Question

A bullet of mass 40 gm is fired the rifle recoil with a 3m/s find the velocity of the bullet when it leaves the rifle if the mass of the rifle is 4 kg
Its urgent ​
Please give with full explanation ​

Answer 1

$\bf\large\red{\underline{ given : - }}$

• Mass of bullet (m₁) = 40grams
• Mass of rifle(m₂) = 4kg
• Velocity of recoil(v₂) = -3m/sec ( minus sign because it is in opposite direction of velocity of bullet.

$\\ \bf\large\red{\underline{to \: find : - }}$

• Velocity of bullet when it leaves the rifle(v₁).

$\\ \bf\large\red{\underline{ concept \: used : - }}$

• Conservation of momentum :-

The amount of momentum remains constant.

Momentum is neither created , nor destroyed.

∴ According to the 2nd Newton's Law of Motion,

total momentum before collision is equal to total momentum after collision when no external force is applied.

$\\ \bf\large\red{\underline{ formula \: used : - }}$

$\boxed{\bf { p = mv}}$

Here ,

• p is momentum.
• m is mass of object.
• v is velocity of object.

$\boxed{\bf { m1u1 + m2u2 = m1v1 + m2v2}}$

Here ,

• m1 ➜ mass of bullet.
• m2 ➜ mass of rifle.
• u1 ➜ initial velocity of bullet.
• u2 ➜ initial velocity of rifle.
• v1 ➜ final velocity of bullet.
• v2 ➜ final velocity of rifle.

$\\ \bf\large\red{\underline{solution : - }}$

Since , rifle and bullet both were initially at rest.

u1 = 0m/sec

u2 = 0m/sec

m1 = 40grams

m1 = 40gramsm2 = 4kg = 4000grams

v2 = -3m/sec

v1 = ?

⇉ Putting in the formula,

↦40(0) + 4000(0) = 40(v1) + 4000(-3)

↦0 + 0 = 40(v1) - 12000

↦40(v1) = 12000

↦v1 = 300 m/sec

Therefore , velocity of bullet when it leaves the rifle is 300m/sec.