Question

A rifle bullet of mass $5\text{g}$ shot into a still target with a velocity of $150{\text{m s}}^{-1}$, comes to rest in the target after penetrating about $30\text{cm}$. What is the resistive force (in $\text{N}$) offered by the target to the bullet?

• A. -187.5

Answer 1

The correct option is A -187.5
Given
The mass of the bullet, $\text{m}=5\text{g}=5\times {10}^{-3}\text{kg}$
The initial velocity of the bullet, $\text{u}=150{\text{m s}}^{-1}$
As the bullet came to rest in the target, its final velocity, $\text{v}=0$
The penetrated length in the target, $\text{s}=30\text{cm}=30\times {10}^{-2}\text{m}$
According to the third equation of motion,
${\text{v}}^2-{\text{u}}^2=-2\text{as}$, as the body decelerates once it hits the target.
$⟹0-(150{)}^2=-2\text{a}\times 30\times {10}^{-2}$

$⟹\text{a}=\frac{-150\times 150}{-60\times {10}^{-2}}$

$⟹\text{a}=37,500{\text{m s}}^{-2}$
The opposing force exerted by the target is given by Newton's second law of motion: Force acting on a body accelerates or decelerates the body. In this case, the opposing force decelerates the bullet, and forces it to come to rest.
$⟹$ Force $\text{F}=\text{-ma}$
$⟹\text{F}=-5\times {10}^{-3}\times 37500$
$⟹\text{F}=-187.5\text{N}$