A 3.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and

Question

A 3.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (-4i + 3j) m/s. What is the impulse exerted on the ball by the wall? A 1.6-kg ball is attached to the end of a 0.40-m string to form a pendulum. this pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80-kg block initially at rest on a horizontal frictionless surface. the speed of the block just after the collision is 3.0 m/s. What is the speed of the ball just after the collision?

Explanation / Answer

a )

the imulse is equal to the change in momentum

so

impulse = momentum after coliision - momentum before collision

= 3 ( - 4 i + 3 j ) - 3 ( 4 i + 3 j )

= -12 i + 9 j - 12 i - 9 j

impulse = - ( 24 i ) N.sec

b )

using conservation of energy

so kinetic energy = potential energy

m g h = 0.5 m v2

v = ( 2 g h )1/2

v = ( 2 X 9.8 x 0.4 )1/2

v = 2.8 m/s

the momentum before collision

= 1.6 X 2.8 = 4.48 kg.m/s

and the same way

after collision the momentum = 0.8 X 3

= 2.4 kg.m/s

the change in momentum = 4.48 - 2.4

= 2.08 kg.m/s

the speed of the ball just after collision is = 2.08 / 1.6 = 1.3 m/s

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