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Elastic And Inelastic Collision - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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.

  • 48 Questions around this concept.

Solve by difficulty

  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 speedv, the average force experienced by each support after a long time is (assume all collisions are elastic) :

 

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 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 2nd mass after collision

 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 :

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.

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 

  1. Equal mass in case of perfectly elastic collision

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

Or, Velocity mutually interchange

  1.  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 \ v_2=\ 2u_1-u_2

  1. 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
  1. 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}

 

  1. 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) 


 

  1.  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}

  1. 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)

 

  1. 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}

 

 

  1. 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. 

  1. 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)

  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}

  1. Loss in kinetic energy

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

 

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

  1. 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

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