Question

Two billiard balls of equal mass are on a level, frictionless surface. The first ball is moving and collides with the second ball, which
was stationary. After the collision, both balls are moving. What is a possible speed for the first ball after the collision?
twice its original speed
one-half its original speed
the same as its original speed
one and one-half times its original speed​

Answer 1

The speed of the first ball after collision will be one half of its original speed.

Option 2.

Explanation:

Let us consider that the two balls be represented as B₁ and B₂ with their masses being M₁ and M₂. Since it is stated that both the balls have equal mass, then M₁=M₂=M. Similarly, the initial velocity of the ball B₁ is represented as V₁ and the initial velocity of ball B₂ is considered as zero since it was stationary.

After collision, the combined velocity is represented as V.

As per conservation of momentum, the total momentum of the balls before collision will be equal to the total momentum after collision.

M₁V₁+M₂(0) = Total momentum before collision

As the balls get sticked together after collision, then

(M₁+M₂)V = Total momentum after collision.

Due to law of conservation of momentum,

M₁V₁+M₂(0)=(M₁+M₂)V

Since M₁=M₂=M

MV₁=2MV

$V = \frac{MV_{1} }{2M} =\frac{V_{1} }{2}$

Thus, the speed of the first ball after collision will be one half of its original speed.