Question

A bullet of mass 10g is fired from a gun of 1kg with a recoil velocity of gun is $5\dfrac{m}{s}$. The muzzle velocity will be.
$A.$ $30\dfrac{{km}}{{\min }}$
$B.$ $60\dfrac{{km}}{{\min }}$
$C.$ $30\dfrac{m}{s}$
$D.$ $500\dfrac{m}{s}$

Answer 1

HINT- In this question we will use the concept of conservation of momentum to calculate the velocity of the bullet. We know that momentum is defined as the product of mass and velocity of the object. It is a vector quantity, possessing a magnitude and a direction. If $m$ is an object’s mass and velocity $v$ , then the object’s momentum is:
$p = mv$ SI unit is kilogram metres per second ($kg\dfrac{m}{s}$ )

Complete step-by-step answer:
The law of conservation of linear momentum states that the momentum of an isolated system remains constant. Momentum is therefore said to be conserved over time; that is, momentum is neither created nor destroyed, only transformed or transferred from one form to another.
Therefore, by conservation of momentum, we know,
${m_1}{v_1} = {m_2}{v_2}$
$\dfrac{{10}}{{1000}}kg \times {v_1} = 1kg \times 5\dfrac{m}{s}$ ($10g = \dfrac{{10}}{{1000}}kg$ )
${v_1} = \dfrac{{1kg \times 5\dfrac{m}{s}}}{{\dfrac{{10kg}}{{1000}}}}$
Velocity of muzzle ${v_1} = 500\dfrac{m}{s}$ which gives option $D.$ as the correct answer

NOTE- So we use the concept of conservation of momentum to find out the correct answer for the above question. For example, that two particles interact. Because of Newton's third law, the forces between them are equal and opposite. If the particles are numbered 1 and 2, the second law states that ${F_1} = \dfrac{{d{p_1}}}{{dt}}$ and ${F_2} = \dfrac{{d{p_2}}}{{dt}}$
Therefore, $\dfrac{{d{p_1}}}{{dt}} = - \dfrac{{d{p_2}}}{{dt}}$ with the negative sign indicating that the forces oppose. Equivalently, $\dfrac{d}{{dt}}({p_1} + {p_2}) = 0$. If the velocities of the particles are ${u_1}$ and ${u_2}$ before the interaction, and afterwards they are ${v_1}$ and ${v_2}$, then ${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$.
This law holds no matter how complicated force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds up to zero. This conservation law applies to all interactions, including collisions and separations caused by explosive forces. It can also be generalized to situations where Newton’s laws do not hold, for example in the theory of relativity and in electrodynamics.