Elastic And Inelastic Collision

A collision is a fundamental aspect of physical interaction, existing in two primary forms: elastic and inelastic. In elastic collisions, both momentum and kinetic energy are conserved, such as with billiard balls or a rubber ball bouncing on a hard surface. These collisions idealize interactions without permanent deformation or heat generation. In contrast, inelastic collisions involve changes in the momentum of the objects post-collision.

This Story also Contains

1. Collision
2. In Perfectly Elastic Collision
3. Special Cases of Head-on Elastic Collision
4. Perfectly Elastic Oblique Collision
5. Head-on Inelastic Collision
6. Perfectly Inelastic collision
7. Collision Between Bullet and Vertically Suspended Block
8. Solved Example Based on Elastic And Inelastic Collision
9. Summary

This article covers Elastic and Inelastic Collisions, including Perfectly Elastic Head-on and Oblique Collisions, Head-on Inelastic Collisions, Perfectly Inelastic Collisions, and Collisions Between a Bullet and a Vertically Suspended Block. Which belongs to the chapter work, energy, and power, which is an important chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than ten questions have been asked on this concept. And for NEET three questions were asked from this concept.

Collision

The interaction between two or more objects is called a collision. And during this interaction strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change.

Stages of Collision

There are three distinct identifiable stages in collision

1. Before the collision-

The interaction forces are zero

2. During the collision-

The interaction forces are very large and these forces act for a very short time. And because of these interaction forces the energy and momentum of the interacting particle change.

3. After the collision-

The interaction forces are zero

In Perfectly Elastic Collision

The law of conservation of momentum and that of Kinetic Energy hold good.

$ \begin{aligned}
& \frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 .. (1) \\
& m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 ..... (2) \\
& \quad m_1, m_2: \text { masses } \\
& \quad u_1, v_1 \text { : initial and final velocity of the mass } m_1 \\
& \quad u_2, v_2 \text { : initial and final velocity of the mass } m_2
\end{aligned}$

From equation (1) and (2)

$ \text { We get, } u_1-u_2=v_2-v_1$ …..(3)

Or, we can say Relative velocity of approach = Relative velocity of separation

And

$ e=\frac{v_2-v_1}{u_1-u_2}=\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}$

So in Perfectly Elastic Collision

e = 1

From equations (1),(2), (3)

We get

$ v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\frac{2 m_2 u_2}{m_1+m_2 } .......(4) $

Similarly,

$ v_2=\left(\frac{m_2-m_1}{m_1+m_2}\right) u_2+\frac{2 m_1 u_1}{m_1+m_2 }........(5)$

Now, let's understand this concept better from the example given below:

Example: A body of mass $m_1$ moving with an unknown velocity of $v_1 \hat{i}$, undergoes a collinear collision with a body of mass $m_2$ moving with a velocity $v_2 \hat{i}$. After collision $m_1$ and $m_2$ move with velocities of $v_3 \hat{i}$ and $v_4 \hat{i}$, respectively. If $m_2=0.5 m_1$ and $v_3=0.5 v_1$ the $v_{1 \text { is : }}$

1) $v_4-\frac{v_2}{2}$
2) $v_4-v_2$
3) $v_4-\frac{v_2}{4}$
4) $v_4+v_2$

Solution:

Perfectly Elastic Collision -

The law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$
\begin{aligned}
& \frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \\
& m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \\
& m_1, m_2: \text { masses }
\end{aligned}
$
$u_1, v_1$ : initial and final velocity of the mass $m_1$
$u_2, v_2$ : initial and final velocity of the mass $m_2$

before collision:

after collision:

$\begin{aligned}
& \left(P_{\text {initial }}\right)_{\text {total }}=\left(P_{\text {final }}\right)_{\text {total }} \\
& m_1 v_1 \hat{i}+m_2 v_2 \hat{i}=m_1 v_3 \hat{i}+m_2 v_4 \hat{i} \\
& m_2=0.5 m_1 \& v_3=0.5 v_1 \\
& \hat{i}\left(m_1 v_1+0.5 m_1 v_2\right)=\hat{i}\left(m_1 v_3+0.5 m_1 v_4\right) \\
& v_1+0.5 v_2=v_3+0.5 v_4 \\
& v_1+0.5 v_2=0.5 v_1+0.5 v_4 \quad\left(v_3=0.5 v_1\right) \\
& 0.5 v_1+0.5 v_2=0.5 v_4 \\
& v_1+v_2=v_4 \\
& v_1=v_4-v_2
\end{aligned}$
Hence, the answer is option (2).

