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Question

A billiard ball moving with a speed of 5m/s collides with an identical ball originally at rest. if the first ball stops after collision, then the second ball will move forward with a speed


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Solution

Step 1: Given

Speed of moving ball initially, v1i=5m/s
Speed of ball in rest initally, v2i=0m/s
Speed of moving the ball after the collision, v1f=0m/s
Speed of resting ball after the collision, v2f

Let m1 and m2 be the mass of the two balls. Since they are identical, m1=m2

Step 2: Formulas used

Since there is no external force involved, the law of conservation of momentum can be used. It is given as,
Pi=Pf
where Pi is initial momentum and Pf is final momentum.

Step 3: Solution

Total initial momentum was,
Pi=m1v1i+m2v2i=5m1+0=5m1

Total final momentum was,
Pf=m1v1f+m2v2f=m1·0+m2v2f=m2v2f

We have,
Pi=Pf5m1=m2v2fv2f=5m/sm1=m2

Therefore, the velocity of the second ball after the collision is 5m/s.


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