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Elastic And Inelastic Collision MCQ - Practice Questions with Answers

Elastic And Inelastic Collision MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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Quick Facts

Perfectly elastic oblique collision is considered one the most difficult concept.
Perfectly Elastic Head on Collision, Head on inelastic collision, Perfectly inelastic collision, Collision Between Bullet and Vertically Suspended Block is considered one of the most asked concept.
34 Questions around this concept.

Solve by difficulty

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss (in Joule) during the collision is :

A

0.34 Joule

Correct Answer

Wrong Answer

B

0.16 Joule

Correct Answer

Wrong Answer

C

1 Joule

Correct Answer

Wrong Answer

D

0.67 Joule

Correct Answer

Wrong Answer

Two particles of masses m₁, m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to a higher level, after absorbing energy (E). If the final velocities of particles are v₁ and v₂ then we must have:

A

$m_1^{2}u_1 +m_2^{2}u_2 - E = m_1^{2}v_1 +m_2^{2}v_2$

Correct Answer

Wrong Answer

B

$\frac{1}{2}m_1u_1^{2}+ \frac{1}{2}m_2u_2^{2}= \frac{1}{2}m_1v_1^{2}+ \frac{1}{2}m_2v_2^{2}-E$

Correct Answer

Wrong Answer

C

$\frac{1}{2}m_1u_1^{2}+ \frac{1}{2}m_2u_2^{2}-E =\frac{1}{2}m_1v_1^{2}+ \frac{1}{2}m_2v_2^{2}$

Correct Answer

Wrong Answer

D

$\frac{1}{2}m_1u_1^{2}+ \frac{1}{2}m_2u_2^{2}+E =\frac{1}{2}m_1v_1^{2}+ \frac{1}{2}m_2v_2^{2}$

Correct Answer

Wrong Answer

A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length L (L>>r) and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed $v$ , the average force experienced by each support after a long time is (assume all collisions are elastic) :

A

$\frac{mv^{2}}{L-nr}$

Correct Answer

Wrong Answer

B

$\frac{mv^{2}}{L-2nr}$

Correct Answer

Wrong Answer

C

$\frac{mv^{2}}{2\left ( L-nr \right )}$

Correct Answer

Wrong Answer

D

zero

Correct Answer

Wrong Answer

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Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be:

A

-0.3 m/s and 0.5 m/s

Correct Answer

Wrong Answer

B

-0.5 m/s and 0.3 m/s

Correct Answer

Wrong Answer

C

0.3 m/s and 0.5 m/s

Correct Answer

Wrong Answer

D

0.5 m/s and -0.3 m/s

Correct Answer

Wrong Answer

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 :

A

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

Correct Answer

Wrong Answer

B

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

Correct Answer

Wrong Answer

C

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

Correct Answer

Wrong Answer

D

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

Correct Answer

Wrong Answer

The question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.

Statement-2: The principle of conservation of momentum holds true for all kinds of collisions.

A

Statement-1 is true, Statement-2 is false.

Correct Answer

Wrong Answer

B

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Correct Answer

Wrong Answer

C

Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

Correct Answer

Wrong Answer

D

Statement-1 is false, Statement-2 is true.

Correct Answer

Wrong Answer

A mass m moves with a velocity $\nu$ and collides inelastically with another identical mass .After collision the first mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion.Find the speed of the 2^nd mass after collision

A

$\frac{2}{\sqrt{3}}v$

Correct Answer

Wrong Answer

B

$\frac{v}{\sqrt{3}}$

Correct Answer

Wrong Answer

C

$v$

Correct Answer

Wrong Answer

D

$3v$

Correct Answer

Wrong Answer

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Concepts Covered - 5

Perfectly Elastic Head on Collision

In Perfectly Elastic Collision,

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

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$ …….(1)

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$ …….(2)

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

From equation (1) and (2)

We get, $u_1-u_2 = v_2 - v_1$ …..(3)

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

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

So in Perfectly Elastic Collision

e = 1 ,

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

We get

$v_1 = (\frac{m_1-m_2}{m_1+m_2})u_1 + \frac{2m_2u_2}{m_1+m_2}$ ……(4)

Similarly, $v_2 = (\frac{m_2-m_1}{m_1+m_2})u_2 + \frac{2m_1u_1}{m_1+m_2}$ ...... (5)

Special cases of head on elastic collision

Equal mass in case of perfectly elastic collision

Then, $v_{1}= u_{2} \: and \: v_{2}= u_{1}$

Or, Velocity mutually interchange

If massive projectile collides with a light target (i.e $m_1>>m_2$ )

Since $m_1>>m_2$ so we use $m_2=0$

Putting $m_2=0$ in equation (4) and (5)

We get $v_1 = u_1\ and\ 2u_1-u_2$

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

Since $m_2>>m_1$

So, the lighter particle recoil with same speed and the massive target particle remain practically at rest.

i.e; $\bar{v}_{2}=\bar u_2$

$\bar{v}_{1}=-\bar{u}_{1}$

Perfectly elastic oblique collision

Let two bodies moving as shown in figure.

