Question

A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of 100 m s -1 . Calculate the velocity recoil of the gun.

Answer 1

☣ Qᴜᴇsᴛɪᴏɴ ☣

➥ A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of 100 m s -1 . Calculate the velocity recoil of the gun.

⛦ Aɴsᴡᴇʀ ⛦

➥ Velocity of gun (v₂) = -0.5 m/s

☄ Gɪᴠᴇɴ ☄

➤ Mass of bullet (m₁)= 100 gm = $\sf{\displaystyle{\frac{100}{1000}}}kg$

➤ Mass of gun (m₂)= 20 kg

➤ Velocity of bullet (v₁)= 100 m/s

⚛ Tᴏ Fɪɴᴅ ⚛

➤ Velocity of gun (v₂) = ?

$\rule{195}{3}$

• m denoted by mass
• u denoted by intial velocity
• v denoted by final velocity

$\rule{195}{3}$

⠀⠀⠀⠀

❏ According To Given Question

⠀⠀⠀⠀

↗ Intial velocity u₁ = u₂ = 0 (Both are at rest)

Applying principal of conservation of momentum

❊ As we know that the formula for finding the conservation of momentum

is ፦

$\displaystyle{\bf{: \implies m_1 \: u_1+m_2 \: u_2=m_1 \: v_1+m_2 \: v_2}}$

⤵ On putting the given value in the formula, we get

$\sf{: \implies \frac{100}{1000} \times 0 + 20 \times 0 = \frac{100}{1000} \times 1000 + 20 \times v_2 } \\ \\ \sf{ : \implies 0 = 10 + 20 \: v_2} \\ \\ \sf{: \implies - 10 = 20 \: v_2} \\ \\ \sf{: \implies v_2 = \frac{ - 10}{20} } \\ \\ \bf{ : \implies \underline{ \boxed{ \bf{ \purple{v_2 = - 0.5 \: m/s}}}}}$

(-) sign indicates that velocity of gun is an opposite direction.

Answer 2

Question:

A bullet of mass 20 g is fired from a gun with a muzzle velocity 100 m/s. Find the recoil of the gun it

its mass is 5 kg.

Required Answer

-0.4m/s

Given data:

• m = 20 g = 0.02 kg
• V = 100 m/s
• M = 5 kg

To find:

• V = ?

Solution:

According to the law of conservation of momentum:

MV+ mv = 0

Putting the values, we get:

⇨5kg × V+ (0.02 kg)x(100 m/s) = 0

or

$\mathcal{⇒5 kg × V = - (0.02 kg)x(100 m/s)}$

$\mathcal{⇒V = \frac{(0.2kg) \times (100m {s}^{ - 1}) }{5kg} }$

⇒0.4m/s

The negative sign indicates that the gun recoils i.e., moves in the backward direction opposite to the motion of the bullet with a velocity of 0.4 m/s.

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