Question

A 94.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 44.0 m/s. If both are initially at rest and if the ice is frictionless,
how far does the player recoil in the time it takes the puck to reach the goal 16.5 m away?

Answer 1

Answer:

0.026 meter

Explanation:

using distance time equation to determine the time for the puck to move 16.5 meters.

distance 'd' = velocity'v' x time't'

16.5 = 44 x t    

t =0.375 second

Here momentum is conserved. Since both objects are initially at rest, the initial momentum is 0.

Next is to determine the puck’s momentum.

Momemtum 'p' = m x v => 0.15 x 44 = 6.6kg⋅m/s

The player momentum is -6.6kg⋅m/s  .

In order to determine the player’s velocity, we'll use p=mv

-6.6 = 94v

v= -0.0702 m/s

The above negative sign represents that the player is moving in the opposite direction of the puck.

Lastly, how far does the player recoil in the time it takes the puck to reach the goal 16.5 m away?

d = v x t = 0.0702 x 0.375= 0.026 meter