Notifications
Clear all
Force and Laws of Motion
1
Posts
2
Users
0
Likes
410
Views
0
01/11/2021 7:56 pm
Topic starter
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 acquires a velocity of 100 m/s. Calculate :
(i) the velocity with which the gun recoils.
(ii) the force exerted on gunman due to recoil of the gun.
1 Answer
0
01/11/2021 8:05 pm
Mass of the gun m1 = 3 kg
Mass of bullet m2 = 30 g = 0.03 kg
Velocity of bullet v2 =100 m/s
(i) According to the law of conservation of momentum
m1 x v1 = m2 x v2
3 x v1 = 0.03 x 100
Recoil velocity v1 = \(\frac{100 \times 0.03}{3}\) = 1 m/s
(ii) Initial velocity of the gun u = 0 m/s
Final velocity of the gun v = 1 m/s
Time t = 0.003 s
Acceleration a = \(\frac{v - u}{t}\) = \(\frac{1 - 0}{0.003}\)
= \(\frac{1000}{3}\) m/s2
Force = m x a = 3 x \(\frac{1000}{3}\) = 1000 N