Perfectly elastic head on collision
Category : JEE Main & Advanced
Let two bodies of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] moving with initial velocities \[{{u}_{1}}\] and \[{{u}_{2}}\] in the same direction and they collide such that after collision their final velocities are \[{{v}_{1}}\] and \[{{v}_{2}}\] respectively.
According to law of conservation of momentum
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] ...(i)
\[\Rightarrow\] \[{{m}_{1}}({{u}_{1}}-{{v}_{1}})={{m}_{2}}({{v}_{2}}-{{u}_{2}})\] ...(ii)
According to 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}\] ...(iii)
\[\Rightarrow\] \[{{m}_{1}}(u_{1}^{2}-v_{1}^{2})={{m}_{2}}(v_{2}^{2}-u_{2}^{2})\] ....(iv)
Dividing equation (iv) by equation (ii)
\[{{v}_{1}}+{{u}_{1}}={{v}_{2}}+{{u}_{2}}\] ...(v)
\[\Rightarrow\] \[{{u}_{1}}-{{u}_{2}}={{v}_{2}}-{{v}_{1}}\] ...(vi)
Relative velocity of separation is equal to relative velocity of approach.
Note :
\[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\]
or \[{{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\]
For perfectly elastic collision, e = 1
\[\therefore\] \[{{v}_{2}}-{{v}_{1}}={{u}_{1}}-{{u}_{2}}\] [As shown in eq. (vi)] For perfectly inelastic collision, e = 0
\[\therefore\] \[{{v}_{2}}-{{v}_{1}}=0\] or \[{{v}_{2}}={{v}_{1}}\]
It means that two body stick together and move with same velocity.
For inelastic collision, 0 < e < 1
\[\therefore\] \[{{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\]
In short we can say that e is the degree of elasticity of collision and it is dimensionless quantity.
Further from equation (v) we get
\[{{v}_{2}}={{v}_{1}}+{{u}_{1}}-{{u}_{2}}\]
Substituting this value of \[{{v}_{2}}\] in equation (i) and rearranging
we get, \[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] ...(vii)
Similarly we get,
\[{{v}_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\] ...(viii)
(1) Special cases of head on elastic collision
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(i) If projectile and target are of same mass i.e. \[{{m}_{1}}={{m}_{2}}\] Since \[{{\upsilon }_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}\] |
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and \[{{\upsilon }_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\] Substituting \[{{m}_{1}}={{m}_{2}}\] we get \[{{\upsilon }_{1}}={{u}_{2}}\] and \[{{\upsilon }_{2}}={{u}_{1}}\] It means when two bodies of equal masses undergo head on elastic collision, their velocities get interchanged. Example : Collision of two billiard balls
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| (ii) If massive projectile collides with a light target i.e. \[{{m}_{1}}>>{{m}_{2}}\] |
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Since \[{{\upsilon }_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] and \[{{\upsilon }_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\] Substituting \[{{m}_{2}}=0\], we get \[{{\upsilon }_{1}}={{u}_{1}}\]and \[{{\upsilon }_{2}}=2{{u}_{1}}-{{u}_{2}}\] Example : Collision of a truck with a cyclist
Before collision
After collision.
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(iii) If light projectile collides with a very heavy target i.e. \[{{m}_{1}}<<{{m}_{2}}\]
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Since \[{{\upsilon }_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] and \[{{\upsilon }_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\] Substituting \[{{m}_{1}}=0\], we get \[{{\upsilon }_{1}}=-\,{{u}_{1}}+2{{u}_{2}}\]and \[{{\upsilon }_{2}}={{u}_{2}}\] Example : Collision of a ball with a massive wall.
