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AP Physics 1- 4.4 Elastic and Inelastic Collisions - Exam Style questions - FRQs- New Syllabus

Elastic and Inelastic Collisions AP  Physics 1 FRQ

Unit 4: Linear Momentum

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question

A student has a cart of mass \( m_c \) and a block of mass \( \dfrac{1}{5}m_c \), as shown in Figure \( 1 \).
• At time \( t=0 \), the cart is moving to the right across a horizontal surface with constant speed \( v_c \), and the student releases the block from rest.
• At \( t=t_1 \), the block collides with and sticks to the top of the cart. The block does not slide on the cart.
• At \( t=t_2 \), the block-cart system continues to move to the right with constant speed \( v_f \).
A.
(i) On the axes shown in Figure \( 2 \), sketch a graph of the magnitude \( p_x \) of the \( x \)-component of the momentum of the block-cart system as a function of time \( t \) from \( t=0 \) until \( t_2 \).
(ii) Derive an expression for the speed \( v_f \) of the block-cart system after time \( t=t_2 \) in terms of \( m_c \), \( v_c \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
(iii) Derive an expression for the change in kinetic energy \( \Delta K \) in the block-cart system from \( t=0 \) to \( t=t_2 \) in terms of \( m_c \), \( v_c \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
B. Consider there a new block is dropped and collides with the top of the cart. The new block slides along the cart during the collision but does not slide off the cart. The time interval from when the new block collides with the cart and moves together with the cart is \( \Delta t \). During \( \Delta t \) there is a frictional force between the new block and the cart.
Indicate whether the \( x \)-component of the momentum of the new block-cart system increases, decreases, or remains constant during \( \Delta t \).
_____ Increases
_____ Decreases
_____ Remains constant
Justify your response.

Most-appropriate topic codes (AP Physics 1):

• Topic \( 4.1 \) — Linear Momentum (Part \( \mathrm{A(i)} \), Part \( \mathrm{B} \))
• Topic \( 4.2 \) — Change in Momentum and Impulse (Part \( \mathrm{B} \))
• Topic \( 4.3 \) — Conservation of Linear Momentum (Part \( \mathrm{A(ii)} \), Part \( \mathrm{B} \))
• Topic \( 4.4 \) — Elastic and Inelastic Collisions (Part \( \mathrm{A(ii)} \), Part \( \mathrm{A(iii)} \))
• Topic \( 3.1 \) — Translational Kinetic Energy (Part \( \mathrm{A(iii)} \))
▶️ Answer/Explanation

A(i)
The graph of \( p_x \) versus \( t \) is a horizontal line above zero from \( t=0 \) all the way to \( t_2 \).

Since there is no net external horizontal force on the block-cart system, the \( x \)-component of momentum remains constant throughout the motion, including during the collision at \( t=t_1 \).

So the sketch should show one continuous, nonzero, horizontal line from \( t=0 \) to \( t=t_2 \).

A(ii)
Use conservation of linear momentum:

\( p_i = p_f \)

Initially, only the cart has horizontal momentum, because the block is released from rest and has no initial horizontal speed. Therefore,

\( p_i = m_c v_c \)

After the collision, the cart and block stick together, so the total mass is

\( m_c + \dfrac{1}{5}m_c = \dfrac{6}{5}m_c \)

Thus,

\( m_c v_c = \left(\dfrac{6}{5}m_c\right)v_f \)

Solving for \( v_f \),

\( v_f = \dfrac{m_c v_c}{(6/5)m_c} = \dfrac{5}{6}v_c \)

Therefore, \( \boxed{v_f=\dfrac{5}{6}v_c} \)

The final speed is smaller than \( v_c \), which makes sense because the cart has to carry extra mass after the perfectly inelastic collision.

A(iii)
Use

\( \Delta K = K_f – K_i \)

Initial kinetic energy:

\( K_i = \dfrac{1}{2}m_c v_c^2 \)

Final kinetic energy:

\( K_f = \dfrac{1}{2}\left(\dfrac{6}{5}m_c\right)v_f^2 \)

Substitute \( v_f=\dfrac{5}{6}v_c \):

\( K_f = \dfrac{1}{2}\left(\dfrac{6}{5}m_c\right)\left(\dfrac{5}{6}v_c\right)^2 \)

\( K_f = \dfrac{1}{2}\left(\dfrac{6}{5}m_c\right)\left(\dfrac{25}{36}v_c^2\right) \)

\( K_f = \dfrac{1}{2}\left(\dfrac{5}{6}\right)m_c v_c^2 = \dfrac{5}{12}m_c v_c^2 \)

Therefore,

\( \Delta K = \dfrac{5}{12}m_c v_c^2 – \dfrac{1}{2}m_c v_c^2 \)

\( \Delta K = \dfrac{5}{12}m_c v_c^2 – \dfrac{6}{12}m_c v_c^2 \)

\( \Delta K = -\dfrac{1}{12}m_c v_c^2 \)

Therefore, \( \boxed{\Delta K=-\dfrac{1}{12}m_c v_c^2} \)

The negative sign shows that kinetic energy decreases, which is expected in a perfectly inelastic collision.

B
\( \boxed{\text{Remains constant}} \)

The friction force between the new block and the cart is internal to the block-cart system. During \( \Delta t \), the friction force on the block and the friction force on the cart are equal in magnitude and opposite in direction, so they do not create a net external horizontal force on the system.

Because only a net external force can change the momentum of the system, the \( x \)-component of the momentum of the new block-cart system remains constant during \( \Delta t \).

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