Question:
A bullet of mass $0.1 \,kg$ is fired with a speed of $100 \,m/s.$ The mass of gun being $50 \,kg$. Then, the velocity of recoil become:
A bullet of mass $0.1 \,kg$ is fired with a speed of $100 \,m/s.$ The mass of gun being $50 \,kg$. 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
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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
$\therefore 0=m_{1} \,v_{1}+m_{2}\, v_{2}$
or $0=0.1 \times 100+50 \times v_{2}$
$\therefore v_{2}=-\frac{10}{50}=-0.2 \,m / s$
Negative sign implies recoil.
Hence, the velocity of recoil is $0.2\, m / s$.
Linear momentum of bullet-gun system is conserved.
i.e.. Initial momentum = Final momentum
$\therefore 0=m_{1} \,v_{1}+m_{2}\, v_{2}$
or $0=0.1 \times 100+50 \times v_{2}$
$\therefore v_{2}=-\frac{10}{50}=-0.2 \,m / s$
Negative sign implies recoil.
Hence, the velocity of recoil is $0.2\, m / s$.
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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