A billiard ball rolling across a table to the right at 2.6 m/s makes a head-on e

Question

A billiard ball rolling across a table to the right at 2.6 m/s makes a head-on elastic collision with an identical ball. The mass of a billiard ball is 36 g.

1)If the second ball is initially at rest, what is the velocity of the first ball after the collision?

2)If the second ball is initially at rest, what is the velocity of the second ball after the collision?

3)If the second ball is initially moving to the left with a velocity of -1.3 m/s, what is the velocity of the first ball after the collision?

4)If the second ball is initially moving to the left with a velocity of -1.3 m/s, what is the velocity of the second ball after the collision?

Explanation / Answer

(a)

the velocity of the first ball after the collision will be

v1 = (m1 - m2) u1 / (m1 + m2) since m1 = m2 = m

v1 = 0m/s

m1 is mass of the first body

m2 is mass of the second body (m1 = m2)

u1 velocity of the first body befiore collision

u2 velocity of the second body befiore collision

v1 velocity of the first body after collision

v2 velocity of the second body after collision

the velocity of the first ball after the collision will be zero

(b)

the velocity of the second ball after the collision will be .

v2 = 2m1 u1 / (m1 + m2) since m1 = m2 = m

= u1

hence as the collision is head-on collision with equal masses

the bodies simply exchange their velocities after collision

(c)

since the second ball is moving towards left and the masses are same then we get

u1 - u2 = v1 + v2.............(1)

since the collision is elastic

u1 - u2 = v2 - v1..................(2)

dividing both the equations we get

v1 + v2 = v2 - v1

the velocity of the first ball after the collision will be zero

the velocity of the second ball after the collision will not change

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