Question

# A bullet of mass $50g$ is moving with a velocity of $500m/s$. It penetrates $10cm$ into a still target and comes to rest. Calculate: the kinetic energy possessed by the bullet, and the average retarding force offered by the target

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Solution

## Step 1: GivenMass of the bullet, $m=50g=0.05kg$Velocity of the bullet, $v=500m/s$Distance penetrated by the bullet, $s=10cm=0.1m$Step 2: Calculate the kinetic energy$KE=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×0.05×{\left(500\right)}^{2}\phantom{\rule{0ex}{0ex}}=6250kg-{m}^{2}/{s}^{2}\phantom{\rule{0ex}{0ex}}=6250J$Step 3: Determine the average retarding forceWork done is given by, $W=Force×Dis\mathrm{tan}ce$In this situation, Work done = Resistance force x distance$W=KE=6250=F×0.1\phantom{\rule{0ex}{0ex}}⇒F=62500N$Therefore,the kinetic energy possessed by the bullet is $6250N$the average retarding force offered by the target is $62500N$

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