Question

Suppose a gangster sprays Superman’s chest with3 gbullets at the rate of 100 bullets/min, and the speed of each bullet is 500 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman’s chest?

Short answer

The magnitude of the average force on Superman’s chest is ${\mathrm{F}}_{\mathrm{avg}}=5.0\mathrm{N}$

Step-by-step solution

Understanding the given information

The mass of each bullet,$\mathrm{m}=3\mathrm{g}\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=3\times {10}^{-3}\mathrm{kg}$

The speed of a bullet is,$\mathrm{v}=500\mathrm{m}/\mathrm{s}$

The rate of bombarding bullets is, 100 bullets/min .

Concept and formula used in the given question

We can use the equation of impulse-momentum theorem to calculate the impulse and then we can use this value of impulse in the equation of average force to calculate its value.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ ∆\mathrm{P}=\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\times \mathrm{t} \end{gathered}$$

Calculation for the magnitude of the average force on Superman’s chest

Suppose, the time of bombardingthebullet is 1 min. So, the total number of bullets shot at is 100.

The equation of the impulse-momentum theorem is,

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ \mathrm{J}=100\times \mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ \mathrm{J}=100\times 3\times {10}^{-3}\mathrm{kg}\times \left(500\mathrm{m}/\mathrm{s}-\left(-500\mathrm{m}/\mathrm{s}\right)\right) \\ \mathrm{J}=300\mathrm{N}·\mathrm{s} \end{gathered}$$

Now,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{∆\mathrm{t}} \\ =\frac{300\mathrm{N}·\mathrm{s}}{60\mathrm{s}} \\ =5.0\mathrm{N} \end{gathered}$$