Inelastic collision is a collision where the kinetic energy is not conserved due to internal friction
Though the kinetic energy is lost in these collision,but they obey law of conservation of Linear Momentum
If the bodies or particles stick together and move together after the collision, the it is called Perfectly inelastic collision
If the bodies or particles does stick together after the collision but there is loss of kinetic energy, the it is inelastic collision
Coefficient of restitution for the inelastic collision lies as $0 \leq e < 1$. e is 0 for perfectly inelastic collision and lies between 0 and 2 for elastic collision
Inelastic collision can be further divided into head on collision (i.e collision in one dimension) and opaque collision (i.e collision in two dimension)
Lets study about these two inelastic collision in one dimension and two dimension further
Head on inelastic collision of two particles which stick together
Consider two particles whose masses are m_{1} and m_{2} .Let u_{1} and u_{2} be the respective velocities before collision
Let both the particles stick together after collision and moves with the same velocity v .Then from law of conservation of linear momentum
m_{1}u_{1} +m_{2}u_{2}=(m_{1}+m_{2})v
or
If we consider second particle to be stationary or at rest then u_{2}=0
then
m_{1}u_{1} =(m_{1}+m_{2})
or
Hence |v| < |u_{1}|
Kinetic energy before collision is
KE_{1}=(1/2)m_{1}u_{1}^{2}
After collison KE of the system is
KE_{2}=(1/2)(m_{1}+m_{2})v^{2}
Hence from equation we come to know that K_{2} < K_{1} hence energy loss would be there after thye collision of the particles
(15) Deflection of an moving particle by a particle at rest ( inelastic collision in two dimension)
Let m_{1} and m_{2} be the two mass particle in a laboratory frame of refrence and m_{1} collide with m_{2} which is initailly is at rest.Let the velocity of mass m_{1} before collison be u_{1} and after the collison it moves with a velocity v_{1} and is delfected by the angle θ_{1} withs its incident direction and
m_{2} after the collision moves with the velocity v_{2} and it is deflected by an angle θ_{2} with its incident direction
From law of conservation of linear momentum ,for components along x-axis
m_{1}u_{1}=m_{1}v_{1}cosθ_{1} +m_{2}v_{2}cosθ_{2} ---(1)
For components along y-axis
0=m_{1}v_{1}sinθ_{1} -m_{2}v_{2}sinθ_{2} --(2)
Analysing above equations we come to know that we have to find values of four unknown quantities v_{1},v_{2},θ_{1},θ_{2} with the help of above three equations which is impossible as we need to have atleast four equations for finding out the values of four unknown quantities .So we cannot predict the variable as they are four of them . However if we measure any two variable ,then other variable can be uniquely determined from the above equation
Example-1
A 10 kg object is moving in +x direction at 30 m/s. An another object of mass 40 g is moving in -x direction at 50 m/s. They collide head on and stick together. Find the final velocity ? Solution
This is a case of Perfectly inelastic collision, applying law of conservation of momentum
Momentum before impact = Momentum After Impact
$(10) \times 30 + (.04) \times (-50) = (10.04)v$
Negative sign is for 40g particle as it is moving in negative direction
v=29.68 m/s