Question 7.SP.30: A 60-g bullet was fired horizontally into a stationary 50-kg......
A 60-g bullet was fired horizontally into a stationary 50-kg sandbag suspended on a rope 900 mm long, as shown in Fig. 7-29. It was calculated from the observed angle θ that the bag with the bullet embedded in it swung to a height of 30 mm. What was the speed of the bullet as it entered the bag?
Step-by-Step
Let
\begin{array}{l}v_{1}=\text { speed of bullet before impact } \\v_{2}=\text { speed of (bag + bullet) after impact }=\sqrt{2 g h}=\sqrt{2(9.8)(0.03)}=0.767 \mathrm{~m} / \mathrm{s}\end{array}
Momentum of system before impact = momentum of system after impact:
0.06 v_{2}+0=(0.06+50) \times 0.767
from which v_{1}=\underline{640 \mathrm{~m} / \mathrm{s}} .
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