Special Cases of Head-on Elastic Collision

1. Equal mass in case of perfectly elastic collision

Then, $v_1=u_2 \text { and } v_2=u_1$

Or, Velocity mutually interchange

2. If a massive projectile collides with a light target $\text { (i.e } m_1>>m_2 \text { ) }$

Since $m_1>>m_2$ so we use $m_2=0$
Putting $m_2=0$ in equation (4) and (5)
We get $v_1=u_1$ and $v_2=2 u_1-u_2$

3. If the target particle is massive in case of elastic collision $\text { (i.e; } m_2>>m_1 \text { ) }$

$\text { Since } m_2>>m_1$

So, the lighter particle recoils with the same speed and the massive target particle remains practically at rest.

i.e; $\bar{v}_2=\bar{u}_2$
$
\bar{v}_1=-\bar{u}_1
$

Perfectly Elastic Oblique Collision

Let two bodies move as shown in the figure.

By the law of conservation of momentum

Along x-axis-

$
m_1 u_1+m_2 u_2=m_1 v_1 \cos \theta+m_2 v_2 \cos \phi ...... (1)
$

Along y-axis-
$
0=m_1 v_1 \sin \theta-m_2 v_2 \sin \phi ............ (2)
$

By the law of conservation of kinetic energy
$
\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 ........(3)
$

In Perfectly Elastic Oblique Collision

Value of e=1

So along the line of impact (here along in the direction of $v_2$)

We apply e=1

And we get $e=1=\frac{v_2-v_1 \cos (\theta+\phi)}{u_1 \cos \phi-u_2 \cos \phi} $ ….. (4)

So we solve these equations (1),(2),(3),(4) to get unknown.

Recommended Topic Video

Special Condition

if $m_1=m_2$ and $u_2=0$
Then, from equation (1), (2) and (3)
We get, $\theta+\phi=\frac{\pi}{2}$

i.e.; after perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle $\theta+\phi$ would be $90^{\circ}$.

Head-on Inelastic Collision

In Inelastic Collision the Law of conservation of momentum holds good but kinetic energy is not conserved.

$
\begin{aligned}
& \frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2 \neq \frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \\
& m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 ........(1)\\
& m_1, m_2: \text { masses }
\end{aligned}
$
$u_1, v_1$ : initial and final velocities of mass $m_1$
$u_2, v_2$ : initial and final velocities of mass $m_2$

In Inelastic Collision (0 < e < 1)

$e=\frac{v_2-v_1}{u_1-u_2}$ ….. (2)

From equations (1),(2)

We get,

$
v_1=\left(\frac{m_1-e m_2}{m_1+m_2}\right) u_1+\frac{(1+e) m_2}{m_1+m_2} u_2 ......(3)
$

Similarly,
$
v_2=\left(\frac{m_2-e m_1}{m_1+m_2}\right) u_2+\frac{(1+e) m_1}{m_1+m_2} u_1 .........(4)
$

Special Case

A sphere of mass m moving with velocity u hits inelastically with another stationary sphere of same mass.

As, $e=\frac{v_2-v_1}{u_1-u_2}$
So, $e=\frac{v_2-v_1}{u}$
or,
$
u e=v_2-v_1
$ ....(5)

By conservation of momentum

As, $m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}$

So $v_2+v_1 = u$ ….(6)

From equation (5) and (6)

$\text { We get, } \frac{v_1}{v_2}=\frac{1-e}{1+e}$

Loss in kinetic Energy

Loss in K.E = Total initial kinetic energy – Total final kinetic energy

$\Delta K . E .=\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\left(\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\right)$ …. (7)

From equation (3), (4) and (7)

We can write, Loss in kinetic energy in terms of e as

$\triangle K . E .=\frac{1}{2}\left(\frac{m_1 m_2}{m_1+m_2}\right)\left(1-e^2\right)\left(u_1-u_2\right)^2$

Perfectly Inelastic collision

In a perfectly inelastic collision, two bodies stick together after the collision, so there will be a final common velocity (v).

