Question

A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s, find the velocity of bullet.

Thanks :) ​

Answer 1

Given :-

• Mass of the bullet (m1) = 5 gm = 0.0005 kg
• Mass of the pistol (m2) = 1.5 kg
• Recoil velocity of the pistol ( v2) = 1.5 m/s
• Initial velocity of the pistol (u2) = 0
• Initial velocity of the bullet (u1) = 0

To Find :-

• Velocity of the bullet ( v2)

Formula used :-

$✯ \: \sf \red {m1u1 +m2u2 = m1v1 +m2v2}$

In the following question, we will use the formula derived by the law of conservation of momentum.

Q. What is "momentum"?

↦The product of mass and velocity is known as momentum. It is denoted by the letter "P".

Q. What is law of conservation of momentum?

↦"Law of conservation of momentum" states that momentum before collision and after collision remains same.

Now, let the velocity of the bullet after collision be p.

According to law of conservation of momentum,

$\sf \implies \:0.005 \times 0 + 1.5 \times 0 = 0.005 \times p + 1.5 \times 1.5 \\ \sf \implies \:0 + 0 = 0.005p + 2.25 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies \:0.005p = - 2.25 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies \:p = \frac{2.25}{0.005} = \frac{2.25}{ \frac{5}{1000} } = \frac{2.25 \times 1000}{5} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies \:\frac{2.25 \times \cancel{1000}}{ \cancel5} = 2.25 \times 200 = 450 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf\pink { \therefore p =velocity \: of \: bullet \: = 450 m/s}$