PHYS101: Introduction to Mechanics

Example 8.5 Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie

(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See Figure 8.9)

Figure 8.9 An ice hockey goalie catches a hockey puck and recoils backward. The initial kinetic energy of the puck is almost entirely converted to thermal energy and sound in this inelastic collision.

Strategy

Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested.

Solution for (a)

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

Conservation of momentum is

$p_{1}+p_{2}=p_{1}^{\prime}+p_{2}^{\prime}$

or

$m_{1} v_{1}+m_{2} v_{2}=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime}$.

Because the goalie is initially at rest, we know $v2=0$. Because the goalie catches the puck, the final velocities are equal, or $v_{1}^{\prime}=v_{2}^{\prime}=v^{\prime}$ . Thus, the conservation of momentum equation simplifies to

$m_{1} v_{1}=\left(m_{1}+m_{2}\right) v^{\prime}$.

Solving for ${v}^{1}$ yields

$v^{\prime}=\frac{m_{1}}{m_{1}+m_{2}} v_{1}$.

Entering known values in this equation, we get

$v^{\prime}=\left(\frac{0.150 \mathrm{~kg}}{0.150 \mathrm{~kg}+70.0 \mathrm{~kg}}\right)(35.0 \mathrm{~m} / \mathrm{s})=7.48 \times 10^{-2} \mathrm{~m} / \mathrm{s}$.

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 $\mathrm{KE}_{\text {int }}$ of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, $\mathrm{KE}_{\text {int }}$ is initially

$\begin{aligned} \mathrm{KE}_{\text {int }} &=\frac{1}{2} m v^{2}=\frac{1}{2}(0.150 \mathrm{~kg})(35.0 \mathrm{~m} / \mathrm{s})^{2} \\ &=91.9 \mathrm{~J} \end{aligned}$.

After the collision, the internal kinetic energy is

$\begin{aligned} \mathrm{KE}_{\text {int }}^{\prime} &=\frac{1}{2}(m+M) v^{2}=\frac{1}{2}(70.15 \mathrm{~kg})\left(7.48 \times 10^{-2} \mathrm{~m} / \mathrm{s}\right)^{2} \\ &=0.196 \mathrm{~J} \end{aligned}$.

The change in internal kinetic energy is thus

$\begin{aligned} \mathrm{KE}_{\text {int }}^{\prime}-\mathrm{KE}_{\text {int }} &=0.196 \mathrm{~J}-91.9 \mathrm{~J} \\ &=-91.7 \mathrm{~J} \end{aligned}$

where the minus sign indicates that the energy was lost.

Discussion for (b)

Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. $\mathrm{KE}_{\text {int }}$ 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. Figure 8.10 shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example 8.6 deals with data from such a collision.

Figure 8.10 An air track is nearly frictionless, so that momentum is conserved. Motion is one-dimensional. In this collision, examined in Example 8.6, 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.

Example 8.6 Calculating Final Velocity and Energy Release: Two Carts Collide

In the collision pictured in Figure 8.10, two carts collide inelastically. Cart 1 (denoted $m1$ carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of $2.00 m/s$. Cart 2 (denoted $m2$ in Figure 8.10) has a mass of 0.500 kg and an initial velocity of $−0.500 m/s$. After the collision, cart 1 is observed to recoil with a velocity of $−4.00 m/s$. (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)?

Strategy

We can use conservation of momentum to find the final velocity of cart 2, because $F_{\text {net }}=0$ (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring.

Solution for (a)

As before, the equation for conservation of momentum in a two-object system is

$m_{1} v_{1}+m_{2} v_{2}=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime}$

The only unknown in this equation is $v_{2}^{\prime}$. Solving for $v_{2}^{\prime}$ and substituting known values into the previous equation yields

$\begin{aligned} v_{2}^{\prime} &=\frac{m_{1} v_{1}+m_{2} v_{2}-m_{1} v_{1}^{\prime}}{m_{2}} \\ &=\frac{(0.350 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})+(0.500 \mathrm{~kg})(-0.500 \mathrm{~m} / \mathrm{s})}{0.500 \mathrm{~kg}}-\frac{(0.350 \mathrm{~kg})(-4.00 \mathrm{~m} / \mathrm{s})}{0.500 \mathrm{~kg}} \\ &=3.70 \mathrm{~m} / \mathrm{s} \end{aligned}$

Solution for (b)

The internal kinetic energy before the collision is

$\begin{aligned} \mathrm{KE}_{\text {int }} &=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \\ &=\frac{1}{2}(0.350 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2}(0.500 \mathrm{~kg})(-0.500 \mathrm{~m} / \mathrm{s})^{2} \\ &=0.763 \mathrm{~J} \end{aligned}$

After the collision, the internal kinetic energy is

$\begin{aligned} \mathrm{KE}_{\text {int }}^{\prime} &=\frac{1}{2} m_{1} v_{1}^{\prime 2}+\frac{1}{2} m_{2} v_{2}^{\prime 2} \\ &=\frac{1}{2}(0.350 \mathrm{~kg})(-4.00 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2}(0.500 \mathrm{~kg})(3.70 \mathrm{~m} / \mathrm{s})^{2} \\ &=6.22 \mathrm{~J} \end{aligned}$

The change in internal kinetic energy is thus

$\begin{aligned} \mathrm{KE}_{\text {int }}^{\prime}-\mathrm{KE}_{\text {int }} &=6.22 \mathrm{~J}-0.763 \mathrm{~J} \\ &=5.46 \mathrm{~J} \end{aligned}$

Discussion

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.