Physics, asked by agrimashakya75, 10 months ago

A bullet of mass 40 g is fixed from a rifel of mass 4 kg if its speed is increased from rest to 50m/s un 10 s calculate the amount of foece exerted​

Answers

Answered by Anonymous
1

Answer:Mass of the rifle (m1) = 4 kg

Mass of the bullet (m2) = 50 g =  0.05 kg

Recoil velocity of the rifle = v1

A bullet is fired with an initial velocity (v2) = 35 m/s

Initially, the rifle is at rest.

Thus, its initial velocity (v) = 0

Total initial momentum of the rifle and bullet system

⇒  (m1+m2) × v = 0

The total momentum of the rifle and bullet system after firing

= m1v1 + m2v2

= (4 × v1 ) + (0.05 × 35)

= 4v1 + 1.75

According to the law of conservation of momentum

Total momentum after the firing = Total momentum before the firing

4v1 + 1.75 = 0

v1= -1.75 / 4

v1 = – 0.4375 m/s

The negative sign indicates that the rifle recoils backwards with a velocity of 0.437m/s

Answered by harshraykwar1332008
0

Explanation:

mass of bullet = 40 grams =40 kg/1000

mass of rifle = 4 kg

initial velocity of bullet = 0 (because bullet is rest)

final velocity of bullet =50 m/s

time taken by bullet =10 s

what is the final velocity of bullet =?

initial momentum = momentum of bullet + momentum of rifle

initial momentum = mbub + mrur

initial momentum = 40/1000*0 + 4*0 = 0

final momentum = momentum of bullet + momentum of rifle

final momentum = mbvb + mrvr

final momentum = 40/1000*vb + 4*50

final momentum = 40/1000*vb + 200

according to Newton third law of motion

pi = pf

0 = 40/1000*vb + 200

-200 = 40/1000*vb

-200000 = 40*vb

vb = -5000 m/s

I hope I am brilliant student

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