4.0 Laws of conservation of linear momentum

The product of mass $(m)$ and velocity $\left( {\overrightarrow v } \right)$ of a particle is defined as its linear momentum $\left( {\overrightarrow p } \right)$

Mathematically, $$\overrightarrow p = m\overrightarrow v $$

According to Newton’s second law of motion, which states that the rate of change of momentum of a particle is proportional to the applied force and takes place in the direction in which the force acts.

Mathematically, $$\overrightarrow F = \frac{{d\overrightarrow p }}{{dt}}$$

If the net external force acting on the system is zero, $$\overrightarrow F = 0$$ or $$\frac{{d\overrightarrow p }}{{dt}} = 0$$ So, $$\overrightarrow p = {\text{constant}}$$

So, when the net external force $\left( {\overrightarrow F } \right)$ acting on the system is zero, the linear momentum of the system remains constant. This is called the law of conservation of linear momentum.

If the net external forces acting on a system of particles is zero.

$${\overrightarrow F _{net}} = {\overrightarrow F _1} + {\overrightarrow F _2} + ... + {\overrightarrow F _n} = 0$$

Then the vector sum of linear momentum of all the particles remains constant. $${\overrightarrow p _1} + {\overrightarrow p _2} + ... + {\overrightarrow p _n} = {\text{constant}}$$

We also know, if the net external force acting on a system of particles is zero, then the centre of mass $(COM)$ of the system is at rest or moves with constant velocity.

$${\overrightarrow v _{COM}} = \frac{{{m_1}{{\overrightarrow v }_1} + {m_2}{{\overrightarrow v }_2} + ... + {m_n}{{\overrightarrow v }_n}}}{{{m_1} + {m_2} + ... + {m_n}}}$$ or $${\overrightarrow v _{COM}} = \frac{{{m_1}{{\overrightarrow v }_1} + {m_2}{{\overrightarrow v }_2} + ... + {m_n}{{\overrightarrow v }_n}}}{M}$$ Also, $$M{\overrightarrow v _{COM}} = {m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow v _2} + ... + {m_n}{\overrightarrow v _n}$$ or $${\overrightarrow p _{COM}} = {\overrightarrow p _1} + {\overrightarrow p _2} + ... + {\overrightarrow p _n}\quad {\text{when }}\left( {{{\overrightarrow F }_{net\;external}} = 0} \right)$$

Question 13. A particle of mass $2kg$ is moving with a velocity $\left( {4\widehat i + 3\overrightarrow j } \right)m/s$. Find the linear momentum associated with the particle.

Solution: Linear momentum $\left( {\overrightarrow p } \right)$ associated with the particle is defined as the product of mass $(m)$ and velocity $\left( {\overrightarrow v } \right)$ of a particle.

Mathematically, $$\begin{equation} \begin{aligned} \overrightarrow p = m\overrightarrow v \\ \overrightarrow p = 2\left( {4\widehat i + 3\widehat j} \right) \\ \overrightarrow p = \left( {8\widehat i + 6\widehat j} \right)Ns \\\end{aligned} \end{equation} $$

Question 14. A ball of mass $3kg$ is fired with an initial velocity of $10m/s$ in the horizontal direction. During its motion, it explodes into 2 parts $A$ & $B$ of masses $1kg$ & $2kg$ respectively. Part $B$ moves perpendicularly with $5m/s$. Find the velocity of part $A$ after explosion.

Solution: As we know that no external force acts on the system, so the linear momentum of the system is conserved.

Therefore, we can write, $$\begin{equation} \begin{aligned} M\overrightarrow v = {m_A}{\overrightarrow v _A} + {m_B}{\overrightarrow v _B} \\ 3\left( {10\widehat i} \right) = 1.{\overrightarrow v _A} + 2\left( {5\widehat j} \right) \\ {\overrightarrow v _A} = \left( {30\widehat i - 10\widehat j} \right)m/s \\\end{aligned} \end{equation} $$

Question 15. A tank of mass $M$ fires a shell of mass $m$ with speed $v_r$ relative to the barrel of the tank, which is inclined at an angle $\theta $ with the horizontal. Find the recoil speed of the tank. The tank is placed over a smooth horizontal surface.

Solution: Initially the whole system was at rest. So, the initial linear momentum of the system is zero.

$${\overrightarrow p _i} = 0\quad ...(i)$$

Now, when the tank fired the shell.

${\overrightarrow v _T}$: Velocity of the tank

${\overrightarrow v _S}$: Velocity of the shell

${\overrightarrow v _ST}$: Velocity of the shell relative to the tank

We know, $$\begin{equation} \begin{aligned} {\overrightarrow v _{ST}} = {\overrightarrow v _S} - {\overrightarrow v _T} \\ {v_r}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) = {\overrightarrow v _S} - {\overrightarrow v _T} \\ {\overrightarrow v _S} = {v_r}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) + {\overrightarrow v _T}\quad ...(ii) \\\end{aligned} \end{equation} $$

In the vertical $y$ direction, normal force from the surface acts as an external force. Therefore, the linear momentum will be conserved only in the horizontal direction.

So, applying conservation of linear momentum in the horizontal direction we get, $$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_x}}} = {\overrightarrow p _{{f_x}}} \\ 0 = M{\overrightarrow v _T} + m{\overrightarrow v _{{S_x}}} \\ 0 = M{\overrightarrow v _T} + m\left( {{v_r}\cos \theta \widehat i + {{\overrightarrow v }_T}} \right) \\ {\overrightarrow v _T} = \frac{{m{v_r}\cos \theta }}{{(M + m)}}\widehat i \\\end{aligned} \end{equation} $$