Finding the impulse of a hockey puck colliding with a wall

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The discussion revolves around calculating the impulse experienced by a hockey puck colliding with a wall at an angle. The puck, weighing 261 g and moving at 5.34 m/s, hits the wall and bounces back with the same speed and angle. The user initially calculated the impulse using only the y-component of velocity but received an incorrect answer, indicating a possible error in their calculations. They also noted that the average force exerted by the puck on the wall can be determined using the impulse and the contact time of 10.9 ms. The conversation highlights the importance of ensuring the correct settings on calculators for accurate results.
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Homework Statement



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.

Homework Equations


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
 
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Check your calculator for the degree/radian setting!
 
Wow I can't believe I missed that, thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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