Question

a bullet of mass 20g is horizontally fired with a velocity 150m/s from a pistol of mass 2kg what is the recoil velocity of the pistol? ​

Answer 1

Given that,

Mass of bullet (m₁) = 20 g

Initial velocity (u₁) = 0 m/s (as the pistol is at rest.)

Final velocity (v₁) = 150 m/s

Mass of pistol (m₂) = 2 kg

Initial velocity of the pistol (u₂) = 0 m/s (as the pistol is at rest.)

To find,

Recoil velocity of the pistol = ?

First of all before doin' any calculations, we've to see that, all the units are in their SI unit or not.

Here, the mass of bullet is given in gram whereas, the SI unit of mass is kilogram. So, we've to convert the gram into kilogram.

As we know,

1 gram = 10⁻³ kilogram

Therefore, 20 g = 20 $\times$ 10⁻³

20 g = 0.02 kg

As the final velocity of bullet is not given, we'll consider it as v₂.

By using law of conservation of mass,

Total momentum before fire = Total momentum after the fire

But, the total momentum before the fire is zero as both of them are in rest.

So, the total momentum after the fire =

m₁v₁ + m₂v₂

(0.02)(150) + (2)(v)

3 + 2v

But, the total momentum after the fire = total momentum before the fire

But the total momentum before the fire is zero, so,

3 + 2v = 0

2v = 0 - 3

2v = - 3

v = $\frac{-3}{2}$

v = $\frac{\cancel{-3}}{\cancel{2}}$

$\boxed{\bold{v = - 1.5\:m/s}}$

Thus, the recoiled velocity of the pistol is - 1.5 m/s.

Answer 2

Answer:

Let the direction in which the bullet is fired be positive.

Therefore,

Mass of Bullet(m)=20 grams=0.02 kg

Mass of Pistol(M)=2 kg

Initial Velocity of Bullet(u)=0 m/s

Initial Velocity of Pistol(U)=0 m/s

Final Velocity of Bullet(v)=150 m/s

Final Velocity of Pistol(V)= Recoil Velocity of Bullet = ?

According to the Law of Conservation of Momentum,

mu + MU = mv + MV

(0.02)(0) + (2)(0) = (0.02)(150) + (2)(V)

0 + 0 = 3 + 2V

-3/2 = V

V = -1.5 m/s

Therefore, the recoil velocity of the pistol will be 1.5 m/s in the opposite direction.(As the sign is negative)