Exemplo de Physics, 8.3 Elastic and Inelastic Collisions

Calculating Velocity: Inelastic Collision of a Puck and a Goalie

Problema e solução em inglês, como no livro original.

Find the recoil velocity of a 70 kg ice hockey goalie who catches a 0.150-kg hockey puck slapped at him at a velocity of 35 m/s. Assume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible (see Figure 8.9).

An illustration shows before and after diagrams of a hockey puck and an ice hockey player on a frictionless ice surface. The hockey puck is labeled m one and the hockey player is labeled m two. Both diagrams are labeled System of Interest. In the before diagram, a velocity vector, v one, points from the puck to the player. The following are shown: p one equals p total, v two equals zero, and net F equals zero. In the after diagram, the hockey puck has made contact with the player. A velocity vector, v prime, points from the hockey player to the right. The equation p one prime plus p two prime equals p total is shown. The equation v one prime plus v two prime equals v prime is also shown.

Figure 8.9 An ice hockey goalie catches a hockey puck and recoils backward in an inelastic collision.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
Momentum is conserved because the net external force on the puck-goalie system is zero. Therefore, we can 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.
Solução
For an inelastic collision, conservation of momentum is

$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'},$
8.8

where v′ is the velocity of both the goalie and the puck after impact. Because the goalie is initially at rest, we know v₂ = 0. This simplifies the equation to

$m_{1} v_{1} = (m_{1} + m_{2}) v^{'} .$
8.9

Solving for v′ yields

$v^{'} = (\frac{m_{1}}{m_{1} + m_{2}}) v_{1} .$
8.10

Entering known values in this equation, we get

$\begin{matrix} v^{'} & = & (\frac{0.150 kg}{70.0 kg + 0.150 kg}) (35 m/s) \\ = & 7.48 \times 10^{- 2} m/s. \end{matrix}$
8.11

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