A classical mechanics problem by Jàfąř ÀĀł

Classical Mechanics Level 1

A 10 g 10\text{ g} bullet is fired from a rifle of mass 5 kg 5\text{ kg} . If the speed of the fired bullet is 880 m/s, what is the approximate recoil speed of the rifle?

1.7 5 3

Jàfąř Àāł by Jàfąř ÀĀł , 21, Syria

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3 solutions

Jafar Badour
Aug 23, 2016

Due to the conservation of momentum, and becuase the frame was in a static state! m 1 v 1 + m 2 v 2 = 0 m_1 v_1 + m_2v_2=0 thus

v 2 = m 1 v 1 m 2 v_2= \frac{m_1v_1}{m_2}

v 2 = 10 ∗ 1 0 − 3 ∗ 880 5 = 1.76 m . s − 1 v_2= \frac{10*10^{-3} *880}{5} = 1.76m.s^{-1}

this was an easy survey problem, keep on the good work (Y).

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Aug 22, 2016

By conservation of momentum, the total momenta of bullet (subscript 1) and rifle (subscript 2) before firing is equal to that after firing. Therefore, we have:

m 1 v 1 + m 2 v 2 = m 1 ( 0 ) + m 2 ( 0 ) = 0 m 2 v 2 = − m 1 v 1 5 v 2 = − 0.01 ( 880 ) ⟹ v 2 = − 17.6 m/s − indicates opposite direction to v 1 ∣ v 2 ∣ ≈ 1.7 \begin{aligned} m_1 v_1 + m_2 v_2 & = m_1(0) + m_2(0) \\ & = 0 \\ m_2v_2 & = -m_1v_1 \\ 5v_2 & = - 0.01(880) \\ \implies v_2 & = \color{#3D99F6}{-} 17.6 \text{ m/s} & \small \color{#3D99F6}{- \text{ indicates opposite direction to }v_1} \\ |v_2| & \approx \boxed{1.7} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Shreyansh Tiwari
Sep 6, 2016

use the law of conservation of momentum

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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