A 0.045-kg golf ball moving at 40 m/s (measured in the Earth reference frame) co

Question

A 0.045-kg golf ball moving at 40 m/s (measured in the Earth reference frame) collides inelastically with a 2.3-kg, heavy-duty plastic flowerpot sitting on a windowsill. The coefficient of restitution for the collision is 0.50. Do this problem by first transforming to the zero-momentum reference frame, where the collision is much simpler, and then transforming back to the Earth reference frame. (Assume that all the action happens in one dimension and that the friction between pot and sill is insignificant.) Call the initial direction of the golf ball's motion the +x direction.

1.) Calculate the x-component of the final velocity of the ball.

2.) Calculate the x-component of the final velocity of the pot.

Explanation / Answer

m1 =0.045 kg , m2 =2.3 kg, u1x =40 m/s, e = 0.5

u1y =0 , u2x =0, u2y =0

From conservation of momentum along x -direction

m1u1x+m2u2x = m1v1x+m2v2x

0.045*40+0 = 0.045v1+2.3v2 .... (1)

e =0.5 = v2-v1/u1-u2

0.5 = v2-v1/(40)

v2 -v1 = 20 ..(2)

By solving (1) and (2) equaitons we get

v1 = -18.85 m/s, v2 = 1.15m/s

x-component of the final velocity of the ball is 18.85 m/s

x-component of the final velocity of the pot =1.15 m/s

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