Question

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

Answer 1

Answer:

• Penetration of bullet into the woode block => 2.25

• Force exerted by the wooden block on the bullet => -50N

[Negative sign shows that force is exerted in the oppo direction of bullet]

Explanation:

Given :

• Mass of the bullet,m=10g=0.01km[1g =0.001]

• Initial velocity of the bullet,u = 150m/s

• Final velocity of the bullet,v=0

• Time,t=0.03

To Find :

• Distance,s = ?

• Magnitude of the force,f = ?

To find Distance(s),we have to find acceleration too.

So,

$\sf{}v=u +at$

Put their values and solve it.

$\Rightarrow \sf{}0=150ms^{-1} +a\times 0.03s$

$\Rightarrow \sf{}-150ms^{-1} =a\times 0.03s$

$\Rightarrow \sf{}-\dfrac{150ms^{-1}}{0.03}=a$

$\Rightarrow\sf{}a=-500ms^{-1}$

Therefore,

$\sf{}s=tu+\dfrac{1}{2}at^2$

Put their values and solve it.

$\Rightarrow \sf{}s=150ms^{-1}\times0.03s+\dfarc{2}{2}(-5000ms^{-1})\times(0.03s)^2$

$\Rightarrow\sf{}s=4.5m-2500ms^{-2}\times0.0009s^2$

$\Rightarrow\sf{}s=4.5m-2.25m$

$\because \sf{}s=2.25m$

Magnitude of force exertrd by the wooden block

We know,

$\sf{}Force=Mass\times Acceleration$

$\Rightarrow \sf{}0.01kg\times -5000m^{-2}$

$\sf{}\because -50N$