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 inmomentum 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