Elastic Collision in One Dimension

Elastic collision in one dimension involves two bodies moving initially along the same straight line, striking each other without losing kinetic energy, and continuing to move along the same straight line after the collision.

Suppose two balls A and B, of masses $m_1$ and $m_2$, are moving initially along the same straight line with velocities $u_1$ and $u_2$ respectively, as shown in the figure below.

When $u_1 > u_2$, the relative velocity of approach before the collision is:

 

 

\[
u_1 – u_2 \tag{1}
\]

When the two balls collide, let the collision be perfectly elastic. After the collision, suppose $v_1$ is the velocity of A and $v_2$ is the velocity of B along the same straight line. When $v_2 > v_1$, the bodies separate after collision.

The relative velocity of separation after collision is

\[
v_2 – v_1 \tag{2}
\]

Linear momentum of the two balls before collision:

\[
m_1 u_1 + m_2 u_2
\]

Linear momentum of the two balls after collision:

\[
m_1 v_1 + m_2 v_2
\]

As linear momentum is conserved in an elastic collision:

\[
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \tag{3}
\]

or,

\[
m_2 (v_2 – u_2) = m_1 (u_1 – v_1) \tag{4}
\]

Total Kinetic Energy of the two balls before collision:

\[
\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \tag{5}
\]

Total Kinetic Energy of the two balls after collision:

\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \tag{6}
\]

As kinetic energy is also conserved in an elastic collision, therefore from (5) and (6):

\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2
\]

or

\[
\frac{1}{2} m_2 \left( v_2^2 – u_2^2 \right) = \frac{1}{2} m_1 \left( u_1^2 – v_1^2 \right)
\]

or

\[
m_2 \left( v_2^2 – u_2^2 \right) = m_1 \left( u_1^2 – v_1^2 \right) \tag{7}
\]

Dividing (7) by (4), we get

$$\dfrac{m_2 \left( v_2^2 – u_2^2 \right)}{m_2 \left( v_2 – u_2 \right)} = \dfrac{m_1 \left( u_1^2 – v_1^2 \right)}{m_1 \left( u_1 – v_1 \right)}$$

\[
\frac{{\left( {{{v}_{2}}+{{u}_{2}}} \right)\left( {{{v}_{2}}-{{u}_{2}}} \right)}}{{{{v}_{2}}-{{u}_{2}}}}=\frac{{\left( {{{u}_{1}}+{{v}_{1}}} \right)\left( {{{u}_{1}}-{{v}_{1}}} \right)}}{{{{u}_{1}}-{{v}_{1}}}}
\]

or

\[
v_2 + u_2 = u_1 + v_1
\]

or

$$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol{{{v}_{2}}-{{v}_{1}}={{u}_{1}}-{{u}_{2}}}}\tag{8}$$

Hence, in one-dimensional elastic collision, the relative velocity of separation after collision is equal to the relative velocity of approach before collision.

From (8),

\[
\frac{v_2 – v_1}{u_1 – u_2} = 1
\]

By definition,

\[
\frac{v_2 – v_1}{u_1 – u_2} = e = 1
\]

Hence, coefficient of restitution (resilience) of a perfectly elastic collision in one dimension is unity.

Calculation of Velocities After Collision:

Velocity of A:

From (8),

\[
v_2 = u_1 – u_2 + v_1 \tag{9}
\]

Putting in (3), we get:

\[
m_1 v_1 + m_2 \left( u_1 – u_2 + v_1 \right) = m_1 u_1 + m_2 u_2
\]

\[
m_1 v_1 + m_2 u_1 – m_2 u_2 + m_2 v_1 = m_1 u_1 + m_2 u_2
\]

\[
v_1 \left( m_1 + m_2 \right) = \left( m_1 – m_2 \right) u_1 + 2 m_2 u_2
\]

$$\bbox[15px, #e4e4e4, border: 2px solid #000000]{{{v}_{1}}=\frac{{\left( {{{m}_{1}}-{{m}_{2}}} \right){{u}_{1}}}}{{{{m}_{1}}+{{m}_{2}}}}+\frac{{2{{m}_{2}}{{u}_{2}}}}{{{{m}_{1}}+{{m}_{2}}}}}\tag{10}$$

Velocity of B:

Put this value of $v_1$ from (10) in (9),

\[
v_2 = u_1 – u_2 + \frac{\left( m_1 – m_2 \right) u_1 + 2 m_2 u_2}{m_1 + m_2}
\]

$$\bbox[15px, #e4e4e4, border: 2px solid #000000]{v_2 = \frac{2 m_1 u_1}{m_1 + m_2} + \frac{\left( m_2 – m_1 \right) u_2}{m_1 + m_2}} \tag{11}$$

Remember:  The expression for $v_2$ can be obtained from (10), the expression for $v_1$, by replacing $m_1$ by $m_2$ and $u_1$ by $u_2$. The reverse is also true, i.e., (10) can also be obtained from (11).

Particular Cases

1. When masses of two bodies are equal, i.e.,

$${{m}_{{1\text{ }}}}=\text{ }{{m}_{2}}\text{ }=\text{ }m$$, say

From (10),

$$ {{v}_{1}}\text{ }=\text{ }\frac{{2\text{ }m\text{ }{{u}_{2}}}}{{2m}}=\text{ }{{u}_{2}}$$

i.e., velocity of A after collision = velocity of B before collision.

From (11),

$$ {{v}_{2}}\text{ }=\text{ }\frac{{2\text{ }{{m}_{1}}\text{ }{{u}_{1}}}}{{2{{m}_{1}}}}=\text{ }{{u}_{1}}$$

i.e., velocity of B after collision = velocity of A before collision.

