Question

Primary SchoolPhysics 5+3 pts


A ball of mass 0.5 kg moving with a velocity of 2 metre per second strikes a wall normally and bounces back with same speed.I f
the time of contact between the ball and the wall is 1 millisecond the average force exerted by the wall on the ball

Answer 1

Hey Dear !

=)

Mass of ball = 0.5 kg

initial speed = 10 m/s

final speed = -10 m/s ( because it rebounds elastically)

Change in momentum= mv - mu =>

∆P = 0.5×(-10) - 0.5×10=>∆P = -10 kg.m/s

Given that contact time = 0.1s

force = ∆P/t = -10/0.1 = -100 N

Thus, the change in momentum is 10 kg.m/s in the opposite direction of its motion and force acted is 100N in the opposite direction of motion.

Hope , It Helps .