Centre of Mass and Conservation of Linear Momentum
7.0 Collision
7.0 Collision
The collision is defined as the redistribution of total momentum of the particles when a large force acts on each colliding particle for a relatively short time.
Let us understand this better with the help of an example.
Consider two blocks $A$ & $B$ of mass $m_1$ & $m_2$ moving with velocities $v_1$ & $v_2$ respectively where $\left( {{v_1} > {v_2}} \right)$ along the same straight line on a smooth horizontal surface. A spring of spring constant $k$ is attached to the block of mass $m_2$ as shown in the figure.
| Description | Diagram |
| As $\left( {{v_1} > {v_2}} \right)$, So, block $A$ is approaching block $B$ |
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Block $A$ starts compressing the spring, which generates tension in the spring. This tension oppose the motion of block $A$ and supports the motion of block $B$ |
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When the spring reaches the maximum compression $\left( {{x_m}} \right)$, then the velocities of both the blocks become same |
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| Due to compression, tension $(T)$ force present in the spring further decreases the velocity of block $A$ and increases the velocities of block $B$ |
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| Finally both the blocks separates and moves independently. Due to the collision the final velocities of block $A$ & $B$ are $v_1^/$ &$v_2^/$ respectively where $\left( {v_1^/ > v_2^/} \right)$. |
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Note: During collision, colliding particle may or may not touch each other.
During collision no external force acts on the system, so the total linear momentum of the system remains conserved. Therefore,
$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ {m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow v _2} = {m_1}{\overrightarrow {v'} _1} + {m_2}{\overrightarrow {v'} _2} \\\end{aligned} \end{equation} $$
Since, this is an example of a perfectly elastic collision, so the total kinetic energy will also be conserved.
This concept will be better understood after discussing types of collision.
Therefore, $$\begin{equation} \begin{aligned} K{E_i} = K{E_f} \\ \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 = \frac{1}{2}{m_1}v_1^{{/^2}} + \frac{1}{2}{m_2}v_2^{{/^2}} \\\end{aligned} \end{equation} $$
Question 20: Two blocks $A$ & $B$ of equal masses $m=2kg$ are lying on a smooth horizontal surface as shown in the figure. A spring of force constant $k=100N/m$ is fixed at one end of block $B$. Block $A$ collides with block $B$ with velocity $v_0=2m/s$. Find the maximum compression of the spring.
Solution: Initial linear momentum $\left( {{{\overrightarrow p }_i}} \right)$,
$${\overrightarrow p _i} = m{v_0} = 2(2) = 4N/s\quad ...(i)$$
At maximum compression block $A$ & $B$ moves with the same velocity $\left( {\overrightarrow v } \right)$
So, the final linear momentum $\left( {{{\overrightarrow p }_f}} \right)$, is given by, $${\overrightarrow p _f} = 2mv\quad ...(ii)$$
As during collision, the linear momentum is conserved. Therefore,
$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ 4 = 2(2)v \\ v = 1m/s\quad ...(iii) \\\end{aligned} \end{equation} $$
Since there is no loss of energy, therefore mechanical energy will be also conserved.
Internal energy $\left( {{E_i}} \right)$, $${E_i} = \frac{1}{2}mv_0^2 = \frac{1}{2}(2){(2)^2} = 4J\quad ...(iv)$$
Final energy $\left( {{E_f}} \right)$, $$\begin{equation} \begin{aligned} {E_f} = \frac{1}{2}2mv_0^2 + \frac{1}{2}kx_m^2 = \frac{1}{2}(4){(1)^2} + \frac{1}{2}(100)x_m^2 \\ {E_f} = 2 + 50x_m^2\quad ...(v) \\\end{aligned} \end{equation} $$
From equation $(iv)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} {E_i} = {E_f} \\ 4 = 2 + 50x_m^2 \\ x_m^2 = \frac{1}{{25}} \\ {x_m} = \frac{1}{5}m \\ {x_m} = 20cm \\\end{aligned} \end{equation} $$
So, the maximum compression in the spring is, $x_m=20cm$.
