Suppose you throw a ball with mass 0.40 kg against a brick wall It hits the wall

Question

Suppose you throw a ball with mass 0.40 kg against a brick wall It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20 m/s. Find the impulse of the force exerted on the ball by the wall. If the ball is in contact with the wall for 0.010 s, find the average force on the ball during the impact. The change in the x component of momentum is Delta Px = Pf,x - Pi,x = mu epsilon f,x - mu epsilon i,x = 8.0 kg . m/s - (-12kg . m/s) = 20 kg . m/s (Note the signs carefully.

Explanation / Answer

change in momentum of lump of clay (Impulse) J = m(V2-V1)

J =0.2 (0-(-30))

J = 6 kg-m/s

now J= F*t

F = J/t

F = 6/0.010

average force F= 600N

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