\[\text{Momentu}{{\text{m}}_{\text{bullet}}}\text{+ Momentu}{{\text{m}}_{\text{block}}}\text{= Momentu}{{\text{m}}_{\text{bullet and block system}}}\]
\[mu+0=(m+M)v\]
\[\therefore \] \[v=\frac{mu}{(m+M)}\] ...(i)
(2) Velocity of bullet : Due to energy which remains in the bullet-block system, just after the collision, the system (bullet + block) rises upto height h.
By the conservation of mechanical energy \[\frac{1}{2}(m+M){{v}^{2}}=(m+M)gh\] Þ \[v=\sqrt{2gh}\]
Now substituting this value in the equation (i) we get \[\sqrt{2gh}=\frac{mu}{m+M}\]
\[\therefore \] \[u=\left[ \frac{(m+M)\sqrt{2gh}}{m} \right]\]
(3) Loss in kinetic energy : We know that the formula for loss of kinetic energy in perfectly inelastic collision
\[\Delta K=\frac{1}{2}\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}\,{{({{u}_{1}}-{{u}_{2}})}^{2}}\]
(When the bodies are moving in same direction.)
\[\therefore \] \[\Delta K=\frac{1}{2}\frac{mM}{m+M}{{u}^{2}}\] [As \[{{u}_{1}}=u\], \[{{u}_{2}}=0\], \[{{m}_{1}}=m\] and \[{{m}_{2}}=M\]]
(4) Angle of string from the vertical
From the expression of velocity of bullet \[u=\left[ \frac{(m+M)\sqrt{2gh}}{m} \right]\] we can get \[h=\frac{{{u}^{2}}}{2g}{{\left( \frac{m}{m+M} \right)}^{2}}\]
From the figure \[\cos \theta =\frac{L-h}{L}=1-\frac{h}{L}\]\[=1-\frac{{{u}^{2}}}{2gL}{{\left( \frac{m}{m+M} \right)}^{2}}\]
or \[\theta ={{\cos }^{-1}}\left[ 1-\frac{1}{2gL}{{\left( \frac{mu}{m+M} \right)}^{2}} \right]\] 
Loss in kinetic energy
\[\Delta K=\left( \frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2} \right)-\frac{1}{2}({{m}_{1}}+{{m}_{2}})v_{comb}^{2}\]
\[\Delta K=\frac{1}{2}\left( \frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{({{u}_{1}}-{{u}_{2}})}^{2}}\]
[By substituting the value of \[{{\upsilon }_{comb}}\]]
(2) When the colliding bodies are moving in the opposite direction
By the law of conservation of momentum
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}(-{{u}_{2}})=({{m}_{1}}+{{m}_{2}}){{v}_{\text{comb}}}\]
(Taking left to right as positive)
\[\therefore \] \[{{v}_{\text{comb}}}=\frac{{{m}_{1}}{{u}_{1}}-{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
when \[{{m}_{1}}{{u}_{1}}>{{m}_{2}}{{u}_{2}}\] then \[{{v}_{\text{comb}}}>0\] (positive)
i.e. the combined body will move along the direction of motion of mass \[{{m}_{1}}\].
when \[{{m}_{1}}{{u}_{1}}<{{m}_{2}}{{u}_{2}}\] then \[{{v}_{\text{comb}}}<0\] (negative)
i.e. the combined body will move in a direction opposite to the motion of mass \[{{m}_{1}}\].
(3) Loss in kinetic energy
\[\Delta K=\] Initial kinetic energy - Final kinetic energy
\[=\left( \frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2} \right)-\left( \frac{1}{2}({{m}_{1}}+{{m}_{2}})\,v_{\text{comb}}^{2} \right)\]
\[=\frac{1}{2}\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{({{u}_{1}}-{{u}_{2}})}^{2}}\] 
\[{{v}_{1}}=e\,{{v}_{0}}\]\[=e\sqrt{2g{{h}_{0}}}\] \[\left[ \text{As }e=\frac{\text{velocity after collision}}{\text{velocity before collision}} \right]\]
(1) First height of rebound : \[{{h}_{1}}=\frac{v_{1}^{2}}{2g}={{e}^{2}}{{h}_{0}}\]
\[\therefore \,\,\,\,{{h}_{1}}={{e}^{2}}{{h}_{0}}\]
(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 \[{{h}_{1}}={{e}^{2}}{{h}_{0}}\]; \[{{h}_{2}}={{e}^{4}}{{h}_{0}}\]]
\[=\sqrt{\frac{2{{h}_{0}}}{g}}\,\,\,\,[1+2e(1+e+{{e}^{2}}+{{e}^{3}}+......)]\]
\[=\sqrt{\frac{2{{h}_{0}}}{g}}\,\left( \frac{1+e}{1-e} \right)\]
\[\therefore \] \[T=\left( \frac{1+e}{1-e} \right)\,\sqrt{\frac{2{{h}_{0}}}{g}}\] 
\[\therefore \] \[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}=\frac{{{v}_{2}}-{{v}_{1}}}{u-0}\]
\[\Rightarrow \] \[{{v}_{2}}-{{v}_{1}}=eu\] ...(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 \[\Delta K=0\] i.e. for perfectly elastic collision, loss of kinetic energy will be zero or kinetic energy remains same before and after the collision. 
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 \[{{90}^{o}}\]. 
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 :
| (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. more...
Collision is an isolated event in which a strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change.
In collision particles may or may not come in real touch e.g. in collision between two billiard balls or a ball and bat, there is physical contact while in collision of alpha particle by a nucleus (i.e. Rutherford scattering experiment) there is no physical contact.
(1) Stages of collision : There are three distinct identifiable stages in collision, namely, before, during and after. In the before and after stage the interaction forces are zero. Between these two stages, the interaction forces are very large and often the dominating forces governing the motion of bodies. The magnitude of the interacting force is often unknown, therefore, Newton?s second law cannot be used, the law of conservation of momentum is useful in relating the initial and final velocities.
(2) Momentum and energy conservation in collision
(i) Momentum conservation : In a collision, the effect of external forces such as gravity or friction are not taken into account as due to small duration of collision \[(\Delta t)\] average impulsive force responsible for collision is much larger than external force acting on the system and since this impulsive force is 'Internal' therefore the total momentum of system always remains conserved.
(ii) Energy conservation : In a collision 'total energy' is also always conserved. Here total energy includes all forms of energy such as mechanical energy, internal energy, excitation energy, radiant energy or even mass energy.
These laws are the fundamental laws of physics and applicable for any type of collision but this is not true for conservation of kinetic energy.
(3) Types of collision : (i) On the basis of conservation of kinetic energy.
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(6) Area under power-time curve gives the work done as \[P=\frac{dW}{dt}\]
\[\therefore \] \[W=\int{P\,dt}\]
\[\therefore \] W = Area under P-t curve
