Question

a bullet of mass 40g is horizontally fired with velocity 200m sec-1 from a pistal of the mass 2 kg .Find out recoil velocity of pistal....​

Answer 1

Answer:

f=mv

f=0.04×200

f=8N

8N=2×v

v=8/2

v=4

recoiling is 4 m per second

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Answer 2

Answer:-

Velocity of pistol= 4 m/s

Explanation:-

To Find:-

Find velocity of pistol.

Solution:-

Given:-

mass of bullet m = 40 g= 0.04 kg

mass of pistol m' = 2 kg

• initial velocity of bullet (u) and pistol(u')

= 0

Final velocity of bullet(v)= 200 m/sg

Let , v' be the recoil velocity of pistol.

Total momentum of bullet and pistol is zero before the fire.(both at rest)

Then,

Total momentum of both after it is fired →

→ (0.04 × 200)+ (2 × v')

→ 8 + 2v'

Hence,

mom. after fire= mom.before fire

→ 8 + 2v' = 0

→ v'= -4 m/s

It is positive.