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This is second of our many quizzes on Elastic collisions. Click on this to access more such competitive quizzes on Mechanics.

Elastic collisions Quiz 2

physics

 Consider two colliding bodies $A$ and $B$, masses ${{m}_{1}}$ and ${{m}_{2}}$, respectively.$A$ is at rest and $B$ has an initial velocity $u $.What will be the final velocity of $A$ and $B$ if the collision is elastic? 

To calculate the formula of final velocities of the colliding objects after collision we use the concept of conservation of momentum and conservation of kinetic energy of the system in an elastic collision. 

  $\text{The kinetic energy and momentum before and after the collision remains the same,}\\ {{m}_{2}}u={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}….(i)\\ \dfrac{1}{2}{{m}_{2}}{{u}^{2}}=\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}+\dfrac{1}{2}{{m}_{2}}{{v}_{2}}^{2} ….(ii)\\ \text{Now, solving equation (i) and (ii) for}\,{{v}_{1}}\,\text{and}\,{{v}_{2}}\,\text{we get,}\\ {{v}_{1}}=\dfrac{2{{m}_{2}}u}{{{m}_{1}}+{{m}_{2}}}\\ {{v}_{2}}=\dfrac{({{m}_{2}}-{{m}_{1}})u}{{{m}_{1}}+{{m}_{2}}}$

The velocity of the centre of mass after an elastic collision, 

There is no external force acting on the system in an elastic collision, since the collision force acting between two bodies during elastic collision is an internal force.

  $\text{The motion of the center of mass of an object is determined by the external forces acting on the system.}\\  \text{The external force is given by,}\\ \sum {{F}_{external}}=\dfrac{dp}{dt}=\dfrac{d(m{{v}_{c.m}})}{dt}\\ \text{In an elastic collision, the collision force acting between two bodies is an internal force}\\ \text{and there is no external force acting on the system.}$<br>Then velocity of the centre of mass after an elastic collision will remain the same.

If two elastic balls with equal masses collide in a two dimensional plane as shown in the figure below, the first ball has a velocity 4 m/s and the second ball is at rest. After collision they scatter at an angle ${{60}^{0}}$ and ${{30}^{0}}$ respectively. What is the velocity of the first ball after collision, if velocity of the second ball is 3 m/s?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/033cc009-9d41-4dc9-b0a2-03690bd7c5945063390685366581213.png"/> 

We make use of the conservation of momentum along the y-direction formula to find the final velocity as there is no y-component of momentum before the collision. 

 $\text{From the figure, consider the conservation of momentum along the y-direction, we get the equation,}\\ {{m}_{1}}{{v}_{1}}\sin {{\theta }_{1}}+{{m}_{2}}{{v}_{2}}\sin {{\theta }_{2}}=0….(i)\\ \text{According to the question,}\\ {{m}_{1}}={{m}_{2}}=m,{{u}_{1}}=4\,m/s,{{u}_{2}}=0\,m/s,\\{{\theta }_{1}}={{60}^{0}},{{\theta }_{2}}={{30}^{0}}\,\text{and}\,{{v}_{2}}=3\,m/s.\\ \text{Putting the above values in equation (i), we get,}\\ \therefore{{v}_{1}}=1.73\,m/s$

Two particles whose mass are in the ratio of $1:2$ and velocity $10$ m/s and $12$ m/s, respectively collide. What is the ratio of the total kinetic energies before and after collision and is the collision elastic or inelastic? 

Kinetic energy of the system before and after an elastic collision is always conserved i.e. there is no loss of kinetic energy during the elastic collision. 

 $\text{For the initial and final velocities of the two particles,}\\ {{v}_{1}}=\dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{1}}+\dfrac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}….(i)\\ {{v}_{2}}=\dfrac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{1}}+\dfrac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}….(ii)\\ \text{According to the question,}\\ {{m}_{1}}=m,{{m}_{2}}=2m,{{u}_{1}}=10\,m/s\,\text{and}\,{{u}_{2}}=12\,m/s.\\ \text{Putting the above values in equation (i) and (ii), we get,}\\ {{v}_{1}}=12.67\,m/s\,\text{and}\,{{v}_{2}}=10.67\,m/s\\ \text{The total kinetic energies of the two particles before and after collision is given by,}\\ {{(K.E.)}_{before}}=\dfrac{1}{2}{{m}_{1}}{{u}_{1}}^{2}+\dfrac{1}{2}{{m}_{2}}{{u}_{2}}^{2}\\<br>\Rightarrow{{(K.E.)}_{before}}=\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}+\dfrac{1}{2}{{m}_{2}}{{v}_{2}}^{2}….(iii)\\ \text{Putting the above values of}\,{{v}_{1}}\,\text{and}\,{{v}_{2}}\,\text{in the above equation, we get,}\\ \therefore\dfrac{{{(K.E.)}_{before}}}{{{(K.E.)}_{after}}}=\dfrac{1}{1}$ 

 If a sphere $A$ moving with initial velocity ‘u’ collides with another sphere $B$ at rest as shown below. Mass of $A$ is one third of the mass of $B$ with scattering angle ${{\theta }_{A}}={{\theta }_{B}}={{45}^{0}}$ and final velocities of $A$ and $B$ are $1.5$ m/s and $2$ m/s respectively. Find the value of the initial velocity ‘u’.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a9123901-f9a7-47bb-8a1e-de0a8098d36a7113731422858729960.png"/> 

Use the fact that in an elastic collision momentum before and after the collision is always conserved.So to find the initial velocity we equate the momentum along the x-direction. 

 $\text{Considering the conservation of momentum along the x-direction, we get the equation,}\\ {{m}_{1}}{{u}_{1}}={{m}_{1}}{{v}_{1}}\cos {{\theta }_{1}}+{{m}_{2}}{{v}_{2}}\cos {{\theta }_{2}}….(i)\\ \text{According to the question,}\\ {{m}_{1}}=\dfrac{m}{3}, {{m}_{2}}=m,{{u}_{1}}=u\,m/s,{{u}_{2}}=0\,m/s,\\ {{\theta }_{1}}={{60}^{0}}, {{\theta }_{2}}={{30}^{0}}, {{v}_{1}}=1.5\,m/s\,\text{and}\,{{v}_{2}}=2\,m/s.\\ \text{Putting the above values in equation (i) and solving, we get,}\\ \therefore{{u}_{1}}=5.3\,m/s$