Question

If a bullet of mass \[5gm\] moving with velocity \[100m/s\], penetrates the wooden block up to \[6cm\]. Then the average force imposed by the bullet on the block is:
\[A.\text{ }8300\text{ }N\]
\[B.\text{ }417\text{ }N\]
\[C.\text{ }830\text{ }N\]
\[D.\text{ }Zero\]

Answer 1

Hint: Using Newton's second law of motion we can find the force imposed by the bullet on block.
Here the bullet is penetrating the wooden block due to which its velocity changes or decreases. This retarding velocity is the reason for acceleration.
Formula used:
\[F=ma\]
\[{{v}^{2}}-{{u}^{2}}=2aS\]
\[\dfrac{1}{2}m{{\left( \Delta v \right)}^{2}}=FS\]

Complete answer:
Given that,
Mass of bullet, \[m=5g=5\times {{10}^{-3}}Kg\]
Initial velocity \[u=100m/s\]
Final velocity of bullet \[v=0\]
Displacement \[S=6cm=6\times {{10}^{-2}}m\]
The bullet velocity starts retarding when it starts penetrating the wooden block,
Hence there is a deceleration due to the change in velocity.
Let, \[deceleration\text{ }of\text{ }bullet\text{ }=-a\]
Then,
We have the third equation of motion which relates
\[\text{the initial velocity u, final velocity v, displacement S, and deceleration -a}\] ,
\[{{v}^{2}}-{{u}^{2}}=2aS\]
Then,
\[a=\dfrac{{{v}^{2}}-{{u}^{2}}}{2S}\]
Substitute the values of\[\text{u, v and S}\]. We get,
\[a=-\dfrac{0-{{100}^{2}}}{2\times 6\times {{10}^{-2}}}\]
\[a=\dfrac{{{10}^{6}}}{12}\]
We know that,
\[F=ma\]
Substituting \[\text{m and a}\] we get,
\[F=5\times {{10}^{-3}}\times \dfrac{{{10}^{6}}}{12}\]
\[F=416.67\]

So, the correct answer is “Option B”.

Additional Information:
Deceleration is the reverse of acceleration. The rate at which an object speeds up is known as acceleration. But deceleration is the rate at which the object slows down. It only applies to objects which slow down and occurs only when a force is applied against the motion of a body. Here the wooden block is imposing an opposing force against the moving body. That is why its velocity is changed.
\[Deceleration=\dfrac{Final\;Velocity - Initial\;Velocity}{Time\;taken}\]
It can be denoted as \[\text{-a}\] where a is the acceleration. The unit of deceleration is the same as that of acceleration \[\left( \text{m/}{{\text{s}}^{2}} \right)\].

Note:
Alternate method to solve the question:
\[Change\text{ }in\text{ }kinetic\text{ }energy\text{ }=\text{ }Work\text{ }done\]
I.e.,
\[\dfrac{1}{2}m{{\left( \Delta v \right)}^{2}}=FS\]
\[m=~mass\text{ }of\text{ }bullet\]
\[\Delta v=change\text{ }in\text{ }velocity\]
\[F=\text{ }force\]
\[S=\text{ }displacement\]
Substituting values of \[\text{m, v and S}\]
We get,
\[\dfrac{1}{2}\times 5\times {{10}^{-3}}\times {{\left( 100-0 \right)}^{2}}=F\times 6\times {{10}^{-2}}\]
\[F=\dfrac{2500}{6}=417N\]