Question

From a rifle of mass 4 kg A bullet of mass 50 gram is fired with an initial velocity of 40 m/s. Calcualte initial recoil velocity of the gun.​

Answer 1

Given :

• Mass of the rifle= 4kg

•Mass of the bullet =50 gram

•Initial velocity =40m/s

To Find :

•The initial recoil velocity of the gun =?

Formula :

• According to the law of conservation of momentum :

$\pink\bigstar\boxed{\sf{m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2} }}$

Solution :

According to the law of conservation of momentum :

$\sf{\bullet \: Total \: momenta \: after \: the \: fire =Total \: momenta \: before \: the \: fire}$

$\implies\sf{m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}}$

Substituting the given values :

$\\ \implies \sf \: 4 \times 0 + 0.05 \times 0 = 4 \times v_{1} + 0.05 \times 40 \\ \\ \implies \sf \: 0 = 4v_{1 }+ 2 \\ \\ \implies \sf \: 4v1 = - 2 \\ \\ \implies \sf \: v_{1} = \frac{ - 2}{4} \\ \\ \implies \sf v_{1}= \cancel \frac{ - 2}{4} \\ \\ \implies \boxed{ \sf{v_{1} = - 0.5}}$

Hence the initial recoil velocity of the gun is 0.5m/s.

Answer 2

Answer:

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