Elastic head on collision and inelastic head on collision
Perfectly elastic head on collision
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 : The ratio of relative velocity of separation and relative velocity of approach is defined as coefficient of restitution.
\(e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\)
or \({{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}}){{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\)
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
(i) If projectile and target are of same mass i.e. m1 = m2
Since \({{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}\) and \({{v}_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\)
Substituting \({{m}_{1}}={{m}_{2}}\) we get
\({{v}_{1}}={{u}_{2}}\) and \({{v}_{2}}={{u}_{1}}\)
It means when two bodies of equal masses undergo head on elastic collision, their velocities get interchanged.
Example : Collision of two billiard balls
(ii) If massive projectile collides with a light target i.e. m1 >> m2
Since \({{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\) and \({{v}_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\)
Substituting \({{m}_{2}}=0\), we get
\({{v}_{1}}={{u}_{1}}\,\, and \,\,{{v}_{2}}=2{{u}_{1}}-{{u}_{2}}\)
Example : Collision of a truck with a cyclist
(iii) If light projectile collides with a very heavy target i.e. m1 << m2
Since\({{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\) and \({{v}_{2}}=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\)
Substituting \({{m}_{1}}=0\), we get
\({{v}_{1}}=-\,{{u}_{1}}+2{{u}_{2}}\,\, and \,\,{{v}_{2}}={{u}_{2}}\)
Example : Collision of a ball with a massive wall.
(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. u2 = 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 : Greater the difference in masses, lesser will be transfer of kinetic energy and vice versa
Transfer of kinetic energy will be maximum when the difference in masses is minimum
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%.
If \({{m}_{2}}=n\,{{m}_{1}}\)then from equation (iii) we get \(\frac{\Delta K}{K}=\frac{4n}{{{(1+n)}^{2}}}\)
Kinetic energy retained by the projectile \({{\left( \frac{\Delta K}{K} \right)}_{\text{Retained}}}=1-\) kinetic energy transferred by projectile
\(\Rightarrow{{\left( \frac{\Delta K}{K} \right)}_{\text{Retained}}}=1-\left[ 1-{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}} \right]={{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\)
(3) Velocity, momentum and kinetic energy of stationary target after head on elastic collision
(i) Velocity of target : We know
\({{v}_{2}}=\left( \frac{{{m}_{\text{2}}}-{{m}_{1}}}{{{m}_{\text{1}}}+{{m}_{2}}} \right)\,{{u}_{2}}+\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{\text{1}}}+{{m}_{2}}}\)
\(\Rightarrow{{v}_{2}}=\frac{2{{m}_{1}}{{u}_{1}}}{{{m}_{\text{1}}}+{{m}_{2}}}\)
\(=\frac{2{{u}_{1}}}{1+{{m}_{2}}/{{m}_{1}}}\,\, As \,\,{{u}_{2}}=0\) 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
Perfectly Elastic Oblique Collision
Let two bodies moving as shown in figure.
By law of conservation of momentum
Along x-axis, \({{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}\cos \theta +{{m}_{2}}{{v}_{2}}\cos \varphi \) ...(i)
Along y-axis, \(0={{m}_{1}}{{v}_{1}}\sin \theta -{{m}_{2}}{{v}_{2}}\sin \varphi \) ...(ii)
By law of conservation of kinetic energy
\(\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\) ...(iii)
In case of oblique collision it becomes difficult to solve problem unless some experimental data is provided, as in these situations more unknown variables are involved than equations formed.
Special condition : If \({{m}_{1}}={{m}_{2}}\,\, and \,\,{{u}_{2}}=0\) substituting these values in equation (i), (ii) and (iii) we get
\({{u}_{1}}={{v}_{1}}\cos \theta +{{v}_{2}}\cos \varphi \) ...(iv)
\(0={{v}_{1}}\sin \theta -{{v}_{2}}\sin \varphi\) ...(v)
and \(u_{1}^{2}=v_{1}^{2}+v_{2}^{2}\) …(vi)
Squaring (iv) and (v) and adding we get
\(u_{1}^{2}=v_{1}^{2}+v_{2}^{2}+2{{v}_{1}}{{v}_{2}}\cos (\theta +\varphi )\) …(vii)
Using (vi) and (vii) we get \(\cos (\theta +\varphi )=0\)
\(\therefore\theta +\varphi =\pi /2\)
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 \({{90}^{o}}\).
Head on Inelastic Collision
(1) Velocity after collision : Let two bodies A and B collide inelastically and coefficient of restitution is e.
Where \(e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}=\frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}\)
\(\Rightarrow{{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\)
\(\therefore{{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\) …(i)
From the law of conservation of linear momentum
\({{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\) …(ii)
By solving (i) and (ii) we get
\({{v}_{1}}=\left( \frac{{{m}_{1}}-e{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\left( \frac{(1+e)\,{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}\)
Similarly \({{v}_{2}}=\left[ \frac{(1+e)\,{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right]\,{{u}_{1}}+\left( \frac{{{m}_{2}}-e\,{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}\)
By substituting e = 1, we get the value of \({{v}_{1}}\,\, and \,\,{{v}_{2}}\) for perfectly elastic head on collision.
