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

`v_0 = 5.89 m/s`

Step-by-step 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

`(1)/(2)mv^2 = mgL(1 - cos\theta)`

`v = √(2gL(1 - cos\theta))`

so we have

`v = √(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`