Question

A billiard ball, labeled 1, moving horizontally strikes another billiard ball, labeled 2, at rest. Before impact, ball 1 was moving at a speed of 3.00 m/s, and after impact it is moving at 0.50 m/s at 50° from the original direction. If the two balls have equal masses of 300 g, what is the possible velocity of the ball 2 after the impact?

Answer 1

Answer:

So after collision 2nd ball will move off with speed 2.70 m/s

Explanation:

When two balls of equal mass collides elastically then after collision the two balls will move away at some angle with respect to each other

Here we know that the first ball is moving with speed 3  m/s initially and after collision it moves off by 50 degree with speed 0.50 m/s

so here we can say by momentum conservation

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} cos50 + m_2v_{2f}cos\theta$

$m_1 = m_2$

so we will have

$3 = 0.50 cos50 + v cos\theta$

$vcos\theta = 2.68 m/s$

Similarly in perpendicular direction we have

$0 = 0.50 sin50 + v sin\theta$

$v sin\theta = -0.38$

so net speed of the ball is given as

$v = \sqrt{0.38^2 + 2.68^2}$

$v = 2.70 m/s$

Answer 2

The possible velocity of the ball 2 is 2.36 m/s in the initial direction of ball 1.

Considering the collision as completely elastic, for elastic collision the momentum before collision and momentum after collision in a direction is always conserved.

Therefore applying the momentum conservation theorem in the direction of ball 1 before collision.

m1u1+m2u2=m1v1cos50°+m2v2

u1+u2=v1cos50°+v2

3+0=0.5cos50°+v2

v2=2.36 m/s