Question

State and proof the law of conservation of momentum. A bullet of mass 4g when fired with a velocity Of 50 m/s, can enter a wall up to a depth of 10 cm. How much will be the average resistance offered by the wall. 5 6. What is the recoil velocity of gun? A bullet of mass 20g is horizontally fired with velocity 150 m/s from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Answer 1

hope it helps you !!!!!

Answer 2

Given:

There is two part in the question.

In Part 1:

Mass of the bullet (m) = 4g = 0.004 kg

Velocity of the bullet (u) = $50\hspace{1mm} ms^{-1}$

It can enter the wall up to depth (s) = 10 cm = 0.1 m

In Part 2:

Mass of the bullet = 20g = 0.02 kg

Velocity = $150 \hspace{1mm} ms^{-1}$ (bullet is fired horizontally)

Mass of the pistol = 2 kg

To Find:

In Part 1:

Average resistance offered by the wall to the bullet. Also says to state and proof law of conservation of momentum.

In Part 2:

Recoil velocity of the pistol

Solution:

Part 1:

Here, initial velocity (u) = $50\hspace{1mm} ms^{-1}$

Final velocity (v) = 0

Now, use formula

$v^{2} - u^{2} = 2as$

And put the respective values in it,

∴ $0^{2} - 50^{2} = 2a\times 0.1$

$=> a\times 0.1m =- 1250 m^{2} s^{-2}$

$=>a =-12500\hspace{1mm} ms^{-2}$

Now, use formula F = ma, to find the value of force

∴ $F = 0.004\hspace{1mm} kg\times (-12500\hspace{1mm} ms^{-2} )$

$=> F = -50\hspace{1mm} N$

Now, we know that every force have equal and opposite reaction, hence force average offered by the wall to the bullet is 50 N.

Law of conservation of momentum says that total momentum of two or more isolated body remains constant until and unless an external force acting on it.

Part 2:

Before we start we have to know about conservation of momentum theorem which says,

∴ Initial momentum of the system = final momentum of the system

Initial velocity of pistol and bullet is 0, so left side of the above relation is 0

Now in case of final momentum let, recoil velocity of the pistol is v then,

Final momentum of the system = $0.02kg\times 150 \hspace{1mm} ms^{-1} +2kg\times v$ = $3kgms^{-1} + 2kg\times v$

Now from conservation of momentum,

$3kgms^{-1} + 2kg\times v = 0$

$v=-\frac{3}{2} ms^{-1}=-1.5 ms^{-1}$

∴Recoil velocity of the pistol is $-1.5ms^{-1}$.