Two carts are colliding on an airtrack (neglect friction). The first cart has a

Question

Two carts are colliding on an airtrack (neglect friction). The first cart has a mass of m1=40 g and an initial velocity of v1=2 m/s. The second cart has a mass of m2=47 g and an initial velocity of v2=-5 m/s. Two experiments are conducted. In the first experiment, the first cart has a final velocity of v1'=-1.11 m/s. What is the velocity of the second cart?

For the second experiment, the bumpers of the carts are modified, but the carts are started with the same initial velocities as before. Now the first cart has a final velocity of v1'=-3.8942 m/s, and the second cart a final velocity of v2'=0.0163 m/s. How much (if any) energy was lost in the collision?

In good approximation, what kind of collision was the second experiment?
Totally inelastic
Elastic
Inelastic, but not totally

Explanation / Answer

(1) In the first experiment, using conservation of momentum

m1v1 + m2v2 = m1v1' + m2v2'

(0.040) (2) + (0.047) (-5) = (0.040) (-1.11) + (0.047) v2'

v2' = -2.35 m/s

(2) In the second experiment

Initial kinetic energy KEi = (1/2)m1v12 + (1/2)m2v22

KEi = (1/2) (0.040) (2)2 + (1/2) (0.047) (-5)2

KEi = 0.6675 J

Final kinetic energy KEf = (1/2)m1v1'2 + (1/2)m2v2'2

KEi = (1/2) (0.040) (-3.8942)2 + (1/2) (0.047) (0.0163)2

KEi = 0.3033 J

Loss in kinetic energy KE = KEi - KEf

KE = 0.6675 - 0.3033

KE = 0.3642 J

This is inelastic collision.

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