Question

A bullet of mass 10g is horizontally fired with a velocity 300m/s from a pistol of mass 6kg. What is the recoil velocity of the pistol?

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Answer 1

Given that:

• Mass of bullet = 10 g
• Final velocity = 300 mps
• Mass of pistol = 6 kg
• Initial velocity of bullet = 0
• Initial velocity of pistol = 0

To calculate:

• Recoil velocity of pistol

Solution:

• Recoil velocity of pistol = -1/2 mps that is -0.5 mps.

Using concepts:

• Law of conservation of momentum.

• Formula to convert g into kg.

Using formulas:

• Law of conservation of momentum:

${\small{\underline{\boxed{\sf{\rightarrow \: m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$ denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.)

• Formula to convert g-kg:

${\small{\underline{\boxed{\sf{\rightarrow \: 1 \: g \: = \dfrac{1}{1000} \: kg}}}}}$

Required solution:

~ Firstly let us convert g-kg!

$:\implies \sf 1 \: g \: = \dfrac{1}{1000} \: kg \\ \\ :\implies \sf 10 \: g \: = \dfrac{10}{1000} \: kg \\ \\ :\implies \sf 10 \: g \: = \dfrac{1}{100} \: kg \\ \\ :\implies \sf 10 \: g \: = 0.01 \: kg \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now by using law of conservation of momentum let us find out the recoil velocity!

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 0.01(0) + 6(0) = 0.01(300) + 6(v_B) \\ \\ :\implies \sf 0 + 0 = 3 + 6v_B \\ \\ :\implies \sf 0 = 3 + 6v_B \\ \\ :\implies \sf 0 - 3 = 6v_B \\ \\ :\implies \sf -3 = 6v_B \\ \\ :\implies \sf \dfrac{-3}{6} \: = v_B \\ \\ :\implies \sf \dfrac{-1}{2} \: ms^{-1} = v_B \\ \\ :\implies \sf -0.5 \: ms^{-1} = v_B$