Question

A 29.0 g ball is fired horizontally with initial speed v0 toward a 100 g ball that is hanging motionless from a 1.0 m long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle θmax=50∘, what was v0?

Answer 1

Answer:

$v_0 = 5.89 m/s$

Explanation:

As we know that the ball swings out to maximum angle of 50 degree

so we will have use energy conservation to find the initial speed of the ball

$\frac{1}{2}mv^2 = mgL(1 - cos\theta)$

$v = \sqrt{2gL(1 - cos\theta)}$

so we have

$v = \sqrt{2(9.81)(1)(1 - cos50)}$

$v = 2.65 m/s$

now we can use momentum conservation for this collision

$m_1v_0 = m_1v_{1f} + m_2v_{2f}$

$29 v_0 = 29v_{1f} + 100(2.65)$

$v_0 = v_{1f} + 9.14$

also we know that it is elastic collision

so we will have

$2.65 - v_{1f} = v_0$

now from above two equations we have

$v_0 = 5.89 m/s$