Question

A bullet of mass 200 g is fire from a gun of mass 10kg with a velocity of 100m/s, calculate the velocity of recoil.​

Answer 1

Answer:

Given :-

• A bullet of mass 200 g is fire from a gun of mass 10 kg with a velocity of 100 m/s.

To Find :-

• What is the velocity of recoil.

Formula Used :-

$\clubsuit$ Law Of Conservation Of Momentum Formula :

$\mapsto \sf\boxed{\bold{\pink{m_1u_1 + m_2u_2 =\: m_1v_1 + m_2v_2}}}$

Solution :-

First, we have to convert mass of bullet (m₁) g into kg :

$\implies \sf Mass_{(Bullet)} =\: 200\: g$

$\implies \sf Mass_{(Bullet)} =\: \dfrac{2\cancel{00}}{10\cancel{00}}\: kg\: \: \bigg\lgroup \sf\bold{\pink{1\: g =\: \dfrac{1}{1000}\: kg}}\bigg\rgroup$

$\implies \sf Mass_{(Bullet)} =\: \dfrac{2}{10}\: kg$

$\implies \sf\bold{\purple{Mass_{(Bullet)} =\: 0.2\: kg}}$

Now, we have to find the velocity of recoil :

Given :

• Mass of bullet (m₁) = 0.2 kg
• Mass of gun (m₂) = 10 kg
• Final Velocity of bullet (v₁) = 100 m/s
• Initial Velocity of bullet (u₁) = 0 m/s
• Initial Velocity of gun (u₂) = 0 m/s

According to the question by using the formula we get,

$\longrightarrow \sf (0.2)(0) + (10)(0) =\: (0.2)(100) + (10)v_2$

$\longrightarrow \sf 0.2 \times 0 + 10 \times 0 =\: 0.2 \times 100 + 10 \times v_2$

$\longrightarrow \sf 0 + 0 =\: 20 + 10v_2$

$\longrightarrow \sf 0 =\: 20 + 10v_2$

$\longrightarrow \sf 0 - 20 =\: 10v_2$

$\longrightarrow \sf - 20 =\: 10v_2$

$\longrightarrow \sf \dfrac{- 2\cancel{0}}{1\cancel{0}} =\: v_2$

$\longrightarrow \sf \dfrac{- 2}{1} =\: v_2$

$\longrightarrow \sf - 2 =\: v_2$

$\longrightarrow \sf\bold{\red{v_2 =\: - 2\: m/s}}$

[ Note : - Here, the negetive sign indicates that the velocity of the gun in the opposite direction. ]

${\small{\bold{\underline{\therefore\: The\: velocity\: of\: recoil\: is\: - 2\: m/s\: .}}}}$

Answer 2

Given :-

• Mass of the bullet ( m₁) = 200 g
• velocity of the bullet (v₁) = 100 m/s
• Mass of gun (m₂) = 10 kg

To Find :-

• Recoil velocity of the gun

Formula used :-

• Conservation of momentum

$\red\longrightarrow \sf \blue{ m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂}$

According to law of conservation of momentum, initial momentum is equal to final momentum.

Solution :-

Since, the gun and the bullet were at rest in the starting, therefore initial velocity of both the objects is 0.

Now,

$➡\sf \: 0 +0\longrightarrow \: 0.2 \times 100 + 10 \times x \\ ➡ \sf \: 0 = 20 + 10x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ➡ \sf \: 10x = - 20 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ➡ \sf \: x = \frac{ - 20}{10} = - 2m/ {s} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bf \therefore the \: recoil \: velocity \: of \: the \: gun \: is - 2m/s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \underline \purple { Hope \: it \: helps \: you!}$