Question

A bullet of mass 10 g is fired horizontally with a velocity of 150 ms-1 from a gun of mass 4 kg. calculate the recoil velocity of gun.

Answer 1

Given:

A bullet of mass 10 g is fired horizontally with a velocity of 150 ms-1 from a gun of mass 4 kg.

To find:

Recoil Velocity

Calculation:

Let recoil Velocity be v ;

Applying Conservation of Momentum;

$\therefore \: \sf{initial \: p = final \: p}$

$= > 0 + 0 = ( \dfrac{10}{100} \times 150) + (4 \times v)$

$= > 0= ( \dfrac{10}{100} \times 150) + (4 \times v)$

$= > 0= 15 + (4 \times v)$

$= > 0= 15 + 4v$

$= > 4v = - 15$

$= > v = - 3.75 \: m {s}^{ - 1}$

Negative recoil velocity shows the fact that the gun recoils in the opposite direction as compared to that of the movement of the bullet.

So, final answer is:

$\boxed{ \sf{v = - 3.75 \: m {s}^{ - 1} }}$

Answer 2

Question:-

A bullet of mass 10 g is fired horizontally with a velocity of 150 ms-1 from a gun of mass 4 kg. calculate the recoil velocity of gun.

Answer:-

Given:

• $\sf{{m}_{1}\:=\:4kg}$ (mass of gun)
• $\sf{{u}_{1}\:=\:0m/s}$ (gun was initially at rest)
• $\sf{{v}_{1}\:=\:?}$ (recoil velocity which we need to find
• $\sf{{m}_{2}\:=\:0.01kg}$ (mass of bullet)
• $\sf{{u}_{2}\:=\:0m/s}$ (bullet was initially at rest)
• $\sf{{v}_{2}\:=\:150m/s}$ (final velocity of bullet)

Applying law of conservation of momentum:-

$\sf{{m}_{1}{v}_{1}\:+\:{m}_{2}{v}_{2}\:=\:{m}_{1}{u}_{1}\:+\:{m}_{2}{u}_{2}}$

=> (4 * v) + (0.01 * 150) = (4 * 0) + (0.01 * 0)

=> v = (-15)/4 m/s

=> v = -3.75 m/s

-ve sign indicates that the gun moves in backward direction

Ans. Recoil Velocity = -3.75 m/s