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A crane drops a 0.30 kg steel ball onto a steel plate. The ball’s speeds just before impact and after are 4.5 m/s and 4.2 m/s, respectively. If the ball is in contact with the plate for 0.030 s, what is the magnitude of the average force that the ball exerts on the plate during impact?

Respuesta :

davidlcastiblanco

Answer:

F=-3N

Explanation:

Newton's 2nd law

Fnet = ∆p/t

where ∆p = mv - mu (change in momentum or impluse).

So  (assuming the ball continues to move in same direction).

∆p = 0.3kg*4.2m/s -0.3kg*4.5m/s

∆p= 1.26 N -1.35 N

∆p = -0.09 Ns

Therefore Fnet

Fnet=∆p/t

Fnet= -0.09N/0.030s

Fnet = -3N

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