Question

A bullet of mass 10 g is fired horizontally with a velocity of 150 ms–1 from a pistol of mass 2 kg. Find the recoil of the pistol.

Answer 1

We have been given that:-

A bullet of mass 10 grams (0.01 Kg) is fired with a velocity of 150 m/s from a pistol of mass 2 Kg.

And we're asked to find the recoil velocity of the pistol.

The total momentum of the system will remain same before and after firing the bullet.

Before the bullet is fired:-

Initially, both the bullet and pistol are at rest. Therefore, their total momentum is zero.

After the bullet is fired:-

Velocity of bullet = 150 m/s

Let the recoil velocity of gun be $v_{final}$

Total momentum = (0.01 x 150) + ( 2 x $v_{final}$)

Since, the total momentum is conserved.

0 = (0.01 x 150) + ( 2 x $v_{final}$)

-1.5 = 2 x $v_{final}$

$v_{final} =\dfrac{-1.5}{2}$

$v_{final} = 0.75\ m/s$      $(In \ opposite \ direction \ to \ the \ bullet)$

Therefore, the recoil of the pistol is 0.75 m/s !

Answer 2

Answer:

•10g=0.010kg

the answer is 0.75m/s

m1v1=m2v2

0.01kg×150m/s=2kg×v2

0.01kg×150m/s/2kg=v2

0.75m/s=v2