Elastic And Inelastic Collision - Practice Questions & MCQ

• In Perfectly Elastic Collision, 

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 \\
& m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \\
& \quad 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 $m_2$
From equation (1) and (2)
We get, $u_1-u_2=v_2-v_1$
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

Similarly,

$
\begin{aligned}
& 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} \\
& 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}
\end{aligned}
$

- Special cases of head on elastic collision
1. Equal mass in case of perfectly elastic collision

Then, $v_1=u_2$ and $v_2=u_1$

Or, Velocity mutually interchange
2. If massive projectile collides with a light target (i.e $m_1>>m_2$ )

Since $m_1 \gg>m_2$ 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 target particle is massive in case of elastic collision (i.e; $m_2>>m_1$ )

Since $m_2 \gg>m_1$
So, the lighter particle recoil with same speed and the massive target particle remain practically at rest.
i.e; $\bar{v}_2=\bar{u}_2$

$
\bar{v}_1=-\bar{u}_1
$
 

• Let two bodies moving as shown in figure.

By 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
$

Along $y$-axis-

$
0=m_1 v_1 \sin \theta-m_2 v_2 \sin \phi
$

By 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
$

And In Perfectly Elastic Oblique Collision
Value of $e=1$
So along line of impact (here along in the direction of $v_2$ )
We apply e=1
And we get $e=1=\frac{v_2-v_1 \cos (\theta+\phi)}{u_1 \cos \phi-u_2 \cos \phi}$
So we solve these equations (1),(2),(3),(4) to get unknown.
- 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 $90^{\circ}$.

1. In Inelastic Collision Law of conservation of momentum hold 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 \\
& 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 $m_2$
2. In inelastic collision ( $0<e<1$ )

$
e=\frac{v_2-v_1}{u_1-u_2}
$

From equations (1),(2)
We get,

$
\begin{aligned}
& 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 \\
& 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
\end{aligned}
$
 

3.  Special case

A sphere of mass $m$ moving with velocity $u$ hits inelastically with another stationary sphere of same mass.
$\mathrm{As}_{\text {, }} e=\frac{v_2-v_1}{u_1-u_2}$

So,

$
e=\frac{v_2-v_1}{u} \quad \text { or, } \quad u e=v_2-v_1
$

By conservation of momentum
As, $\quad m_1 u_1+m_2 u_2=m_1 v_1+m_2$
So $v_2+v_1=u$
From equation (5) and (6)
We get, $\frac{v_1}{v_2}=\frac{1-e}{1+e}$
4. Loss in kinetic energy

Loss in K.E $=$ Total initial kinetic energy - Total final kinetic energy

$
\triangle 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)
$

From equation (3), (4) and (7)
We can write, Loss in kinetic energy in terms of $\mathbf{e}$ 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
$
 

In 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 . 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 . 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

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

A block of mass M suspended by vertical thread.

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

Let after the collision bullet gets embedded in block.

And, the combined system raised upto height h where the string makes an angle   with the vertical. 

1. Common velocity of system just after the collision (V)

Here, 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}
$

2. 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)

$
\begin{aligned}
& \text { We get } V=\sqrt{2 g h}=\frac{m u}{m+M} \\
& u=\left(\frac{m+M}{m}\right) \sqrt{2 g h}
\end{aligned}
$

3. Loss in kinetic energy

Loss of kinetic energy in perfectly inelastic collision when $\left(u_2=0, u_1=0\right)$ is given by

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

4.  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)
   $