

\(mgh = \frac{1}{2}mv^{2}\)

10 (L) = .sv                                                                 v2 = 20L

                                                                                        \(V=\sqrt{20L}=2\sqrt{SL}\)

(b)

(c)

FT = ma

FT – Fg = 3M.

(d)

Xy= Xoy + Voy t+ 1/2 at2

-2L = 0 + 0 t+ 1/2 (-10) t2

-1.5 m = -5t2

3 = t2

\(t = \sqrt{\frac{3}{10}}=0.548 seconds\)

(e)

\(V = \frac{\Delta x}{\Delta t}= \frac{4L}{\sqrt{\frac{3}{10}}}=\frac{3}{\sqrt{\frac{3}{10}}}=5.477 m/s\)

(f) 

P1 + P2 = P1f + P2f

m1v1 + m2v2 = m1v1f + m2v2f

\((3m)\left ( \sqrt{20(.70)} \right )+(m)(0)=(3m)(x)+(m)(5.477)\)

11.619 + 0 = 3x + 5.477

6.142 = 3x

x = 2.047 m/s

(g)

\(\frac{1}{2}3MV^{2}=3Mgh\)

\(\frac{1}{2}(2.047)^{2}=10 h\)

H = 0.210 m

\(cos(\theta )=\frac{A}{H}\)

\(\theta = -cos^{-1}\left ( \frac{A}{H} \right )\)

\(\theta = cos^{-1}\left ( \frac{.75-H}{.75} \right )=43.897^{0}\)

Question

Two carts are on a horizontal, level track of negligible friction. Cart 1 has a sensor that measures the force exerted on it during a collision with cart 2, which has a spring attached. Cart 1 is moving with a speed of v0 = 3.00 m/s 0 toward cart 2, which is at rest, as shown in the figure above. The total mass of cart 1 and the force sensor is 0.500 kg, the mass of cart 2 is 1.05 kg, and the spring has negligible mass. The spring has a spring constant of k = 130 N/m.The data for the force the spring exerts on cart 1 are shown in the graph below. A student models the data as the quadratic fit F = (3200 N/s2 )t2 – (500 N/s )t .

(a) Using integral calculus, calculate the total impulse delivered to cart 1 during the collision.
(b)
i. Calculate the speed of cart 1 after the collision.
ii. In which direction does cart 1 move after the collision?
____ Left ____ Right
____ The direction is undefined, because the speed of cart 1 is zero after the collision.
(c)
i. Calculate the speed of cart 2 after the collision.
ii. Show that the collision between the two carts is elastic.
(d)
i. Calculate the speed of the center of mass of the two-cart–spring system.
ii. Calculate the maximum elastic potential energy stored in the spring.

Answer/Explanation

Ans:

(a)

\(F = \frac{dp}{dt}\)                                                                                 \(\frac{3200 + 3}{3} – \frac{500+2}{2}\int_{t=0}^{t = 0.16}\)

dp = Fdt

\(\Delta P = \int_{t_{0}}^{tf}Fdt\)                                                      \(\frac{3200(.16)^{3}}{3}-\frac{500(.16)^{1}}{2}\)

\(=\int_{0}^{0.16}\left ( 3200t^{2}-500t \right )dt\)                     = -2.03  kg. m. s-1

(b) i.

ΔP = mΔv = -2.03

\(\Delta V = \frac{-2.03}{0.500}=-4.06 m/s\)

Vf = V0 + ΔV = 3.00 – 4.06 = -1.06 → 1.06 m/s

ii.

   X    Left

(c) i.

By Newton’s 3rd law of conservation of momentum:

ΔP1 = -ΔP2

ΔP2 = +2.03 = m2ΔV2

ΔV2 = \(\frac{+2.03}{1.05}=1.933 m/s\)

Vf2 = V02 + ΔV2 = 0 + 1.933 = 1.933 m/s

ii.

KE0 : KEf (elastic condition)

KE0 : \(\frac{1}{2}(0.500)(3)^{2}=2.25\)

KEf : \(\frac{1}{2}(0.500)(1.06)^{2}+\frac{1}{2}(1.05)(1.933)^{2}=2.24\)

2.25 ≈ 2.24, thus elision is elastic

(d) i. 

∑P = ∑m (Vcom)

\(V_{com}=\frac{\sum P}{\sum m}=\frac{m_{2}v_{1}}{m_{1}+m_{2}}=\frac{0.500(3)}{0.500+1.05}=0.968 m/s\)

ii.

\(U_{max}=\frac{1}{2}Kx_{max}\)

\(F_{max}=Kx_{max}, \)         \(x_{max} = \frac{F_{max}}{K}\approx \frac{21}{130}=.161\)

\(U_{max} = \frac{1}{2}(130)(0.161)=10.465 J\)

Question

A block of mass 2M rests on a horizontal, frictionless table and is attached to a relaxed spring, as shown in the figure above. The spring is nonlinear and exerts a force f(x) = – Bx3, where B is a positive constant and x is the displacement from equilibrium for the spring. A block of mass 3M and initial speed v0 is moving to the left as shown.
(a) On the dots below, which represent the blocks of mass 2M and 3M, draw and label the forces (not components) that act on each block before they collide. Each force must be represented by a distinct arrow starting on, and pointing away from, the appropriate dot.

The two blocks collide and stick to each other. The two-block system then compresses the spring a maximum distance D, as shown above. Express your answers to parts (b), (c), and (d) in terms of M, B, v0 , and physical constants, as appropriate.
(b) Derive an expression for the speed of the blocks immediately after the collision.
(c) Determine an expression for the kinetic energy of the two-block system immediately after the collision.
(d) Derive an expression for the maximum distance D that the spring is compressed.
(e)
i. In which direction is the net force, if any, on the block of mass 2M when the spring is at maximum compression? 
____  Left ____Right ____ The net force on the block of mass 2M is zero.
Justify your answer.
ii. Which of the following correctly describes the magnitude of the net force on each of the two blocks when the spring is at maximum compression?
____ The magnitude of the net force is greater on the block of mass 2M.
____ The magnitude of the net force is greater on the block of mass 3M.
____ The magnitude of the net force on each block has the same nonzero value.
____ The magnitude of the net force on each block is zero.
Justify your answer.
____
____
(f) Do the two blocks, which remain stuck together and attached to the spring, exhibit simple harmonic motion after the collision?
____ Yes ____ No
Justify your answer.

Answer/Explanation

Ans:

(a)

(b)

3MV0 = 5MV

\(\frac{3}{5}V_{0} = V\)

(c)

\(KE = \frac{1}{2}MV^{2}\)

\(= \frac{1}{2}SM\left ( \frac{3}{5}V_{0} \right )^{2}\)

\(KE = \frac{9}{10}M{V_{0}}^{2}\)

(d)

\(W = \int -Bx^{3}dx = \frac{Bx^{4}}{4}\)

\(\frac{9}{10}M{V_{0}}^{2} = \frac{BD^{4}}{4}\)

\(\sqrt[4]{\frac{3.6M{V_{0}}^{2}}{B}}= D\)

(e) i.

    √    Right

The block will start accelerating to the right due to the force of the spring of the maximum distance, so the net force is to the right.

ii.

    √    The magnitude of the net force is greater on the block of mass 3M.

The acceleration is constant, but the net force on the 3M block will be greater since 3Ma > 2Ma

(f)

    √    No

Sma the surface is frictionless, while the 2M block will follow SMH, the 3M block will continue to the right with a constant acceleration

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