Question

A bullet of mass 40g horizontally fired with a velocity 300 m/ s from a pistol of mass 4

kg.What is the recoil velocity of the pistol?​

Answer 1

Given, v = -0.4 m/s

Mass of the bullet, m = 40 g = 0.04 kg

Speed of the bullet, v = 20 m/s

Mass of the pistol, m' = 2 kg

Let p is the initial momentum of the system and p' is the final momentum.

The initial momentum of the bullet + pistol is equal to zero because they both are at rest.

When the bullet is fired, final momentum is given by :

p' = 0.04 × 20 + 2 × v

= 0.04 × 20 + 2 × v

According to the conservation of momentum,

0 = 0.04 × 20 + 2 × v=0.04×20+2×v

v = -0.4 m/s

Hence, the recoil velocity of the pistol is

0.4 m/s.

Hope its helpful for you.

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Answer 2

Provided that:

• Mass of bullet = 40 gram
• Final velocity = 300 m/s
• Mass of pistol = 4 kg
• Initial velocity of bullet = 0
• Initial velocity of pistol = 0

Don't be confused!

• Initial velocity came as zero for both bullet and pistol because they hadn't peform any function till firing as the both are at rest!

To calculate:

• Recoil velocity of pistol

Solution:

• Recoil velocity of pistol = -3 m/s

Using concepts:

• Formula to convert g-kg
• Conservation of momentum

Using formulas:

• 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 grams into kilograms as the SI unit of mass is kilogram!

→ 1 g = 1/1000 kg

$:\implies \sf 40 \: g \: = \dfrac{40}{1000} \: kg$

$:\implies \sf 40 \: g \: = \dfrac{4}{100} \: kg$

$:\implies \sf 40 \: g \: = 0.04 \: kg$

${\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let's find out the recoil velocity of the pistol by using law of conservation of momentum!

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 0.04(0) + 4(0) = 0.04(300) + 4(v_B) \\ \\ :\implies \sf 0 + 0 = 0.04(300) + 4(v_B) \\ \\ :\implies \sf 0 = 0.04(300) + 4(v_B) \\ \\ :\implies \sf 0 = 012.00 + 4(v_B) \\ \\ :\implies \sf 0 = 12 + 4(v_B) \\ \\ :\implies \sf 0 - 12 = 4(v_B) \\ \\ :\implies \sf -12 = 4(v_B) \\ \\ :\implies \sf \dfrac{-12}{4} = \: v_B \\ \\ :\implies \sf -3 = \: v_B \\ \\ :\implies \sf v_B \: = -3 \: ms^{-1} \\ \\ :\implies \sf Recoil \: velocity \: = -3 \: ms^{-1}$

Negative sign tell us about the direction in which the pistol would recoil is opposite to that of bullet.