Question

A bullet of mass 20 kg is horizontally fired with a velocity 150m per second from a pistol of mass 2 kg what is the recoil velocity of the pistol ?​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Correct \: Question}}} \mid}}$

A bullet of mass 20 grams is horizontally fired with a velocity 150m per second from a pistol of mass 2 kg what is the recoil velocity of the pistol ?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of Bullet (m) = 20 grams.
• Velocity of the Bullet after Firing (u) = 150 m/s.
• Mass of The Pistol (M) = 2 kg.
• Let the recoil velocity of the Pistol be "v".

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From Law of Conservation of momentum,

Initial Momentum = Final Momentum.

$\large{\boxed{\tt P_i = P_f}}$

But Before Firing, the whole system (Gun & Bullet) is at Rest.

Therefore, Initial momentum will be zero.

Then the equation becomes,

$\large{\leadsto {\underline{\underline{\tt P_f = 0}}}}$

Now, Formulating the Values of Final Momentum.

$\large{\tt \leadsto mu + Mv = 0}$

• M = Mass of Pistol.
• m = Mass of Bullet.
• u = Velocity of bullet.
• v = Velocity of bullet.

Substituting the values

$\large{\tt \leadsto 20 \times 10^{-3} \times 150 + 2 \times v = 0}$

[Here Mass of bullet (m) is taken as 20 grams = 20 × 10⁻³ Kg.]

$\large{\tt \leadsto 20 \times 150 \times 10^{-3} + 2v = 0}$

$\large{\tt \leadsto 3000 \times 10^{-3} + 2v = 0}$

$\large{\tt \leadsto 3\times 10^3 \times \dfrac{1}{10^3}+ 2v = 0}$

$\large{\tt \leadsto 3 \times \cancel{10^3} \times \dfrac{1}{\cancel{10^3}}+ 2v = 0}$

$\large{\tt \leadsto 3 \times 1 + 2v = 0}$

$\large{\tt \leadsto 3 + 2v = 0}$

$\large{\tt \leadsto 2v = - 3}$

$\large{\tt \leadsto v = \dfrac{- 3}{2}}$

$\large{\tt \leadsto v = \cancel{\dfrac{- 3}{2}}}$

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

So, The Recoil velocity of the Pistol is 1.5 m/s.

Note:-

• Negative sign of Recoil velocity indicates that the recoil velocity of the Pistol is opposite to the Motion of the Bullet.

$\rule{300}{1.5}$

Answer 2

Answer:

1.5m/s

Explanation:

Mass of bullet=M₁=20g=0.02kg

Mass of gun=M₂=2kg

velocity of bullet initially at rest=u₁=0m/s

velocity of gun initially=u₂=0m/s

velocity of bullet after firing=v₁=150m/s

velocity of gun after firing=v₂

according to law of conservation of momentum,

M₁u₁+M₂u₂=M₁v₁+M₂v₂

=0.02×0+2×0=0.02×150+2×v₂

=0+0=3+2v₂

= -3=2v₂

v₂=$\frac{-3}{2}$

v₂=-1.5m/s

the recoil velocity of the bullet is 1.5m/s

Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet