A 0.105-kg hockey puck moving at 20 m/s is caught and held by a 77-kg goalie at rest. With what speed does the goalie slide on the ice?
One would not expect a goalie in front of the net to NOT have his skates parallel in a "free to slide" position. A goalie needs to be able to push his body forward and from side to side on short notice. They must want you to apply conservation of momentum and assume no friction, but that is a poor assumption in this situation. Here is what they expect you to use, even though is it is not accurate:
(Initial puck speed)(Puck mass)
= (combined mass)*(final goalie velocity)
To answer this question, we can apply the law of conservation of momentum. According to this law, the total momentum before the catch should be equal to the total momentum after the catch.
Before the catch, the momentum of the hockey puck can be calculated as the product of its mass (0.105 kg) and its velocity (20 m/s), which gives:
Momentum of hockey puck before catch = 0.105 kg * 20 m/s
After the catch, both the hockey puck and the goalie will move together. The momentum of the goalie can be calculated as the product of the goalie's mass (77 kg) and the velocity we need to find, which gives:
Momentum of goalie after catch = 77 kg * v (where v is the velocity we need to find)
Since momentum is conserved before and after the catch, we can equate the two expressions:
0.105 kg * 20 m/s = 77 kg * v
Now, we can solve for v by dividing both sides of the equation by 77 kg:
v = (0.105 kg * 20 m/s) / 77 kg
Calculating this expression gives us the speed at which the goalie slides on the ice after catching the hockey puck.