First slide

Law of conservation of momentum and it's applications

Question

A gun of mass 20 kg has bullet of mass 0.1 kg in it. The gun is free to recoil 804 J of recoil energy are released on firing the gun. The speed of bullet (ms^-1) is

Moderate

A

$\sqrt{804 \times 2010}$
B

$\sqrt{\frac{2010}{804}}$
C

$\sqrt{\frac{804}{2010}}$
D

$\sqrt{804 \times 4 \times 10^{3}}$

Solution

Here, $m_{1} = 20 kg$
$m_{2} = 0 .1 kg$
where, v₁ = velocity of recoil of gun
and v₂ = velocity of bullet.
As, $m_{1} v_{1} = m_{2} v_{2}$ ( $∵$ momentum is conserved)
$v_{1} = \frac{m_{2}}{m_{1}} v_{2} = \frac{0 .1}{20} v_{2} = \frac{v_{2}}{200}$
Recoil energy of gun $= \frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} \times 20 {(\frac{v_{2}}{200})}^{2}$
$804 = \frac{10 v_{2}^{2}}{4 \times 10^{4}} = \frac{v_{2}^{2}}{4 \times 10^{3}}$
$v_{2} = \sqrt{804 \times 4 \times 10^{3}} {ms}^{- 1}$

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