Question

A bullet of mass 10 g travelling horizontally with a velocity of 150 m s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.​

Answer 1

Given:

Mass of bullet,m= 10g= 0.01 kg

Initial velocity of bullet, u= 150 m/s

Final velocity of bullet, v= 0 m/s

(Since, the bullet finally comes to rest)

Time taken by bullet, t= 0.03 s

To Find:

The distance of penetration of the bullet into the block and the magnitude of the force exerted by the wooden block on the bullet

Solution:

We know that,

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to third equation of motion for constant acceleration,

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

t is time taken

• Expression of force F is given by

$\red{\boxed{\bf{F=ma}}}$

where,

m is mass

a is acceleration

$\rule{190}{1}$

Let the acceleration of bullet while penetrating the wooden block be a

So, on applying first equation of motion on bullet, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=150+a(0.03)}$

$\longrightarrow\rm{-150=0.03a}$

$\longrightarrow\rm{0.03a=-150}$

$\longrightarrow\rm{a=\dfrac{-150}{0.03}}$

$\longrightarrow\rm{a=-5000\ m/s^{2}}$

Here, negative shows that direction of acceleration is opposite to direction of motion of bullet

Now, on applying third equation of motion on bullet, we get

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{(0)^{2}-(150)^{2}=2(-5000)s}$

$\longrightarrow\rm{-22500=-10000s}$

$\longrightarrow\rm{-10000s=-22500}$

$\longrightarrow\rm{s=\dfrac{-22500}{-10000}}$

$\longrightarrow\rm\green{s=2.25\ m}$

$\rule{190}{1}$

Let the force exerted by bullet on wooden block be F

So,

$\longrightarrow\rm{F=ma}$

$\longrightarrow\rm{F=0.01\times(5000)}$

$\longrightarrow\rm\green{F=50\ N}$

Hence, the distance of penetration of the bullet into the block is 2.25 m and the magnitude of force exerted by the wooden block on the bullet is 50 N.

Answer 2

$\bigstar\:\:\rm\blue{GIVEN}\:\:\bigstar$

• Mass of Bullet=10g=0.01kg.

• Initial Velocity of the bullet= 150m/s.

• Time= 0.03s.

• Final Velocity= 0m/s( As it comes to rest)

$\bigstar\:\:\rm\red{TO\:FIND}\:\:\bigstar$

• The distance of Penetration of the Bullet into the wooden Block.

• Force exerted by the wooden Block on the bullet.

$\bigstar\rm\blue{FORMULAE\:USED}\bigstar$

• ${\boxed{\bf{\blue{v=u+at}}}}$

V= Final Velocity

u=initial Velocity

a=acceleration

t= time

• ${\boxed{\bf{\blue{v^2-u^2=2as}}}}$

Where,

V= Final Velocity

u=initial Velocity

a=acceleration

s=Distance.

• ${\boxed{\bf{\blue{F=ma}}}}$

Where,

F= Force

M= Mass

a= acceleration.

Using 1st equation of motion to find acceleration

$\implies\rm{v=u+at}$

$\implies\rm{0=150+a\times{0.03}}$

$\implies\rm{a=\dfrac{-150}{0.03}}$

$\implies\rm\pink{a=-5000m/s^{2}}$

Now, Using third equation of motion to find Distance.

$\implies\rm{v^2-u^2=2as}$

$\implies\rm{(0)^2-(150)^2=2\times{(-5000)}(S)}$

$\implies\rm{-22500=-10000s}$

$\implies\bf{s=\dfrac{\cancel{-22500}}{\cancel{-10000}}}$

$\implies\bf\blue{S=2.25m}$

Now,

$\implies\rm{F=ma}$

$\implies\rm{F=0.01\times{-5000}}$

$\implies\rm\red{F=-50N}$

Negative Sign implies that Force is acting on opposite direction of motion.