Question

5)A 0.50 kg hockey puck is at rest on ice when you hit it with a hockey stick, applying a force of 100 N for

0.10 seconds. The puck then slides across the ice, where the coefficient of kinetic friction is 0.20. After
sliding for 4 seconds, the puck collides with a 0.80 kg box of donuts. The puck immediately comes to rest
after the collision. How far will the box of donuts slide before coming to rest if the coefficient of friction
between the box and the ice is 0.30.

Answer 1

F t = m Δv impulse delivered = change in momentum

Δv = 100 * .1 / .5 = 20 m/s original speed of puck

KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck

E = μ m g d energy lost by puck

Ff = μ m g = m a deceleration of puck due to friction

a = μ g = 9.8 * .2 = 1.96 m/s^2

v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box

m v2 = M V conservation of momentum when puck strikes box

V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision

KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision

KE = μ M g d energy lost by box in sliding distance d

d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides