A 261 g hockey puck is sliding on ice with a speed of 5.34 m/s hits a wall at an angle of 27.3° to the wall and bounces back at the same angle with the same speed.
a) If the wall and the hockey puck are in contact for 10.9 ms, find the magnitude of the impulse on the puck.
b) Calculate the average force exerted by the puck on the wall.
J = F*∆t = ∆p
p = m*v
a = ∆v/∆t
The Attempt at a Solution
a) I assumed that velocity in the x is not changing, and thus ∆Vx = ∆px = 0. Based on that, I assumed that any impulse/change in momentum, can be calculated from the y-component.
Jy = J = m*∆Vy
= m(Vfy - Viy) _____ Viy = -Vfy
= m (2Vfy)
=2*m*Vfy = (2)(0.261)(5.34 sin 27.3)
J = 2.306 kg*m/s
however, it says that this answer is incorrect. I'm not sure where i'm going wrong, as only the initial and final velocities in the y are changing, and thus Jy = J.
b) I'm assuming that the average force can be calculated via J = F*∆t, as i don't have the correct magnitude of the impulse yet, I haven't been able to calculate it correctly