A hockey player strikes the 0.25 kg puck with an impulse force given by F(t) =10

Question

A hockey player strikes the 0.25 kg puck with an impulse force given by F(t) =100sin(50t) N for the duration of 0.04pi seconds. The puck hits the wall and bounces back as shown. the coefficient of friction between the puck and the ice is 0.02 and the coefficientt of restitution between the puck and the wall is 0.85. The distance between the puck and the wall is 5 meters.

What is the speed of the puck after applying the force F?

What is the speed of the puck just before the impact with the wall?

What is the magnitude and direction of bouncing puck after the impact with the wall?

PLEASE HELP

Explanation / Answer

1) Using Impulse

Integral F.dt from 0 to 0.04 pi

2 Cos(50t) = 2 * (Cos(50*0.04*3.14)- 1) = 0.000010146

Impulse = change in momentum

so 0.000010146 = mv

v= 0.000010146/0.25 = 0.000040584 m/s = 4.058 *10^(-5) m/s

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