1. When the colliding bodies are moving in the same direction

By the law of conservation of momentum

$\begin{aligned}
& m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\
& v=\frac{m_1 u_1+m_2 u_2}{\left(m_1+m_2\right)}
\end{aligned}$

Loss in kinetic energy

$ \begin{aligned}
& \Delta K \cdot E=\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\left(\frac{1}{2}\left(m_1+m_2\right) V^2\right) \\
& \Delta K \cdot E=\frac{1}{2}\left(\frac{m_1 m_2}{m_1+m_2}\right)\left(u_1-u_2\right)^2
\end{aligned}$

2. When the colliding bodies are moving in the opposite direction

By the law of conservation of momentum

$\begin{aligned}
& m_1 u_1+m_2\left(-u_2\right)=\left(m_1+m_2\right) v \\
& v=\frac{m_1 u_1-m_2 u_2}{m_1+m_2}
\end{aligned}$

If v is positive then the combined body will move along the direction of motion of mass $m_1$

If v is negative then the combined body will move in a direction opposite to the motion of mass $m_1$

Loss in kinetic energy

$ \begin{aligned}
& \Delta K \cdot E=\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\left(\frac{1}{2}\left(m_1+m_2\right) V^2\right) \\
& \Delta K \cdot E=\frac{1}{2}\left(\frac{m_1 m_2}{m_1+m_2}\right)\left(u_1+u_2\right)^2
\end{aligned}$

Understand this concept better from the solved example given below:

Example: Two particles of equal mass $\mathrm{m}$ have respective initial velocities $u \hat{i}$ and $u\left(\frac{\hat{i}+\hat{j}}{2}\right)$ They collide completely inelastically. The energy lost in the process is:

1) $\frac{3}{8} m u^2$
2) $\sqrt{\frac{2}{3}} m u^2$
3) $\frac{1}{3} m u^2$
4) $\frac{1}{8} m u^2$

Solution:

Conserving Momentum-

$m \cdot u \hat{i}+m\left(\frac{u}{2} \hat{i}+\frac{u}{2} \hat{j}\right)=2 m\left(u_1 \hat{i}+u_2 \hat{j}\right)$

On solving

$
u_1=\frac{3 u}{4} \text { and } u_1=\frac{u}{4}
$

Change in K.E
$
\begin{aligned}
& {\left[\frac{1}{2} m u^2+\frac{1}{2} m\left(\frac{u}{2} \sqrt{2}\right)^2\right]-\left[\frac{1}{2}(2 m)\left(\frac{9 u^2}{16}+\frac{u^2}{16}\right)\right]} \\
& =\frac{3 m u^2}{4}-\frac{5 m u^2}{8}=\frac{m u^2}{8}
\end{aligned}
$

Hence, the answer is the option (4).

Collision Between Bullet and Vertically Suspended Block

A block of mass M is suspended by a vertical thread.

A bullet of mass m is fired horizontally with velocity u in the block.

After the collision bullet gets embedded in the block. And, the combined system is raised up to height h where the string makes an angle $\theta$ with the vertical.

The Common Velocity of the System Just After the Collision (V)

Here, the system is (block + bullet)

$\mathrm{P}=$ momentum
$
\begin{aligned}
& P_{\text {bullet }}+P_{\text {block }}=P_{\text {system }} \\
& m u+0=(m+M) V \\
& V=\frac{m u}{m+M}
\end{aligned}
$ .....(1)

The Initial Velocity of the Bullet in Terms of h

By the conservation of mechanical energy

(T.E of system ) Just after collision $=(T . E$ of system $)$ At height $h$
$
\begin{aligned}
& \frac{1}{2}(m+M) V^2=(m+M) g h \\
& V=\sqrt{2 g h}
\end{aligned}
$

Equating (1) and (2)
We get $V=\sqrt{2 g h}=\frac{m u}{m+M}$
$
u=\left(\frac{m+M}{m}\right) \sqrt{2 g h}
$

Loss in Kinetic Energy

Loss of kinetic energy in a perfectly inelastic collision when given by

$\Delta K \cdot E=\frac{1}{2}\left(\frac{m M}{m+M}\right) u^2$

Value of angle $\theta$

From $u=\left(\frac{m+M}{m}\right) \sqrt{2 g h}$
We can write
$
h=\left(\frac{u^2}{2 g}\right)\left(\frac{m}{m+M}\right)^2
$

And from figure
$O r, \cos \theta=\frac{L-h}{L}=1-\frac{h}{L}=1-\left(\left(\frac{u^2}{2 g L}\right)\left(\frac{m}{m+M}\right)^2\right)$

