(C)A bullet of mass 20 g is horizontally fired with a velocity 150 m/s

1.	(C)A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from apistol of mass 2 kg. What is the recoil velocity of the pistol? (3)
Answer» $\mathtt{\huge{\underline{\bf{\red{Answer :-}}}}}$ GIVEN :- A bullet of mass 20 g is HORIZONTALLY FIRED with a velocity 150 m/s from a pistol of mass 2 kg. To Find :- The recoil velocity of the pistol. Solution :- A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from a pistol of mass 2 kg. Mass of bullet = 20 g = 0.02 kg Mass of pistol = 2 kg Velocity of bullet = 150 m/s. Total mass = M(b) + M(p) = (0.02 + 2) kg = 2.02 kg Intial velocity is 0. Hence, Initial momentum = 2.02 × 0 = 0 Final momentum = M1V1 + m2v2 = 0.02 × 150 + 2 × v2 ...(ii) We know, Initial momentum = Final momemtum ➝ 0 = 0.02 × 150 + 2 × v2 ➝0 = $\dfrac{2}{100}$ × 150 + 2 × v2 ➝ 0 = 3 + 2v2 ➝ -3 = 2v2 ➝ v2 = 0.5m/s ____________________________________

(C)A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from apistol of mass 2 kg. What is the recoil velocity of the pistol? (3)

Answer»

$\mathtt{\huge{\underline{\bf{\red{Answer :-}}}}}$

A bullet of mass 20 g is HORIZONTALLY FIRED with a velocity 150 m/s from a pistol of mass 2 kg.

A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from a pistol of mass 2 kg.

Total mass = M(b) + M(p) = (0.02 + 2) kg = 2.02 kg

Hence, Initial momentum = 2.02 × 0 = 0

Final momentum = M1V1 + m2v2

= 0.02 × 150 + 2 × v2 ...(ii)

We know, Initial momentum = Final momemtum

➝ 0 = 0.02 × 150 + 2 × v2

➝0 = $\dfrac{2}{100}$ × 150 + 2 × v2

➝ 0 = 3 + 2v2

➝ -3 = 2v2

➝ v2 = 0.5m/s

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