Question

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

Step-by-step explanation:

Initial momentum of the system is 0 as it is at rest.

After firing-

MV=mv

⟹V=

M

mv

=

2

0.02×150

⟹V=1.5m/s

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Answer 2

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

Mass of pistol, m2 = 2 kg

Initial velocity of the bullet (u) and pistol (uz) = 0

Final velocity of the bullet, v1 = +150m s1

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

ns) = (0.02 kg x 150 m s) + (2 kg x v ms

= (3 + 2v) kg m s

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.