Elastic And Inelastic Collision MCQ - Practice Questions with Answers

• In Perfectly Elastic Collision, Í

Law of conservation of momentum and that of Kinetic Energy hold good.

$$\begin{array}{rl}& \frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ & m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ & m_1,m_2:\text{ masses }\end{array}$$

$u_1,v_1$ : 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

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

So in Perfectly Elastic Collision  

                                                                    e = 1, 

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

                     We get

$$\begin{array}{rl}& v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\frac{2m_2{u}_2}{m_1+m_2}\dots \dots \\ & v_2=\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2+\frac{2m_1{u}_1}{m_1+m_2}\end{array}$$

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 $m_1\gg m_{2\text{ ) }}$

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 $2u_1-u_2$
3. If target particle is massive in case of elastic collision (i.e; $m_2\gg m_{1\text{ ) }}$

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; ${\overline{v}}_2={\overline{u}}_2$

$${\overline{v}}_1=-{\overline{u}}_1$$
 

• Let two bodies moving as shown in figure.

By the law of conservation of momentum

Along $x$-axis-

$$m_1{u}_1+m_2{u}_2=m_1{v}_1\cos \theta +m_2{v}_2\cos \varphi$$

Along $y$-axis-

$$0=m_1{v}_1\sin \theta -m_2{v}_2\sin \varphi$$

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 $e=1=\frac{v_2-v_1\cos (\theta +\varphi )}{u_1\cos \varphi -u_2\cos \varphi }$
So we solve these equations (1),(2),(3),(4) to get unknown.

• Special condition

$$\text{ if }{m}_1=m_2\text{ and }{u}_2=0$$

Then, from equation (1), (2) and (3)
We get, $\theta +\varphi =\frac{\pi }{2}$
i.e; after perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle $\theta +\varphi$ would be ${90}^{\circ }$.

1. In Inelastic Collision Law of conservation of momentum hold good but kinetic energy is not conserved .

$\begin{array}{rl}& \frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2\ne \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ & m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ & m_1,m_2:\text{ masses }\\ & u_1,v_1:\text{ initial and final velocities of mass }{m}_1\\ & u_2,v_2:\text{ initial and final velocities of mass }{m}_2\end{array}$

2. In inelastic collision  (0 < e < 1)

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

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

Similarly,

$$\begin{array}{rl}{v}_1& =\left(\frac{m_1-e{m}_2}{m_1+m_2}\right)u_1+\frac{(1+e)m_2}{m_1+m_2}{u}_2\\ v_2& =\left(\frac{m_2-e{m}_1}{m_1+m_2}\right)u_2+\frac{(1+e)m_1}{m_1+m_2}{u}_1\end{array}$$
 

3.  Special case

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

As. $e=\frac{v_2-v_1}{u_1-u_2}$
So, $e=\frac{v_2-v_1}{u}$ or, $ue=v_2-v_1$
By conservation of momentum
As, $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2$
So $v_2+v_1=u$
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

$$\mathrm{\Delta }K.E.=(\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2)-(\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2)$$

From equation (3) (4) and (7)
We can write, Loss in kinetic energy in terms of e as

$$\mathrm{△}K.E.=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right)(1-e^2){(u_1-u_2)}^2$$
 

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 

- By the law of conservation of momentum

$$\begin{array}{rl}& m_1{u}_1+m_2{u}_2=(m_1+m_2)v\\ & v=\frac{m_1{u}_1+m_2{u}_2}{(m_1+m_2)}\end{array}$$

- Loss in kinetic energy

$$\begin{array}{rl}& \mathrm{\Delta }K\cdot E=(\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2)-\left(\frac{1}{2}(m_1+m_2)V^2\right)\\ & \mathrm{\Delta }K\cdot E=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right){(u_1-u_2)}^2\end{array}$$

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

• - By the law of conservation of momentum

  $$\begin{array}{rl}& m_1{u}_1+m_2(-u_2)=(m_1+m_2)v\\ & v=\frac{m_1{u}_1-m_2{u}_2}{m_1+m_2}\end{array}$$

  
  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

  $$\begin{array}{rl}& \mathrm{\Delta }K\cdot E=(\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2)-\left(\frac{1}{2}(m_1+m_2)V^2\right)\\ & \mathrm{\Delta }K\cdot E=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right){(u_1+u_2)}^2\end{array}$$
   

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 the block.

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

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

Here, system is (block + bullet) 

$$\mathrm{P}=\text{ momentum }$$

$$\begin{array}{rl}& P_{\text{bullet }}+P_{\text{block }}=P_{\text{system }}\\ & mu+0=(m+M)V\\ & V=\frac{mu}{m+M}\end{array}$$
 

                         .....(1)

2. Initial velocity of the bullet in terms of h

By the conservation of mechanical energy

(T.E of system ) Just after collision $=(T\cdot E$ of system $)$ At height

$$\begin{array}{rl}& \frac{1}{2}(m+M)V^2=(m+M)gh\\ & V=\sqrt{2gh}\end{array}$$

Equating (1) and (2)
We get $V=\sqrt{2gh}=\frac{mu}{m+M}$

$$u=\left(\frac{m+M}{m}\right)\sqrt{2gh}$$
 

3. Loss in kinetic energy

Loss of kinetic energy in a perfectly inelastic collision when $(u_2=0)$ is given by

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

4. Value of angle $\theta$

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

$$h=\left(\frac{u^2}{2g}\right){\left(\frac{m}{m+M}\right)}^2$$

And from figure

$$\text{ Or, }\cos \theta =\frac{L-h}{L}=1-\frac{h}{L}=1-\left(\left(\frac{u^2}{2gL}\right){\left(\frac{m}{m+M}\right)}^2\right)$$