Question

find the velocity of a bullet of mass 10 grams which ks fired from a gun of mass 2.5kg. the recoil velocity of gun is 2.5m/s​

Answer 1

Answer:

1.5m/s

Explanation:

mass of bullet,m1 =20g(=0.02kg)

mass of pistol ,m2=2kg

initial velocity of the bullet,v1 =+150m s-1

let v be the recoil velocity of the pistol.

the total momentum of the pistol and bullet is zero before the fire.

(since both are at rest)

total momentum of the pistol and bullet after it is fired is = (0.02kg×150m s -1 )+(2kg× v m s -1)

=(3+2v) kg m s -1

total momentum after the fire =total momentum before the fire

3+2v =0

=v=-1.5/s

thus, the recoil velocity of the pistol is 1.5m/s

Answer 2

Answer:

$\bigstar{\bold{Velocity\:of\:bullet=625\:m/s}}$

Explanation:

$\Large{\underline{\underline{\it{Given:}}}}$

• Mass of bullet (m₂) = 10 g = 0.01 kg
• Mass of gun (m₁) = 2.5 kg
• Recoil velocity (v₁) = 2.5 m/s

$\Large{\underline{\underline{\it{To\:Find:}}}}$

• Velocity of the bullet (v₂)

$\Large{\underline{\underline{\it{Solution:}}}}$

→ Here by conservation of momentum, the momentum of the gun and before firing + momentum of gun and bullet after firing = 0

→ Hence,

  m₁ v₁ +  m₂ v₂ = 0

→ Here we have to find v₂ , that is velocity of the bullet.

  v₂ = - m₁ v₁/m₂-----(1)

→ Recoil velocity of gun = -2.5 m/s

→ Here recoil velocity is taken as negative since the recoil velocity is in the direction opposite to that of the bullet.

→ Substitute the given datas in equation 1

  v₂ = - ( 2.5 × -2.5)/0.01

  v₂ = 6.25/0.01

  v₂ = 625 m/s

→ Hence velocity of the bullet is 625 m/s

$\boxed{\bold{Velocity\:of\:bullet=625\:m/s}}$

$\Large{\underline{\underline{\it{Notes:}}}}$

→ Momentum can neither be created nor be destroyed.

→ The total momentum in an isolated system is always 0.

   m₁ v₁ + m₂ v₂ + m₃ v₃ +..... = 0