A bullet of mass 20 gram is horizontally fired with a velocity 150 metre per second from a pistol of mass 2kg what is the recoil velocity of the pistol
Answers
Answer: Mass of bullet, m1 = 20g (= 0.02 kg)
Mass of pistol, m2 = 2 kg
Initial velocity of the bullet (u1) and pistol (u2) = 0
Final velocity of the bullet, v1 = +150m s-1
Let v be the recoil velocity of the pistol.
The total momentum of the pistol and bullet is zero before the fire. (Since both are at rest)
Total momentum of the pistol and bullet after it is fired is
= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)
= (3 + 2v) kg m s-1
Total momentum after the fire = Total momentum before the fire
3 + 2v = 0
→v = -1.5 m/s
Thus, the recoil velocity of the pistol is 1.5 m/s.
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Answer:
mass1=0.02kg
u1=0m/s
v1=150m/s
mass2=2kg
u2=0m/s
v2=v.
∴by the law of conservation of momentum,
m1u1+m2u2=m1v1+m2v2
0.02*0+2*0=0.02*150+2*v
⇒0+0=3+2v
⇒3=2v
⇒3/2=v
⇒1.5=v
∴recoil velocity = 1.5 m/s
Explanation:
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