Question:

A bullet of mass 0.1kg0.1 \,kg is fired with a speed of 100m/s.100 \,m/s. The mass of gun being 50kg50 \,kg. Then, the velocity of recoil become:

Updated On: Jun 23, 2024
  • 0.05 m/s
  • 0.5 m/s
  • 0.1 m/s
  • 0.2 m/s
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

In absence of any external force, linear momentum of a system is conserved..
Linear momentum of bullet-gun system is conserved.
i.e.. Initial momentum = Final momentum
0=m1v1+m2v2\therefore 0=m_{1} \,v_{1}+m_{2}\, v_{2}
or 0=0.1×100+50×v20=0.1 \times 100+50 \times v_{2}
v2=1050=0.2m/s\therefore v_{2}=-\frac{10}{50}=-0.2 \,m / s
Negative sign implies recoil.
Hence, the velocity of recoil is 0.2m/s0.2\, m / s.
Was this answer helpful?
0
0

Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration