More often you see the objects undergo collision like a billiard balls collides, a man accidentally dashes another man.
These collisions are basically categorized into two types:1. Elastic Collision2. Inelastic Collision Sometimes you could notice bodies sticking together after collision like a balls collides a wall and breaks, a car getting smashed by a heavy loaded truck etc. This is what inelastic collision is! Here energy gets transferred from a body to the other however the momentum gets conserved. |

In an Inelastic collision the kinetic energy is not
conserved. The total momentum of two bodies remains the same but the
initial kinetic energy is converted into heat. Here the particles stick together on impact and hence the loss of kinetic energy takes place. Common velocity of the body after collision is given by

m

or

v

When second body is at rest v_{1}v_{1}+ m_{2}v_{2}= (m_{1}+ m_{2})v_{f}or

v

_{f}= $\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$_{2 }= 0 then

v

_{f }= $\frac{m_1}{m_1 + m_2}$ v_{1}Consider two bodies of masses

By law of conservation of momentum

**m**_{1}and**m**_{2}with initial velocity**v**and_{1}**v**_{2}along a straight line. After collision if these two bodies stick on together the common velocity is**v**then the collision is perfectly inelastic. The momentum before collision is equal to the momentum after collision._{}By law of conservation of momentum

**m**

_{1}v_{1}+ m_{2}v_{2}= (m_{1}+ m_{2})v_{2}

Since mass m

_{2}is at rest v_{2}= 0. Hence**m**_{1}v_{1}= (m_{1}+ m_{2}) v_{}The final speed is**v =**

**$\frac{m_1 v_1}{m_1 + m_2}$**

The total momentum before collision = m

_{1}v

_{1}+ m

_{2}v

_{2}

The total momentum after collision = (m

_{1}+ m

_{2})v

If the momentum is conserved then

**p**

_{before Collision}= p_{after Collision}so

**m**

_{1}v_{1}_{ }

**= (m**

_{1}+ m_{2})v_{}Hence v_{}= $\frac{m_1 v_1}{m_1 + m_2}$_{ }This is Momentum Inelastic Collision formula

Let us take a body of mass m

The loss of kinetic energy is the kinetic energy after collision minus the kinetic energy before collision i.e.,

K.E

= $\frac{1}{2}$ m

= $\frac{1}{2}$ m

= $\frac{1}{2}$ $\frac{m_1 v_1^2 (m_1 + m_2) - m_1^2 v_1^2 }{m_1 + m_2}$

= $\frac{1}{2}$ $\frac{m_1^2 v_1^2 + m_1 m_2 v_1^2 - m_1^2 v_1^2}{m_1 + m_2}$

= $\frac{m_1 m_2}{m_1 + m_2}$ v

The velocities after a one dimensional Collision equation is given by_{1}moving with velocity v_{1}and mass m_{2}moving with velocity v_{2}undergoes the inelastic Collision as shown in fig. Then their kinetic energy is given by energy before Collision = $\frac{1}{2}$ m_{1}v_{1}^{2}and energy after Collision = $\frac{1}{2}$ (m_{1}+ m_{2}) v_{}^{2}.The loss of kinetic energy is the kinetic energy after collision minus the kinetic energy before collision i.e.,

K.E

_{i}- K.E_{f}= $\frac{1}{2}$ m_{1}v_{1}^{2}- $\frac{1}{2}$ (m_{1}+ m_{2}) v^{2}= $\frac{1}{2}$ m

_{1}v_{1}^{2}- $\frac{1}{2}$ (m_{1}+ m_{2}) $(\frac{m_1 v_1}{m_1 + m_2})^2$= $\frac{1}{2}$ m

_{1}v_{1}^{2}- $\frac{1}{2}$ $\frac{m_1^2 v_1^2}{m_1 + m_2}$= $\frac{1}{2}$ $\frac{m_1 v_1^2 (m_1 + m_2) - m_1^2 v_1^2 }{m_1 + m_2}$

= $\frac{1}{2}$ $\frac{m_1^2 v_1^2 + m_1 m_2 v_1^2 - m_1^2 v_1^2}{m_1 + m_2}$

= $\frac{m_1 m_2}{m_1 + m_2}$ v

_{1}^{2}.v

v

M

M

U

U

_{a}= $\frac{c_r M_b (U_b - U_a) + M_a U_a + M_b U_b}{M_a + M_b}$v

_{b}= $\frac{c_r M_b (U_a - U_b) + M_a U_a + M_b U_b}{M_a + M_b}$Where C_{r}= Coefficient of restitutionM

_{a}= Mass of body aM

_{b}= Mass of body bU

_{a}= Initial Velocity of body aU

_{b}= Initial velocity of body b.Lets us consider the inelastic Collision in 2 dimension. If v

_{i}is the initial velocity of mass m_{1}colliding with mass m_{2}having velocity v_{2}. It will be having two dimension x and y given as:The Inelastic Collision in x-direction is

**m**

_{1}v_{1ix}+ m_{2}v_{2ix}= (m_{1}+ m_{2}) v_{fx}**m**

_{1}v_{2iy}+ m_{2}v_{2iy}= (m_{1}+ m_{2}) v_{fy}- Crash taking place between the two vehicles
- A Bullet hitting the target
- A man in hurry clashes with another guy coming in front of him
- A boy hitting the ball