Question

from a rifle of mass 4 kg, a bullet of mass 50 gm is fired with an initial velocity of 35 m/s . calculate the initial recoil velocity of the rifle​

Answer 1

Answer:

Mass of rifle (M) = 4kg

Mass of bullet (m) = 50g = 0.5 kg

Initial velocity of bullet (u1) = 35m/s

Momentum,

M×u2 = m×u1

4 ×u2 = 0.5 ×35

u2 = 4.375

Therefore,

Recoil Velocity = 4.375m/s.

Answer 2

Answer:

Given :-

• A rifle of mass 4 kg, a bullet of mass 50 gm is fired with an initial velocity of 35 m/s.

To Find :-

• What is the initial recoil velocity of the rifle.

Formula Used :-

$\clubsuit$ Law of Conservation of Momentum Formula :

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

where,

• m₁ = Mass of rifle
• u₁ = Initial Velocity of rifle
• m₂ = Mass of bullet
• u₂ = Initial Velocity of bullet
• v₁ = Final Velocity of rifle
• v₂ = Final Velocity of bullet

Solution :-

Given :

➳ Mass of rifle (m₁) = 4 kg

➳ Mass of bullet (m₂) = 50 gm = - 0.05 kg

➳ Initial Velocity of bullet (u₂) = 35 m/s

➳ Final Velocity of rifle (v₁) = 0 m/s²

➳ Final Velocity of bullet (v₂) = 0 m/s²

According to the question by using the formula we get,

$\implies \sf\bold{\purple{m_1u_1 + m_2u_2 =\: m_1v_1 + m_2v_2}}$

$\implies \sf (4)u_1 + (- 0.05)(35) =\: (4)(0) + (0.05)(0)$

$\implies \sf (4 \times u_1) + (- 0.05 \times 35) =\: (4 \times 0) + (0.05 \times 0)$

$\implies \sf 4u_1 + (- 1.75) =\: 0 + 0$

$\implies \sf 4u_1 - 1.75 =\: 0$

$\implies \sf 4u_1 =\: 1.75$

$\implies \sf u_1 =\: \dfrac{1.75}{4}$

$\implies \sf\bold{\red{u_1 =\: 0.44\: m/s}}$

$\therefore$ The initial recoil velocity of the rifle is 0.44 m/s .