Question

Find the recoil velocity of the gun having mass equal to 5kg. If a bullet of 25g acquires the velocity of 500 m/s after firing of the gun.

Answer 1

By using law of conservation of momentum equation we will get
Recoil velocity = v2 = m1v1 / m2
v2 = (25 * 10^-3 * 500) / 5
v2 = 2.5 m/s

Recoil velocity of gun is 2.5 m/s

Answer 2

Answer:

The recoil velocity of the gun exists equal to 2.5 m/s.

Explanation:

Here given that,

Mass of bullet (m1) = 25 gm = 0.025 kg

Velocity of bullet before firing (u1) = 0

Velocity of bullet after firing (v1) = 500 m/s

Mass of gun (m2) = 5 kg

Velocity of gun before firing, (u2) = 0

The velocity of the gun after firing =?

It is known that,

$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$

$\Rightarrow 0.025 \mathrm{~kg} \times 0+5 \mathrm{~kg} \times 0=0.025 \mathrm{~kg} \times 500 \mathrm{~m} / \mathrm{s}+5 \mathrm{~kg} \times v_{2}$

$\Rightarrow 0=12.5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}+5 \mathrm{~kg} \times v_{2}$

$\Rightarrow 5 \mathrm{~kg} \times v_{2}=-12.5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

Hence,

$\Rightarrow v_{2}=\frac{-12.5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}}{5 \mathrm{~kg}}$

$\Rightarrow v_{2}=-2.5 \mathrm{~m} / \mathrm{s}$

Therefore, the recoil velocity of the gun exists equal to 2.5 m/s. Here negative (- ve) sign indicates that the gun moves in the opposite direction of the bullet.

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