By law of conservation of momentum

Along x-axis-

$m_1u_1+m_2u_2 = m_1v_1cos\theta+m_2v_2cos\phi$ ….. (1)

Along y-axis-

$0 = m_1v_1sin\theta-m_2v_2sin\phi$ …..(2)

By law of conservation of kinetic energy

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$ ….(3)

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 $e=1=\frac{v_2-v_1cos(\theta+\phi)}{u_1cos\phi-u_2cos\phi}$ ….. (4)

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 $90^{0}$ .

Head on inelastic collision

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

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}\neq \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}v_2$ …… (1)

$m_{1},m_{2}: masses$

$u_{1},v_{1}: initial \: and \: final\: velocities\: of \: mass \: m_{1}$

$u_{2},v_{2}: initial \: and \: final\: velocities\: of \: mass \: m_{2}$

In inelastic collision (0 < e < 1)

$e = \frac{v_2-v_1}{u_1-u_2}$ ….. (2)

From equations (1),(2)

We get,

$v_1 = (\frac{m_1-em_2}{m_1+m_2})u_1+\frac{(1+e)m_2}{m_1+m_2}u_2$ ……(3)

Similarly, $v_2 = (\frac{m_2-em_1}{m_1+m_2})u_2+\frac{(1+e)m_1}{m_1+m_2}u_1$ ...... (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, $ue = v_2-v_1$ ....(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)

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

$\bigtriangleup K.E. = (\frac{1}{2}m_1u^2_1+\frac{1}{2}m_2u^2_2)-(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2)$ …. (7)

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

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

$\bigtriangleup K.E. = \frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(1-e^2)(u_1-u_2)^2$

Perfectly inelastic collision

In Perfectly Inelastic Collision - Two bodies stick together after the collision ,so there will be a final common velocity (v)

When the colliding bodies are moving in the same direction

By the law of conservation of momentum

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

Loss in kinetic energy

$\begin{aligned} \Delta K . E &=\left(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\right)-\left(\frac{1}{2}\left(m_{1}+m_{2}\right) V^{2}\right) \\ \Delta K . E &=\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\left(u_{1}-u_{2}\right)^{2} \end{aligned}$

When the colliding bodies are moving in the opposite direction

By the law of conservation of momentum

$\\ {m_{1} u_{1}+m_{2}\left(-u_{2}\right)=\left(m_{1}+m_{2}\right) v} \\ \\ {v=\frac{m_{1} u_{1}-m_{2} u_{2}}{m_{1}+m_{2}}}$

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

$\\ {\Delta K . E=\left(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\right)-\left(\frac{1}{2}\left(m_{1}+m_{2}\right) V^{2}\right)} \\ \\ {\Delta K . E=\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\left(u_{1}+u_{2}\right)^{2}}$

Collision Between Bullet and Vertically Suspended Block

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 $\theta$ with the vertical.

Common velocity of system just after the collision (V)

Here, system is (block + bullet)

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

Initial velocity of the bullet in terms of h

By the conservation of mechanical energy

$\\ {(T . E \text { of system }) \text { Just after collision }=(T . E \text { of system) At height } h} \\\\ {\frac{1}{2}(m+M) V^{2}=(m+M) g h} \\ \\ {V=\sqrt{2 g h}} \\ \\ {\text { Equating }(1) \text { and }(2)} \\ \\ {\text { We get } V=\sqrt{2 g h}=\frac{m u}{m+M}}$

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

Loss in kinetic energy

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

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

Value of angle $\theta$

$\\ {\text { From } u=\left(\frac{m+M}{m}\right) \sqrt{2 g h}} \\\\ {\text { We can write }} \\\\ {h=\left(\frac{u^{2}}{2 g}\right)\left(\frac{m}{m+M}\right)^{2}} \\\\{\text { And from figure }} \\\\ Or,\ {\cos \theta=\frac{L-h}{L}=1-\frac{h}{L}=1-\left(\left(\frac{u^{2}}{2 g L}\right)\left(\frac{m}{m+M}\right)^{2}\right)}$

Study it with Videos

Perfectly Elastic Head on Collision

Perfectly elastic oblique collision

Head on inelastic collision

Perfectly inelastic collision

Collision Between Bullet and Vertically Suspended Block

Study other Related Concepts

Elastic And Inelastic Collision Current Topic

Work Done By A Constant Force

Work Done By Variable Force

Conservative And Non-conservative Forces

Basics Of Energy And Its Various Forms

Potential Energy

Law Of Conservation Of Energy

Vertical Circular Motion

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