Hence, when two bodies of equal masses undergo elastic collision in one dimension, their velocities are just interchanged.

 

 

This principle is particularly significant in the operation of nuclear reactors. In these reactors, fast-moving neutrons are decelerated by causing them to collide with the nuclei of a moderating substance. Since hydrogen nuclei (protons) have a similar mass to neutrons, when these collisions occur, the neutrons lose energy and slow down, transferring some of their kinetic energy to the protons. As a result, materials rich in hydrogen, such as heavy water or paraffin, are often used as effective moderators in nuclear reactors to achieve this slowing effect.

2. When the target body B is initially at rest

i.e., $u_2 = 0$

From (10),

\[
v_1 = \frac{(m_1 – m_2) u_1}{m_1 + m_2} \tag{12}
\]

From (11),

\[
v_2 = \frac{2 m_1 u_1}{m_1 + m_2} \tag{13}
\]

This leads to three cases further:

(a) When masses of two bodies are equal

i.e., $m_1 = m_2$

From (12),

\[
v_1 = 0
\]

From (13),

\[
v_2 = u_1
\]

i.e., body A comes to rest and body B starts moving with the initial velocity of A. Obviously, in such a collision, 100% kinetic energy of A is transferred to the body B. This is shown in the above figure.

(b) When body B at rest is very heavy

i.e., $m_2 \gg m_1$, i.e., $m_1$ can be ignored compared to $m_2$.

Putting $m_1 = 0$ in (12) and (13), we obtain:

\[
v_1 = -u_1
\]

\[
v_2 = 0
\]

Hence, when a light body A collides against a heavy body B at rest, A rebounds with its own velocity, and B continues to be at rest. This is what happens when a ball rebounds to the same height from which it was thrown, on striking a floor.

(c) When body B at rest has negligible mass

i.e., $m_2 \ll m_1$, i.e., $m_2$ can be ignored compared to $m_1$.

Putting $m_2 = 0$ in (12) and (13), we get:

\[
v_1 = u_1
\]

\[
v_2 = 2u_1
\]

Hence, when a heavy body A undergoes an elastic collision with a light body B at rest, the body A keeps on moving with the same velocity of its own and the body B starts moving with double the initial velocity of A.

Remember: A head-on elastic collision between two balls moving in opposite directions is shown in the above figure.

---

Example 1: A body of mass 2 kg makes an elastic collision with another body at rest and continues to move in the original direction with a speed equal to one third of its original speed. Find the mass of the second body.

Solution:

Here,

$m_1 = 2 \, \text{kg}, v_2 = 0, v_1 = \frac{u_1}{3}, m_2 = ?$

In a perfectly elastic collision in one dimension:

\[
v_1 = \frac{(m_1 – m_2) u_1}{m_1 + m_2} + \frac{2 m_2 u_2}{m_1 + m_2}
\]

\[
\frac{u_1}{3} = \frac{(2 – m_2) u_1}{2 + m_2}
\]

\[
\frac{1}{3} = \frac{2 – m_2}{2 + m_2}
\]

\[
2 + m_2 = 6 – 3 m_2
\]

\[
4 m_2 = 4 \, \Rightarrow m_2 = 1 \, \text{kg}.
\]

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Elastic Collision in Two Dimensions or Oblique Collision

Elastic collision in two dimensions occurs when two bodies initially traveling along same straight line collides without any loss of kinetic energy and after collision move along different directions in a plane.

Let’s  consider that there are two masses $ \displaystyle {{m}_{1}}$ and $ \displaystyle {{m}_{2}}$ of bodies $ \displaystyle A$ $ \displaystyle B$ moving initially along the X-axis with velocities $ \displaystyle {{u}_{1}}$ and $ \displaystyle {{u}_{2}}$ respectively.

When the condition $u_1 > u_2$ holds, the two bodies come into contact and collide. After the collision, body A moves with a velocity $v_1$ at an angle $\theta$ with respect to the X-axis, while body B moves with velocity $v_2$ at an angle $\phi$ with the X-axis, as depicted in figure below.

 

 

Since this is an elastic collision, the total kinetic energy remains conserved.

Total kinetic energy after the collision is equal to the total kinetic energy before the collision i.e.

$\displaystyle \frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}$ $\displaystyle =\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}\ldots \left( {\text{14}} \right)$

\[
\Rightarrow m_1 v_1^2 + m_2 v_2^2 = m_1 u_1^2 + m_2 u_2^2 \tag{15}
\]

In an elastic collision, momentum is also conserved. Along the X-axis, this means that the total linear momentum before and after the collision remains the same:

$ \displaystyle {{m}_{1}}{{v}_{1}}\cos \theta +{{m}_{2}}{{v}_{2}}\cos \phi $ $ \displaystyle ={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\ldots \left( {\text{16}} \right)$

Since both bodies are initially moving along the X-axis, their initial linear momentum along the Y-axis is zero. After the collision, the total Y-axis momentum becomes:

\[
m_1 v_1 \sin \theta – m_2 v_2 \sin \phi = 0 \tag{17}
\]

With the three equations (15), (16), and (17), there are four unknowns to calculate: $v_1$, $v_2$, $\theta$, and $\phi$. Since solving all four unknowns isn’t feasible with just these three equations, we need to measure at least one parameter experimentally — such as the final velocities ($v_1$, $v_2$) or the angles ($\theta$, $\phi$) of bodies A and B after the collision.