Question

# 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.

A

450 ms1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
400 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
500 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
600 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 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.

Suggest Corrections
12
Related Videos
Conservation of Momentum
PHYSICS
Watch in App
Explore more