Question

find the recoil velocity of a gun having mass equal to 5kg, if a bullet of 25gm acquires the velocity of 500m/s after firing from the gun​

Answer 1

GivEn:

• Mass of Bullet, $\sf m_1$ = 25 gm = 0.025 kg

• Velocity of bullet before firing, $\sf u_1$ = 0 m/s

• Mass of Gun, $\sf m_2$ = 5 kg

• Velocity of bullet after firing, $\sf v_2$ = 500 m/s

• Velocity of bullet before firing, $\sf u_2$ = 0 m/s

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To find:

• Velocity of gun after firing?

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SoluTion:

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As we know that,

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$\star\;{\boxed{\sf{\purple{Momentum,\;p = mv}}}}$

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${\underline{\sf{\bigstar\;Momentum\;before\;firing\;:}}}$

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$:\implies\sf m_1 u_1 + m_2 u_2$

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$:\implies\sf 0.025 \times 0 + 5 \times 0$

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$:\implies\sf 0 + 0$

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$:\implies\bf \red{0\;kg\; m/s}$

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${\underline{\sf{\bigstar\;Momentum\;after\;firing\;:}}}$

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$:\implies\sf m_1 v_1 + m_2 v_2$

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$:\implies\sf 0.025 \times 500 + 5 + v_2$

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$:\implies\sf 12.5 + 5 + v_2$

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$:\implies\bf \red{12.5 + 5 v_2\; kg\;m/s}$

━━━━━━━━━━━━━━━

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As we know that,

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$\star\;\sf Momentum\;before\;firing = Momentum\;after\;firing$

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$:\implies\sf 0 = 12.5 + 5 v_2$

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$:\implies\sf - 12.5 = 5 v_2$

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$:\implies\sf v_2 = \cancel{ \dfrac{- 12.5}{5}}$

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$:\implies{\underline{\boxed{\bf{\blue{- 2.5\;m/s}}}}}\;\bigstar$

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$\therefore$ Thus, the recoil velocity of gun or velocity after firing is - 2.5 m/s.

Here, -ve sign shows that the gun moves in direction opposite to that of bullet.