A bullet of mass 20g is horizontally fired with a velocity 150m.s1 from a pistol of mass 2kg. what is the recoil velocity of pistol
Answers
Answered by
57
from law of conservation of momentum,
mass of bullet*velocity of bullet=mass of gun*velocity of gun
therefore,
150*0.02=2*x,
i.e.,
velocity of pistol=150*0.02*(1/2)
=150*0.01
=150*(1/100)
=3/2 m/s
mass of bullet*velocity of bullet=mass of gun*velocity of gun
therefore,
150*0.02=2*x,
i.e.,
velocity of pistol=150*0.02*(1/2)
=150*0.01
=150*(1/100)
=3/2 m/s
Answered by
61
By the law of conservation of linear momentum, we have
Momentum of bullet fired = Momentum of recoiling
p1 = p2
m1×v1 = m2×v2
0.02kg×150 = 2kg×v2
(150×2)÷(100×2) = v2
300÷200 = v2
So, v2 = 3/2 = 1.5 m/s
hope it helps......... :-)
Momentum of bullet fired = Momentum of recoiling
p1 = p2
m1×v1 = m2×v2
0.02kg×150 = 2kg×v2
(150×2)÷(100×2) = v2
300÷200 = v2
So, v2 = 3/2 = 1.5 m/s
hope it helps......... :-)
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