Question

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

Answer 1

m1 = 0.02kg

m2 = 2 kg

u1 = 0 m/s

u2 = 0 m/s

v1 = 150 m/s

v2 = ?

m1u1 + m2u2 = m1v1 + m2v2

(0.02)(0) +( 2)(0) = (0.02)(150) + (2)(v2)

0 = 3 + (2)(v2)

2 × v2 = -3

v2 = -1.5 m/s

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Answer 2

AnswEr:

• Mass ($\sf{m}_{1}$) of bullet = 20 g . ( convert g into kg)

$\therefore$ $\sf{0.02kg}$

• Mass ($\sf{m}_{2}$) of pistol = 2 kg.

↝ Initial velocity is inside the gun and is not moving.

$\therefore$$\sf\underline\blue{\:\:\:\:\:Initial\:\:Momentum\:\:\::-}$

$\rightarrow \tt \: 2.02 \times 0 \qquad \: (i)$

Let the velocity of the pistol be $\sf{V}_{2}$ and

Velocity of bullet = 150

$\therefore$$\sf\underline\green{\:\:\:\:\:Final\:momentum\:\:\:\::-}$

$\sf m_1v_1 + m_2 + v_2 \\ \\ \sf \: = 0.02 \times 150 \times 2v_2 \qquad \: (ii)$

We know that, Initial momentum = final momentum

$\therefore \sf \frac{(0.02 \times 150)}{100} + 2V_2 = 0 \qquad \: (from \: i \: and \: ii) \\ \\ \sf \implies \: 3 + 2V_2 = 0 \\ \\ \implies \sf \: 2V_2 = 0 - 3 \\ \\ \implies \sf \: 2V_2 = - 3 \\ \\ \implies \sf \: V_2 = - 1.5 \: ms - {}^{1}$

Here, (-) sign indicates that gun recoils in opposite direction that of the bullet.