Question

A 0.500 kg ice puck, moving east with a speed of 5.10 m/s , has a...

A 0.500 kg ice puck, moving east with a speed of 5.10 m/s , has a head-on collision with a 0.800 kg puck initially at rest.

1)Assuming a perfectly elastic collision, what will be the speed of the 0.500 kg object after the collision?
2)What will be the direction of the 0.500 kg object after the collision?
3)What will be the speed of the 0.800 kg object after the collision?
4)What will be the direction of the 0.800 kg object after the collision?





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Answer #1

a)

The velocity in head on elastic collision is given by

v1 = (u1(m1 - m2) + 2m2u2) / m1 + m2

v1 = (5.10(0.500 - 0.800) + 2*0.800*0) / (0.500 + 0.800)

v1 = -1.18 m/s

b)

direction is towards west.

c)

speed of second object..

v2 = u2(m2 - m1) + 2m1u1 / (m1 + m2)

v2 = 0*(0.800 - 0.500) + 2*0.500*5.1 / (0.500 + 0.800)

v2 = 3.92 m/s

d)

direction of second object will be towards east.

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