Question

a 38g bullet is fired horizontally with a speed of 180m/s at a 5kg sandbag suspended on a light rod 1.5m high forming a pendulum that is free to spring. To what maximum angle will the pendulum swing to?

Answer 1

Answer on Question #77167 Physics / Other

A $m = 38$ g bullet is fired horizontally with a speed of $v = 180 \, \text{m/s}$ at a $M = 5 \, \text{kg}$ sandbag suspended on a light rod $l = 1.5 \, \text{m}$ high forming a pendulum that is free to spring. To what maximum angle will the pendulum swing to?

Solution:

Using the law of conservation of momentum we get

$$mv = (M + m)u$$

The law of conservation of energy gives

$$\frac{(M + m)u^2}{2} = (M + m)gh$$

So

$$h = \frac{u^2}{2g} = \left(\frac{m}{M + m}\right)^2 \frac{v^2}{2g}$$

The maximum angle

$$\begin{aligned} \theta &= \arccos \frac{l - h}{l} = \arccos \left(1 - \left(\frac{m}{M + m}\right)^2 \frac{v^2}{2gl}\right) = \\ &= \arccos \left(1 - \left(\frac{0.038}{5 + 0.038}\right)^2 \frac{180^2}{2 \times 9.8 \times 1.5}\right) = 20{}^\circ \end{aligned}$$

Answer: $20{}^\circ$

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