A bullet of mass 20 g is fired from a gun weighing 5 km .if the velocity of the velocity of the bullet is 500meter/s what will be the speed of recoiling gun?
Answers
Answer:
2 m/sec
Explanation:
mass of bullet=20g=0.02kg
u=500m/sec
mass of gun=5kg
u=0,v=??
momentum of bullet before firing=0.02*500=10kgm/s
momentum of gun before firing is 0 as u is 0
total momentum before firing=0+10 = 10 kgm/s
momentum of bullrt after firing is 0 as v is o
recoil velocity of gun = v
momentum of gun after firing=m*v=5*v
total momentum after firing is 5v
we know that according to law of conservation of linear momentum
total p before firing=total p after firing
10=5v
v=10/5=2m/s
therefore velocity of recoiling gun is 2m/s
........hope it helps you..
Answer:
2 m/s
Explanation:
m1 = mass of bullet = 20 g = 0.02 kg
u1 = 0 m/s (body at rest)
v1 = 500 m/s
m2 = mass of gun = 5 kg
u2 = 0 m/s (body at rest)
v2 = ?
(m1 x u1) + (m2 x u2) = (m1 x v1) + (m2 x v2)
(0.02 x 0) + (5 x 0) = (0.02 x 500) + (5 x v2)
0 = 10 + 5v2
5v2 = -10
v2 = -10/5 = -2 m/s
Therefore, recoiling speed will be 2 m/s