Question

A bowling ball is initially at rest. A Ping-Pong ball moving in the +z direction hits the bowling ball and bounces off it, traveling back in the –z direction. Consider a time interval $\mathbf{∆}\mathit{t}$ extending from slightly before to slightly after the collision. (a) In this time interval, what is the sign of $\mathbf{∆}{\mathit{P}}_{\mathbf{z}}$ for the system consisting of both balls? (b) In this time interval, what is the sign of $\mathbf{∆}{\mathit{P}}_{\mathbf{z}}$ for the system consisting of the bowling ball alone?

Short answer

a)$\mathbf{∆}{\mathit{P}}_{\mathbf{z}}$ for the system consisting of both balls is zero.

b)$\mathbf{∆}{\mathit{P}}_{\mathbf{z}}$for the system consisting of the bowling ball alone is positive

Step-by-step solution

Significance of conservation of momentum

The conservation of momentum is a fundamental rule of physics that asserts that when no external forces are acting on a system, its momentum remains constant.

The sign ∆Pz  for the system consisting of both balls

Part a)

$∆P_z=0$

Because there should be a conversion of momentum when the ping-pong ball bounces backward at a much higher velocity than the bowling ball travelling in the +z direction at a much slower velocity, and the overall momentum is considered only because the mass is more different

The sign ∆Pz for the system consisting of bowling ball alone

Part b)

$∆P_z>0$

The change in velocity for ping pong ball is greater as compare to the bowling ball.

Because having no momentum in the z-direction, that has some small positive momentum.

Momentum difference for bowling ball role="math" localid="1653980302269" $$\begin{gathered} =P_2-P_1 \\ =m_2{v}_2-m_2{v}_1 \end{gathered}$$

As $v_2<v_1$, the change will be negative.

Thus, the$\mathbf{∆}{\mathit{P}}_{\mathbf{z}}$ for the system consist of bowling ball alone is negative.