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(2) Kinetic energy transfer during head on elastic collision
Kinetic energy of projectile before collision \[{{K}_{i}}=\frac{1}{2}{{m}_{1}}u_{1}^{2}\]
Kinetic energy of projectile after collision \[{{K}_{f}}=\frac{1}{2}{{m}_{1}}v_{1}^{2}\]
Kinetic energy transferred from projectile to target \[\Delta K=\] decrease in kinetic energy in projectile
\[\Delta K=\frac{1}{2}{{m}_{1}}u_{1}^{2}-\frac{1}{2}{{m}_{1}}v_{1}^{2}\] \[=\frac{1}{2}{{m}_{1}}(u_{1}^{2}-v_{1}^{2})\]
Fractional decrease in kinetic energy
\[\frac{\Delta K}{K}=\frac{\frac{1}{2}{{m}_{1}}(u_{1}^{2}-v_{1}^{2})}{\frac{1}{2}{{m}_{1}}u_{1}^{2}}\]\[=1-{{\left( \frac{{{v}_{1}}}{{{u}_{1}}} \right)}^{2}}\] ...(i)
We can substitute the value of \[{{v}_{1}}\] from the equation
\[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
If the target is at rest i.e.\[{{\upsilon }_{2}}=0\] then \[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}\]
From equation (i) \[\frac{\Delta K}{K}=1-{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\] ...(ii)
or \[\frac{\Delta K}{K}=\frac{4{{m}_{1}}{{m}_{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}\] ...(iii)
or \[\frac{\Delta K}{K}=\frac{4{{m}_{1}}{{m}_{2}}}{{{({{m}_{1}}-{{m}_{2}})}^{2}}+4{{m}_{1}}{{m}_{2}}}\] ...(iv)
Note :
i.e. \[{{m}_{1}}-{{m}_{2}}=0\] or \[{{m}_{1}}={{m}_{2}}\] then \[\frac{\Delta K}{K}=1=100%\]
So the transfer of kinetic energy in head on elastic collision (when target is at rest) is maximum when the masses of particles are equal i.e. mass ratio is 1 and the transfer of kinetic energy is 100%.
\[\Rightarrow \] \[{{\left( \frac{\Delta K}{K} \right)}_{\text{Retained}}}=\]\[1-\left[ 1-{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}} \right]\]\[={{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\]
(3) Velocity, momentum and kinetic energy of stationary target after head on elastic collision
(i) Velocity of target : We know
\[{{v}_{2}}=\left( \frac{{{m}_{\text{2}}}-{{m}_{1}}}{{{m}_{\text{1}}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{\text{1}}}+{{m}_{2}}}\]
\[\Rightarrow \] \[{{v}_{2}}=\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{\text{1}}}+{{m}_{2}}}\]
\[=\frac{2{{u}_{1}}}{1+{{m}_{2}}/{{m}_{1}}}\] As \[{{\upsilon }_{2}}=0\] and
Assuming \[\frac{{{m}_{2}}}{{{m}_{1}}}=n\]
\[\therefore \] \[{{v}_{2}}=\frac{2{{u}_{1}}}{1+n}\]
(ii) Momentum of target : \[{{P}_{2}}={{m}_{2}}{{v}_{2}}\]\[=\frac{2n{{m}_{1}}{{u}_{1}}}{1+n}\] \[\left[ \text{As }{{m}_{2}}={{m}_{1}}n\text{ and }{{v}_{2}}=\frac{2{{u}_{1}}}{1+n} \right]\]
\[\therefore \] \[{{P}_{2}}=\frac{2{{m}_{1}}{{u}_{1}}}{1+(1/n)}\]
(iii) Kinetic energy of target : \[{{K}_{2}}=\frac{1}{2}{{m}_{2}}v_{2}^{2}\]\[=\frac{1}{2}n\,{{m}_{1}}{{\left( \frac{2{{u}_{1}}}{1+n} \right)}^{2}}\]\[=\frac{2\,{{m}_{1}}u_{1}^{2}n}{{{(1+n)}^{2}}}\]
\[=\frac{4({{K}_{1}})n}{{{(1-n)}^{2}}+4n}\] \[\left[ \text{ As }{{K}_{1}}=\frac{1}{2}{{m}_{1}}u_{1}^{2} \right]\]
(iv) Relation between masses for maximum velocity, momentum and kinetic energy
| Velocity | \[{{\upsilon }_{2}}=\frac{2{{u}_{1}}}{1+n}\] |
For \[{{\upsilon }_{2}}\] to be maximum n must be minimum i.e. \[n=\frac{{{m}_{2}}}{{{m}_{1}}}\to 0\]\[\therefore \]\[{{m}_{2}}<<{{m}_{1}}\] |
Target should be very light. |
| Momentum | \[{{P}_{2}}=\frac{2{{m}_{1}}{{u}_{1}}}{(1+1/n)}\] |
For \[{{P}_{2}}\] to be maximum, (1/n) must be minimum or n must be maximum. i.e. \[n=\frac{{{m}_{2}}}{{{m}_{1}}}\to \infty \]\[\therefore \] \[{{m}_{2}}>>{{m}_{1}}\] |
Target should be massive. |
| Kinetic energy | \[{{K}_{2}}=\frac{4{{K}_{1}}\,n}{{{(1-n)}^{2}}+4n}\] |
For \[{{K}_{2}}\] to be maximum \[{{(1-n)}^{2}}\] must be minimum. i.e. \[1-n=0\,\Rightarrow \,n=1=\frac{{{m}_{2}}}{{{m}_{1}}}\]\[\therefore \] \[{{m}_{2}}={{m}_{1}}\] |
Target and projectile should be of equal mass. |
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