A basket ball of mass 0.14kg is moving horizontally at a speed of 42 m/s when it is struck by a bat.It leaves the bat in a direction at an angle 35° above its incident path and with a speed of 50m/s.(1)Find the impulse of the force exerted on the ball. (2) Assuming the collision lasts for 1.5ms.what is the average force?(3) Find the change in momentum of the bat.

a) impulse = change in momentumI.e 0.14*50*cos35 - 0.14*42=5.74-5.88 = -0.14 B) force = impulse /time = -0.14/1.5 =0.09 NC) as there was no external force the momentum should be conserved. Let m be mass of bat, and u be initial vel of bat and v be final vel. of bat. Conserving momentum in horizontal direction, mu+0.14*42=mv+o.14*50*cos35Ie. mv-mu=o.14*42-0.14*50*cos35=0.14 Nt