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

Answer 1

Answer:

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