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
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
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
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