Perfectly Inelastic Collision
In such types of collisions, the bodies move independently before collision but after collision as a one single body.
(1) When the colliding bodies are moving in the same direction
By the law of conservation of momentum
\({{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{v}_{\text{comb}}}\)
\(\Rightarrow{{v}_{\text{comb}}}=\frac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\)
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 vcomb]
(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
DK = Initial kinetic energy – Final kinetic energy
\(=\left( \frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2} \right)-\left( \frac{1}{2}({{m}_{1}}+{{m}_{2}})\,v_{\text{comb}}^{2} \right)\)
\(=\frac{1}{2}\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{({{u}_{1}}-{{u}_{2}})}^{2}}\)
Collision Between Bullet and Vertically Suspended Block
A bullet of mass m is fired horizontally with velocity u in block of mass M suspended by vertical thread. After the collision bullet gets embedded in block. Let the combined system raised upto height h and the string makes an angle q with the vertical.
(1) Velocity of system
Let v be the velocity of the system (block + bullet) just after the collision.
Momentumbullet + Momentumblock = Momentumbullet 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 \(\frac{1}{2}(m+M){{v}^{2}}=(m+M)gh\Rightarrow v=\sqrt{2gh}\)
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 \(\theta ={{\cos }^{-1}}\left[ 1-\frac{1}{2gL}{{\left( \frac{mu}{m+M} \right)}^{2}} \right]\)