A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right with a velocity of magnitude 0.200 m/s. Let +x be to the right. (Since the collision is head-on, all the motion is along a line. )

Find the magnitude of the velocity of 30.0-g marble after the collision.

Find the magnitude of the velocity of 10.0-g marble after the collision.

Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for 30.0-g marble

Calculate the change in momentum for 10.0-g marble.

Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for 30.0-g marble.

Calculate the change in kinetic energy for 10.0-g marble.

Explanation / Answer

Here ,

let the final velocity of 10 gm marble is v1

final velocity of 30 gm is v2

Using conservation of momentum

10 * v1 + 30 * v2 = 30 * 0.2 - 0.4 * 10 -----(1)

for the elastic collision

as coeffcicient of restitution , e = 1

v1 - v2 = 0.60

solving 1 and 2

v1 = 0.5 m/s

v2 = -0.1 m/s

the magnitude of velocity of 30 gm after the collision is -0.1 m/s

magnitude of velocity of 10 gm after the collision is 0.5 m/s

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change in momentum for 30 g marble = (-0.1 * 30)- (30 * 0.2)

change in momentum for 30 g marble = - 0.009 Kg.m/s

the change in momentum for 30 g marble is - 0.009 Kg.m/s

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change in momentum at 10 gm = 10 * 0.5 - (-10 * 0.4)

change in momentum at 10 gm = 0.009 Kg.m/s

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