Question

a bolwing ball of mass 2.00 kg strikes a stationary pin of mass 5.00 x 10^2 g. the collision lasts for .60s after which the pin moves off with a velocity of 12.0 m/s [w]
a)accel of pin during the collision
b)force exerted by bowling ball on the pin
c)the accel of the bowling ball during collision

Answer 1

a) acceleration = (Pin velocity change)/(0.60 s)
= 20 m/s^2 = a

b) F = M(pin)*a = 0.500 kg*20 m/s^2
= 10 Newtons
The forces F on ball and pin are equal and opposite, so
c) a(ball) = F/M(ball) = 10/2 = 5 m/s^2