A bullet of 5 g is fired from a pistol of 1.5 kg. If the recoil speed of pistol is 1.5 ms−1, find the velocity of bullet.
450 ms−1
Given:
Mass of the bullet:
m1=5g=5×10−3 kg
Mass of the pistol, m2=1.5 kg
Recoil speed of the pistol, v2=1.5 ms−1
Let the velocity of the bullet be v1
Before firing, the bullet and pistol are in rest. Hence,
Initial speed of bullet, u1=0
Initial speed of pistol, u2 = 0
From law of conservation of momentum, initial and final momentums are equal.
m1u1+m2u2=m1v1+m2v2
0=(5×10−3×v1)+(1.5×1.5)
v1=−450 ms−1
Thus, speed of the bullet is 450 ms−1.
Negative sign shows that bullet moves in the direction opposite to the direction of recoiling of the gun.