A 2.55 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0

Question

A 2.55 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0 with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.225 s, what is the average force exerted on the ball by the wall?

x-component
  N
y-component
N

A 2.55 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0½ with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.225 s, what is the average force exerted on the ball by the wall? x-component N y-component N

Explanation / Answer

v1-v2 = (v0*sin(60))-(-v0*sin(60)) = 8.66+8.66 = 17.32 m/s
Then you will take the change in velocity over the time to find the acceleration:
(v1-v2)/t = (17.32)/(.225) = 76.978 m/s^2

After this, to find the force, you will use F=ma:
F=m*a = (2.55)*(76.978) = 196.294 N   

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