Calculate the velocity of an elastic collision?

Question

A stationary billard ball of mass 0.17kg is struck by an identical ball moving at 4.0m/s. After the collision, the second ball moves off at 60.0 degrees to the left of the original direction. The stationary ball moves off at 30.0 degrees to the right of the second ball's original direction. What is the velocity of each ball after the collision?

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Answer 1 · 2018-02-04T06:53:17

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As,no other external force is acting on the system,we can apply law of conservation of momentum.

Suppose after the collision,the stationary ball moves with velocity $v 1$ $v1$ and the other ball moves with $v 2$ $v2$

So,along original direction of the 2nd ball,applying law of conservation of momentum we get,

$0 + m \cdot 4 = m \cdot v 1 cos 30 + m \cdot v 2 cos 60$ $0+m*4 = m*v1 cos 30+m*v2 cos 60$

or, $4 = v 1 cos 30 + v 2 cos 60$ $4= v1 cos 30 +v2 cos 60$ ....1

Along perpendicular direction to the motion of the 2nd ball,taking rightside positive,we get,

$0 = m \cdot v 1 sin 30 - m \cdot v 2 sin 60$ $0=m*v1 sin 30 - m*v2 sin 60$

or, $v 1 = \sqrt{3} v 2$ $v1=sqrt(3) v2$

putting in 1...we get, $v 2 = 2 \frac{m}{s}$ $v2=2 m/s$

And , $v 1 = 2 \sqrt{3} \frac{m}{s}$ $v1=2sqrt(3) m /s$