Question

120kg द्रव्यमान की बंदूक से एक 0.04 kg द्रव्यमान की गोली दांगी जाती हैं यदि गोली की बंदूक के मुख से निकलने की गति 90m/sहै तो बंदूक की प्रतिघाती गति क्या होगी

Answer 1

Answer:

$120 + 0.04 \\ 120.04 \\ \\ 90 \times \frac{5}{18} \\ 25 \\ = 120.04 \times 25 = 3001.$

here is ur answer.

Answer 2

The retardant speed of the gun is 0.03 m/s.

Explanation:

A 0.04 kg mass shot is fired from a gun of 120 kg mass. If the speed of the gun's mouth is 90m / s, what will be the retardant speed of the gun

Given that,

Mass of the bullet, m = 0.04 kg

Mass of the gun, m' = 120 kg

Speed of the gun, v = 90 m/s

We need to find the retardant speed of the gun. It is a case of conservation of momentum as :

$0=mv+m'v'$ (as initial momentum is 0)

$v'=\dfrac{-mv}{m'}$

$v'=\dfrac{-0.04\times 90}{120}$

v' = -0.03 m/s

So, the retardant speed of the gun is 0.03 m/s. Hence, this is the required solution.

Learn more,

Conservation of momentum

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