Question

A bullet of mass $25 kg$ moving with a speed of $100 m/s$ pierces a bag full of send kept adjacent to a wall. The bullet stops in the bag after $0.05 second$, find:-
1) The acceleration of the bullet (assuming uniform)
2) The force exerted by send bag on the bullet
3) The distance covered by the bullet before coming to the rast

Answer 1

Hint
Try to get the acceleration and the distance covered by the bullet using the equation of motion and then try to find the force.

Complete Step By Step Solution
Now we can write the equation of motion as
$\text{v}\ \text{=}\ \text{u}\ + \text{at}$
Where, $\text{V}\to $ is the final velocity
$u\to $ is the initial velocity
$a\to $ is the acceleration
$\text{t}\to $ is the time taken
Since the bullet is at rest its final velocity is equal to 0.
$\therefore \ \text{V}\ \text{=}\ \text{0}$
Now initial velocity and the time taken is given $i.e$
$u=100{m}/{s\ \ \ \ \ \text{and}\ \ \ \text{t}\ \text{=}\ \text{0}\text{.05sec}}\;$
1). Acceleration can be written as
     $\text{a}\ \text{=}\ \dfrac{\text{V}\ \text{-}\ \text{u}}{\text{t}}$
     $=\dfrac{0-100}{0.05}$
     $=\dfrac{-100}{0.05}$
     $a=-2000{m}/{{{s}^{2}}}\;$
The acceleration of the bullet is negative, it indicates that the bullet was decelerating.
2).The force exerted by sand bag on the bullet is given by
          $\text{F}\ \text{=}\ \text{ma}$
Where $m$ is mass of the body.
Now substituting the value of mass and acceleration.
$\text{F}\ \text{=}\dfrac{25}{1000}\times -2000$
$=-\text{50N}$
The force is negative and it is exerted in the opposite dissection of motion of the bullet.
3).Now to find the distance covered by the bullet before coming to rest can be given as
          $s\ =\ ut+\dfrac{1}{2}a{{t}^{2}}$
Now substituting value of u,a, and t in above equation we get
          $\text{S}\ \text{=}\ \text{100}\times \text{0}\text{.05+}\dfrac{1}{2}\times \left( -2000 \right)\times {{\left( 0.05 \right)}^{2}}$
          $\text{s}\ \text{=}\ \text{5-1}{{\text{0}}^{3}}\times 25\times {{10}^{-4}}$
          $\text{S}\ \text{=}\ \text{5-2}\text{.5}$
          $\text{S}\ \text{=}\ \text{2}\text{.5m}$
So the distance covered by the bullet before coming to rest is $2.5m$.

Note
You can find the distance travelled of the bullet using another equation of motion given as ${{\text{v}}^{2}}={{u}^{2}}+2as$
Where $v$ is final velocity, $u$ is initial velocity and $a$ is acceleration.