A 95.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 49.0 m/s. If both are initially at rest and if the ice is frictionless, how far (in m) does the player recoil in the time it takes the puck to reach the goal 12.5 m away

0.019 meter

Explanation:

Let’s use the following equation to determine the time for the puck to move 15 meters.

Firstly, let us solve for the time the puck takes to move 12.5m

d = v * t

12.5= 49 * t    

t = 12.5/49 second

t= 0.25sec

also we know that momentum M=mv

the momentum of the puck

M = 0.15 *49 = 7.35

this means that the players momentum is  -7.35.

we can solve for the velocity of the player as

M=mv

-7.35 = 95 * v  

v = -0.077 m/s  ( "The negative sign means the player is moving in the opposite direction of the puck".)

when the puck move 15 meters. the player will move

d = v * t =0.077 * 0.25 = 0.019 meter

this means that the temperature of both substances will be equal

i think the answer would be reserve.