Question

In a Little Sticks hockey game at the community ice rink, a 0.102 kg hockey puck moving at 26.4 m/s is caught and held by a 40.9 kg goalie who is at rest. With what speed (in cm/s) does the goalie slide on the ice after the catch?

Answer 1

Approximately
`6.57\; {\rm cm \cdot s^(-1)}`.

Step-by-step explanation:

When an object of mass
`m` travels at a velocity of
`v`, the momentum
`p` of that object would be
`p = m\, v`.

Before the catch:

• Velocity of the hockey puck:
  `26.4\; {\rm m\cdot s^(-1)}`. Mass of the hockey puck:
  `0.102\; {\rm kg}`. Momentum of the hockey would be
  `(26.4\; {\rm m\cdot s^(-1)})\, (0.102\; {\rm kg}) = 2.6928\; {\rm kg\cdot m\cdot s^(-1)}`.
• Velocity of the goalie:
  `0\; {\rm m\cdot s^(-1)}`. Momentum of the goalie would be
  `0\; {\rm kg\cdot m\cdot s^(-1)}`.

Therefore, the total momentum of the hockey and the goalie before the catch was:

`2.6928\; {\rm kg\cdot m\cdot s^(-1)} + 0\; {\rm kg \cdot m\cdot s^(-1)} = 2.6928\; {\rm kg\cdot m\cdot s^(-1)}`

The goalie and the hockey move at the same velocity after the catch. Let
`v` denote that velocity. The total momentum of them would be:

`\begin{aligned}& m(\text{goalie})\, v + m(\text{hockey})\, v \\ =\; & (m(\text{goalie})+ m(\text{hockey}))\, v \end{aligned}`.

Assume that momentum is conserved during the catch. Hence:

`\begin{aligned}& (m(\text{goalie})+ m(\text{hockey}))\, v \\ &= (\text{momentum after catch}) \\ &= (\text{momentum before catch}) \\ &= 2.9628\; {\rm kg \cdot m\cdot s^(-1)}\end{aligned}`.

Rearrange the equation to find
`v`, the velocity of the goalie and the hockey after the catch:

`\begin{aligned}v &= \frac{2.9628\; {\rm kg \cdot m\cdot s^(-1)}}{m(\text{goalie})+ m(\text{hockey})} \\ &= \frac{2.9628\; {\rm kg \cdot m\cdot s^(-1)}}{0.102\; {\rm kg} + 40.9\; {\rm kg}} \\ &\approx 0.0657\; {\rm m\cdot s^(-1)}\end{aligned}`.

Apply unit conversion:

`\begin{aligned}v &\approx 0.0657\; {\rm m\cdot s^(-1)} = 6.57\; {\rm cm\cdot s^(-1)}\end{aligned}`.