First slide

Vertical circular motion

Question

A pendulum consists of a wooden bob of mass m and length I . A bullet of mass m₁ is fired towards the pendulum with a speed v_{1 .}The bullet emerges out of the bob with a speed $\frac{v_{1}}{3}$ , and the bob completes a motion along a vertical circle. then v₁ is

Moderate

A

$(\frac{m}{m_{1}}) \sqrt{(5 g l)}$
B

$\frac{3}{2} {(\frac{m}{m_{1}})}^{} \sqrt{(5 gl)}$
C

$\frac{2}{3} (\frac{m_{1}}{m}) \sqrt{(5 gl)}$
D

$(\frac{m_{1}}{m}) \sqrt{(gl)}$

Solution

Applying the law of conservation of momentum
$mv = m_{1} \cdot (v_{1} - \frac{v_{1}}{3}) ∴ v = \frac{m_{1}}{m} (\frac{2 v_{1}}{3}) To describe a vertical circle, v should be \sqrt{(5 gl)} . Hence \sqrt{(5 gl)} = \frac{m_{1}}{m} (\frac{2}{3} v_{1}) v_{1} = (\frac{m}{m_{1}}) (\frac{3}{2}) \sqrt{(5 gl)}$

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