Question

Example 9.6 A bullet of mass 20 g is
horizontally fired with a velocity
150 m s' from a pistol of mass 2 kg.
What is the recoil velocity of the pistol?​

Answer 1

Explanation:

According to law of conservation of linear momentum,

MV = mv.

M is mass of pistol = 2 kg.

m is mass of bullet = 2 g = 2 (10 ^-3) kg.

v = velocity of bullet = 150 m/s.

V is recoil velocity of pistol .

V= (m/M) v = [ 2(10^-3)/(2)] (150) = 0.15 m/s...

Answer 2

$\bf\huge{\underline{\underline{ANSWER}}}$

Final 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= (0.02 kg x 150 m s-1) + (2 kg x 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 m/s

Thus, the recoil velocity of the pistol is 1.5 m/s.

hope it helps.........✌️

please follow me if you can.........xd