a gun of mass 3 kg fires a bullet of mass 30 g. the bullet takes 0.003 s to move through the barrel of the gun and aqquires a velocity of 100 m/ s. calculate
a) the velocity with which the gun recoils
b) the force exerted on gun man due to recoil of the gun
Answers
Answer:
Given that,
Mass of the gun = m_1 = 3 kg
Mass of the bullet = m_2 = 30 g = 0.03 kg
Time taken by the bullet to love through the barrel of the gun = t = 0.003 s
Acquired velocity by the bullet = v_2 = 100 m/s
We need to Find : a) velocity by which gin recoils and b) force exerted on gun man due to recoil of the gun
Solution :
We know by law of conservation of momentum that,
m_1v_1 = m_2v_2
Hence, Substituting the given values,
⇒ 3*v_1 = 0.03*100
⇒ 3v_1 = 3
⇒ v_1 = 3/3
⇒ v_1 = 1 m/s
Now, initial velocity of the gun = 0 m/s
And the final velocity = 1 m/s
Time = 0.003 s
We know, Acceleration = v-u/t
⇒ Acceleration = 1-0/0.003
⇒ Acceleration = 1/0.003
⇒ Acceleration = 333.33 m/s²
We know, Force = ma = 3*333.33 = 999.99 = 1000 N
Hence, a) velocity by which gin recoils = 1 m/s² b) force exerted on gun man due to recoil of the gun = 1000 N.
ANSWER :
- The velocity with which the gun recoil is
- The force exerted on gun man due to recoil of the gun is 1000 N.
GIVEN :
- Mass of gun = 3 kg.
- Mass of bullet = 30 g.
- Time, t = 0.003 sec.
- Velocity, v = 100 m/s.
TO CALCULATE :
- The velocity with which the gun recoils.
- The force exerted by the gun man due to recoil of the gun.
FORMULA :
SOLUTION :
Net force on gun and bullet, system is zero.