Solved Example Based on Elastic And Inelastic Collision

Example 1: A bullet of mass m moving with a velocity v strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height h then the initial velocity of the bullet is:

1) $\frac{m}{m-M} \sqrt{2 g h}$
2) $\frac{m}{m+M} \sqrt{2 g h}$
3) $\frac{m+M}{m} \sqrt{2 g h}$
4) $\frac{m-M}{m} \sqrt{2 g h}$

Solution:

The common velocity of the system just after the collision (V)

Here, the system is (block + bullet)

$ \begin{aligned}
& \mathrm{P}=\text { momentum } \\
& P_{\text {bullet }}+P_{\text {block }}=P_{\text {system }} \\
& m u+0=(m+M) V \\
& V=\frac{m u}{m+M}
\end{aligned} $ .....(1)

And, the initial velocity of the bullet in terms of h. By the conservation of mechanical energy

(T.E of system ) Just after collision $=(T . E$ of system) At height $h$
$
\begin{aligned}
& \frac{1}{2}(m+M) V^2=(m+M) g h \\
& V=\sqrt{2 g h}
\end{aligned}
$

Equating (1) and (2)
We get $V=\sqrt{2 g h}=\frac{m u}{m+M}$
$
u=\left(\frac{m+M}{m}\right) \sqrt{2 g h}
$

So, the answer is -

$\frac{m+M}{m} \sqrt{2 g h}$

Example 2: A ball of mass ' $m$ ' moving with velocity ' $v$ ', collides inelastically with another identical ball. After the collision, the 1st ball moves with velocity $v / \sqrt{2}$ in a direction perpendicular to the initial direction of motion. Find the speed of the second ball after the collision:

1) $\frac{\sqrt{3}}{2} v$
2) $\frac{1}{2} v$
3) $\sqrt{\frac{3}{2}} v$
4) $\sqrt{2} v$

Solution:

Momentum conservation along X-axis

$
\begin{aligned}
& m v=m v_1 \cos \theta \\
& v=v_1 \cos \theta \ldots(1)
\end{aligned}
$
along $y$-axis
$
\begin{aligned}
& 0=m v / \sqrt{2}-m v_1 \sin \theta \\
& \frac{v}{\sqrt{2}}=v_1 \sin \theta \ldots(2) \\
& v^2+\frac{v^2}{2}=v_1^2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
& v_1^2=\frac{3 v^2}{2} \Rightarrow v_1=\sqrt{\frac{3}{2}} v
\end{aligned}
$

Hence, the answer is option (3).

Example 3: Two particles A and B of equal mass M are moving with the same speed v as shown in the figure. They collide completely inelastically and move as a single particle C. The angle θ that the path of C makes with the X-axis is given by :

1) $\tan \Theta=\frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}$
2) $\tan \Theta=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}$
$\tan \Theta=\frac{1-\sqrt{2}}{\sqrt{2}(1+\sqrt{3})}$
4) $\tan \Theta=\frac{1-\sqrt{3}}{1+\sqrt{2}}$

Solution:

As we know,

Inelastic Collision -

The law of conservation of momentum holds good but kinetic energy is not conserved

i.e

$
\begin{aligned}
& \frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2 \neq \frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \\
& \text { but } m_1 u_1+m_2 u_2=m_1 v_1+m_2
\end{aligned}
$
where
$
m_1, m_2: \text { masses }
$
$u_1, v_1$ : initional and final velocities of mass $m_1$
$u_2, v_2$ : initional and final velocities of mass $m_2$

So apply the Law of conservation of momentum

(i) Along x-direction

$m v \sin 30^{\circ}-m v \sin 45^{\circ}=2 m v^{\prime} \cos \theta \Rightarrow v^{\prime} \cos \theta=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)$

(ii) Along y-direction

$m v \cos 30^{\circ}+m v \cos 45^{\circ}=2 m v^{\prime} \sin \theta \Rightarrow v^{\prime} \sin \theta=\frac{1}{2}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\right)$

Divide (i) with (ii):

$\tan \theta=\frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}$

Hence, the answer is option (1).

Summary

Grasping the basics of elastic and inelastic collisions is key to understanding how objects interact and transfer energy. Real-world examples, like billiard games, often illustrate these principles. In inelastic collisions, objects can lose a lot of energy or stick together. Knowing these concepts is vital for explaining how things behave when they collide.