A 120g ball moving to the right at 4.5m/s catches up and collides with a 400g ba

Question

A 120g ball moving to the right at 4.5m/s catches up and collides with a 400g ball that is moving to the right at 1.0m/s .

A) If the collision is perfectly elastic, what is the speed of the 120g ball after the collision?

B) If the collision is perfectly elastic, what is the direction of motion of the 120 ball after the collision?

C) If the collision is perfectly elastic, what is the speed of the 400g ball after the collision?

D) If the collision is perfectly elastic, what is the direction of motion of the 400g ball after the collision?

Explanation / Answer

1/m1v^2 = 1/m2v1^2 + 1/2m2v2^2

1/2*.12*(4.5)^2 = 1/2*.4v1^2 + 1/2*.12*v2^2

1.215 = .2v1^2 + .06v2^2

Also,

conservation of momentum

.12*4.5 = .4v1 + .12v2

v2 = (.54 - .4v1)/.12 = 4.5 - 3.33v1

=> 1.215 = .2v1^2 + .06(4.5 - 3.33v1)^2

v1 = .16m/s

v2 = 4m/s (approx)

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