Question

A 110 g ball moving to the right at 4.0 m/s catches up and collides with a 410 g ball that is moving to the right at 1.0 m/s .

If the collision is perfectly elastic, what is the speed of the 110 g ball after the collision?

If the collision is perfectly elastic, what is the direction of motion of the 110 g ball after the collision?

If the collision is perfectly elastic, what is the speed of the 410 g ball after the collision?

If the collision is perfectly elastic, what is the direction of motion of the 410 g ball after the collision?

Answer #1

Let the speed of 110g after collision be v1 m/s and 410 g be v2 m/s

by the conservation of momentum we have

m1u1 + m2u2 = m1v1 + m2v2

110*4 + 410*1 = 110v1 + 410v2

1.1v1 + 4.1v2 = 8.5 --------(1)

velocity of approach = velocity of recess

v2 - v1 = u1 - u2

v2 - v1 = 4 - 1

v2 - v1 = 3 --------(2)

multiply (2) by 1.1 and adding to (1) we have

5.2v2 = 11.8

v2 = 2.27 m/s

v1 = 2.27 - 3 = -0.73 m/s

speed of 110g after collision was 0.73 m/s direction of motion is to the left

speed of 410 g after collision was 2.27 m/s direction of motion is to the right

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