(2) Ratio of velocities after inelastic collision : A sphere of mass m moving with velocity u hits inelastically with another stationary sphere of same mass.
\(\therefore e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}=\frac{{{v}_{2}}-{{v}_{1}}}{u-0}\)
\(\Rightarrow{{v}_{2}}-{{v}_{1}}=eu\) …(i)
By conservation of momentum :
Momentum before collision = Momentum after collision
\(mu=m{{v}_{1}}+m{{v}_{2}}\)
\(\Rightarrow{{v}_{1}}+{{v}_{2}}=u\) …(ii)
Solving equation (i) and (ii) we get \({{v}_{1}}=\frac{u}{2}(1-e)\)
and \({{v}_{2}}=\frac{u}{2}(1+e)\)
\(\therefore\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1-e}{1+e}\)
(3) Loss in kinetic energy :
Loss in K.E. (DK) = Total initial kinetic energy – Total final kinetic energy
\(= \left( \frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2} \right)-\left( \frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2} \right)\)
Substituting the value of \({{v}_{1}}\,\, and \,\,{{v}_{2}}\) from the above expressions
Loss (DK) = \(\frac{1}{2}\left( \frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,(1-{{e}^{2}})\,{{({{u}_{1}}-{{u}_{2}})}^{2}}\)
By substituting e = 1 we get DK = 0 i.e. for perfectly elastic collision, loss of kinetic energy will be zero or kinetic energy remains same before and after the collision.
Rebounding of Ball After Collision With Ground
If a ball is dropped from a height h on a horizontal floor, then it strikes with the floor with a speed.
\({{v}_{0}}=\sqrt{2g{{h}_{0}}}\) [From \({{v}^{2}}={{u}^{2}}+2gh]\)
and it rebounds from the floor with a speed
\({{v}_{1}}=e\,{{v}_{0}}=e\sqrt{2g{{h}_{0}}}\) \(\left[ \text{As }e=\frac{\text{velocity after collision}}{\text{velocity before collision}} \right]\)
(1) First height of rebound :\({{h}_{1}}=\frac{v_{1}^{2}}{2g}={{e}^{2}}{{h}_{0}}\)
\(\therefore\) h1 = e2h0
(2) Height of the ball after nth rebound : Obviously, the velocity of ball after nth rebound will be
\({{v}_{n}}={{e}^{n}}{{v}_{0}}\)
Therefore the height after nth rebound will be
\({{h}_{n}}=\frac{v_{n}^{2}}{2g}={{e}^{2n}}{{h}_{0}}\)
\(\therefore{{h}_{n}}={{e}^{2n}}{{h}_{0}}\)
(3) Total distance travelled by the ball before it stops bouncing
\(H={{h}_{0}}+2{{h}_{1}}+2{{h}_{2}}+2{{h}_{3}}+...={{h}_{0}}+2{{e}^{2}}{{h}_{0}}+2{{e}^{4}}{{h}_{0}}+2{{e}^{6}}{{h}_{0}}+...\)
\(H={{h}_{0}}[1+2{{e}^{2}}(1+{{e}^{2}}+{{e}^{4}}+{{e}^{6}}....)]\)
\(={{h}_{0}}\left[ 1+2{{e}^{2}}\left( \frac{1}{1-{{e}^{2}}} \right) \right]\)
\(\left[ \text{As}\,\,\,1+{{e}^{2}}+{{e}^{4}}+....=\frac{1}{1-{{e}^{2}}} \right]\)
\(\therefore H={{h}_{0}}\left[ \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right]\)
(4) Total time taken by the ball to stop bouncing
\(T={{t}_{0}}+2{{t}_{1}}+2{{t}_{2}}+2{{t}_{3}}+..=\sqrt{\frac{2{{h}_{0}}}{g}}+2\sqrt{\frac{2{{h}_{1}}}{g}}+2\sqrt{\frac{2{{h}_{2}}}{g}}+..\)
\( =\sqrt{\frac{2{{h}_{0}}}{g}}\,\,\,\,[1+2e+2{{e}^{2}}+......]\) [As \({{h}_{1}}={{e}^{2}}{{h}_{0}}; {{h}_{2}}={{e}^{4}}{{h}_{0}}\)]
\(=\sqrt{\frac{2{{h}_{0}}}{g}}\,\,\,\,[1+2e(1+e+{{e}^{2}}+{{e}^{3}}+......)] \)
\(=\sqrt{\frac{2{{h}_{0}}}{g}}\,\left( \frac{1+e}{1-e} \right)\)
\(\therefore T=\left( \frac{1+e}{1-e} \right)\,\sqrt{\frac{2{{h}_{0}}}{g}}\)