# 8.5 Inelastic collisions in one dimension  (Page 2/7)

 Page 2 / 7

Solution for (a)

Momentum is conserved because the net external force on the puck-goalie system is zero.

Conservation of momentum is

${p}_{1}+{p}_{2}={p\prime }_{1}+{p\prime }_{2}$

or

${m}_{1}{v}_{1}+{m}_{2}{v}_{2}={m}_{1}{v\prime }_{1}+{m}_{2}{v\prime }_{2}.$

Because the goalie is initially at rest, we know ${v}_{2}=0$ . Because the goalie catches the puck, the final velocities are equal, or ${v\prime }_{1}={v\prime }_{2}=v\prime$ . Thus, the conservation of momentum equation simplifies to

${m}_{1}{v}_{1}=\left({m}_{1}+{m}_{2}\right)v\prime .$

Solving for $v\prime$ yields

$v\prime =\frac{{m}_{1}}{{m}_{1}+{m}_{2}}{v}_{1}.$

Entering known values in this equation, we get

$v\prime =\left(\frac{\text{0.150 kg}}{\text{70.0 kg}+\text{0.150 kg}}\right)\left(\text{35}\text{.0 m/s}\right)=7\text{.}\text{48}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$

Discussion for (a)

This recoil velocity is small and in the same direction as the puck’s original velocity, as we might expect.

Solution for (b)

Before the collision, the internal kinetic energy ${\text{KE}}_{\text{int}}$ of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, ${\text{KE}}_{\text{int}}$ is initially

$\begin{array}{lll}{\text{KE}}_{\text{int}}& =& \frac{1}{2}{\text{mv}}^{2}=\frac{1}{2}\left(0\text{.}\text{150 kg}\right){\left(\text{35}\text{.0 m/s}\right)}^{2}\\ \text{}& =& \text{91}\text{.}\text{9 J}\text{.}\end{array}$

After the collision, the internal kinetic energy is

$\begin{array}{lll}{\text{KE}\prime }_{\text{int}}& =& \frac{1}{2}\left(m+M\right){v}^{2}=\frac{1}{2}\left(\text{70}\text{.}\text{15 kg}\right){\left(7\text{.}\text{48}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)}^{2}\\ & =& \text{0.196 J.}\end{array}$

The change in internal kinetic energy is thus

$\begin{array}{lll}{KE\prime }_{\text{int}}-{\text{KE}}_{\text{int}}& =& \text{0.196 J}-\text{91.9 J}\\ & =& -\text{91.7 J}\end{array}$

Discussion for (b)

Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. ${\text{KE}}_{\text{int}}$ is mostly converted to thermal energy and sound.

During some collisions, the objects do not stick together and less of the internal kinetic energy is removed—such as happens in most automobile accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. [link] shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. [link] deals with data from such a collision. An air track is nearly frictionless, so that momentum is conserved. Motion is one-dimensional. In this collision, examined in [link] , the potential energy of a compressed spring is released during the collision and is converted to internal kinetic energy.

Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports—a lightweight bat (such as a softball bat) cannot hit a hardball very far.

The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the “sweet spot” on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.

## Take-home experiment—bouncing of tennis ball

1. Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your friend’s hand during the collision. Explain your observations and measurements.
2. The coefficient of restitution $\left(c\right)$ is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the speeds after and before the collision. A perfectly elastic collision has a $c$ of 1. For a ball bouncing off the floor (or a racquet on the floor), $c$ can be shown to be $c=\left(h/H{\right)}^{1/2}$ where $h$ is the height to which the ball bounces and $H$ is the height from which the ball is dropped. Determine $c$ for the cases in Part 1 and for the case of a tennis ball bouncing off a concrete or wooden floor ( $c=0\text{.}\text{85}$ for new tennis balls used on a tennis court).

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