Question

A bullet of mass 0.02 kilogram is fired from a gun weighing 7.5 kilogram .if the initial velocity of the bullet is 200 metre per second. calculate the speed with which the gun recoils.

Answer 1

Given :-

• Mass of bullet = 0.02 kg
• Velocity of bullet = 200 m/s
• Mass of gun = 7.5 kg

To Find :-

• The speed with which the gun recoils = ?

Solution :-

• M = mass of the gun
• V = velocity of the gun
• m = mass of the bullet
• v = velocity of the bullet

As no external force is acting on the system the momentum remains conserved.

By applying law of conservation of momentum we get :

• Pi = Pf

Where, Pi is Initial momentum and Pf is Final momentum.

Initially the system (gun + bullet) is at rest

→ Pi = 0

→ 0 = Pf

→ 0 = MV + mv

→ - mv = MV

→ V = - mv ÷ M

→ V = - 0.02 × 200 ÷ 7.5

→ V = -4 ÷ 7.5

→ V = - 0.53 m/s

Therefore,the speed with which the gun recoils is -0.53 m/s.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

A bullet of mass 0.02 kilogram is fired from a gun weighing 7.5 kilogram .if the initial velocity of the bullet is 200 metre per second. calculate the speed with which the gun recoils.

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• Mass of gun = 7.5 kg
• Mass of bulet = 0.02 kg
• Speed of bulet = 200 m/s

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• The speed with which the gun recoils.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

By the law of conservation of momentum we get ➜

☞ Momentum P = mv, must be if there are no external forces acting on the system. Since there are only internal forces acting between the bullet and the gun momentum must been conserved.

⚪Mass of bullet × Speed of bullet = Mass of gun × Recoil velocity of gun.

Let x means Recoil velocity of gun

⚪0.02 × 200 = 7.5 × x

⚪4 = 7.5 × x

⚪4 ÷ 7.5 × x

⚪0.53 m/s = x

⚪x = 0.53 m/s

$\large\orange{\texttt{0.53m/s is Recoil velocity}}$

@Itzbeautyqueen23

Hope it's helpful

Thank you :)