A pendulum consists of a wooden bob of mass mand of length l. A bullet of mass `m_1` is fired towards the pendulum with a speed `v_1`. The bullet emerges out of the bob with a speed `v_1//3` and the bob just completes motion along a vertical circle. Then `v_1` is:

Options

A. `(m/(m_1)) sqrt(5 gl)`
B. `3/2 (m/(m_1)) sqrt(5 gl)`
C. `2/3 ((m_1)/m) sqrt(5 gl)`
D. `((m_1)/(m))sqrt(gl)`

Answer

B

Explanation

Applying the law of conservation of momentum,
`mv = m_1(v_1 - (v_1)/3) " or " v = (m_1)/m xx (2v_1)/3`
To describe a vertical circle v should be `sqrt(5 gl)` . Hence, `sqrt(5 gl) = (m_1)/(m) xx (2v_1)/(3)`
or `v_i = (m/(m_1)) 3/2 sqrt(5 gl)`