In an Inelastic collision the total kinetic energy or momentum is conserved or not?

In an inelastic collision, the total K.E is not conserved during collision as well as even after the collision. But the total linear Momentum of the two balls is conserved.

Inelastic Collisions: Understanding Key Concepts,Examples and Applications

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Inelastic Collision

An inelastic collision is a collision in which the total kinetic energy of the system is not conserved, even though momentum remains conserved. During inelastic collisions, kinetic energy is transformed into heat or sound, or deformation of the colliding objects. Inelastic collisions frequently occur in real-world situations, such as during car accidents where vehicles crumple and absorb impact energy, or in sports where players or balls deform upon impact. Understanding these collisions is crucial for analyzing real-life impact scenarios and developing effective safety measures across various fields.

1.0Definition of Inelastic Collision

An inelastic collision is a type of collision in which the total kinetic energy of the system is not conserved, even though momentum is conserved.
In an inelastic collision, a portion of the initial kinetic energy is converted into heat, sound, or the deformation of the colliding objects.
The extent of energy loss varies, with perfectly inelastic collisions being a specific case in which the colliding objects stick together and move as one combined mass after the collision.
Coefficient of Restitution for inelastic collisions is 0<e<1
Coefficient of Restitution for perfectly inelastic collisions e=0

2.0Examples of Inelastic Collision

Some common examples of inelastic collisions are:

Car Accidents: In car crashes, vehicles crumple and deform, absorbing a significant amount of kinetic energy. This deformation converts kinetic energy into heat and sound, illustrating a real-world example of an inelastic collision.
Drop of a Soft Object: Dropping a rubber ball onto a hard surface results in the ball deforming and then rebounding with less height. The kinetic energy lost during deformation.

3.0How to solve problem on inelastic collision

Finding final velocities just after collision:

Solution: Let just after collision:

Momentum conservation:

$P_{i} = 4 \times 2 - 2 \times 1 = 6 kg m / s$

$P_{f} = 4 V_{1} + 2 V_{2}$

$P_{i} = P_{f}$

$6 = 4 V_{1} + 2 V_{2}$

$2 V_{1} + V_{2} = 3$ .…………(1)

$\frac{V _{2} - V _{1}}{2 - ( - 1 )} = \frac{1}{2}$

$2 (V_{2} - V_{1} = 3)$

$(V_{2} - V_{1}) = 1.5$

$V_{2} = 2 m / s V_{1} = 0.5 m / s$

4.0Perfectly Inelastic Collision In One Dimension

When the two colliding objects stick together and move as a single body with a common velocity after the collision, the collision is perfectly inelastic.

A body of mass m1moving with velocity u1collides head-on with another body of mass m2 at rest. After the collision ,the two bodies move together with a common velocity v.

As the linear momentum is conserved, so

$m_{1} u_{1} + m_{2} \times 0 = (m_{1} + m_{2}) v$

$v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}} = v_{common} = v_{com}$

The common velocity is equal to velocity of com.

5.0Loss of Kinetic Energy in Perfectly Inelastic Collision

$Δ K = K_{i} - K_{f}$

$= \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} [m_{1} + m_{2}] V_{co m}^{2}$

$= \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} [m_{1} + m_{2}] [\frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}]^{2}$

$= \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} [\frac{m _{1}^{2} u _{1}^{2} + m _{2}^{2} u _{2}^{2} + 2 m _{1} m _{2} u _{1} u _{2}}{m _{1} + m _{2}}]$

$= \frac{1}{2} [\frac{( m _{1} + m _{2} ) m _{1} u _{1}^{2} + m _{2} ( m _{1} + m _{2} ) u _{2}^{2} - m _{1}^{2} u _{1}^{2} - m _{2}^{2} u _{2}^{2} - 2 m _{1} m _{2} u _{1} u _{2}}{m _{1} + m _{2}}]$

$= \frac{1}{2} [\frac{m _{1}^{2} u _{1}^{2} + m _{1} m _{2} u _{1}^{2} + m _{2} m _{1} u _{2}^{2} + m _{2}^{2} u _{2}^{2} - m _{1}^{2} u _{1}^{2} - m _{2}^{2} u _{2}^{2} - 2 m _{1} m _{2} u _{1} u _{2}}{m _{1} + m _{2}}]$

$= \frac{1}{2} \frac{m _{1} m _{2} u _{1}^{2} + m _{1} m _{2} u _{2}^{2} - 2 m _{1} m _{2} u _{1} u _{2}}{m _{1} + m _{2}}$

$= \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} [u_{1}^{2} + u_{2}^{2} - 2 u_{1} u_{2}]$

$= \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} [u_{1} - u_{2}]^{2}$

$= \frac{1}{2} μ μ_{re l}^{2}$

$Δ K E = \frac{1}{2} μ μ_{re l}^{2}$

Where $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ is reduced mass $u_{re l} = u_{1} - u_{2}$ is initial relative velocity of approach

This is the maximum possible loss in KE during collision.

6.0Collision With A Fixed Surface

Image showing collision with a fixed surface

$e = \frac{v _{1}}{v _{0}} \Rightarrow v_{1} = e v_{0}$

7.0Sample Questions on Inelastic Collision

Q1. A particle with mass m and velocity v collides with a stationary particle of mass 2m, and they stick together upon impact. What is the resulting speed of the combined mass after the collision?

Sol. On applying conservation of momentum

$P_{i} = P_{f}$

$m \times v - 2 m \times 0 = (m + 2 m) V_{System}$

$m v = 3 m V_{System}$

$V_{System} = \frac{v}{3} m / s$

Q2. A body of 2 Kg mass having velocity 3 m/s collides with a body of 1 kg mass moving with a velocity of 4 m/s in the opposite direction. After a collision both bodies stick together and move with a common velocity. Find the velocity?

Sol. $v_{System} = \frac{m _{1} v _{1} - m _{2} v _{2}}{m _{1} + m _{2}} = \frac{2 \times 3 - 1 \times 4}{3} = \frac{2}{3} m / s$

Q3. A ball is dropped to the ground from a height of 2m. The coefficient of restitution is 0.6.To what height will the ball rebound?

problems on dropping ball for inelastic collision

Sol. Here note that the initial instant is not just before collision, so we have to calculate speed just before collision.

As the ball falls to the ground its potential energy m g h_1 changes into Kinetic Energy

$\Rightarrow m g h_{1} = \frac{1}{2} m v_{1}^{2}$ ……………(1)

After rebounding its Kinetic Energy $\frac{1}{2} m v_{2}^{2}$ changes into potential Energy $m g h_{2}$

$\Rightarrow m g h_{2} = \frac{1}{2} m v_{2}^{2} = \frac{1}{2} m (e v_{1})^{2}$ ……(2)

Dividing equation (2) by (1),

$\frac{h _{2}}{h _{1}} = (e)^{2}$

$\Rightarrow \frac{h _{2}}{2} = (0.6)^{2} = 0.36$

$\Rightarrow h_{2} = 0.72 m$

Q5. Two bodies of the same mass are moving with the same speed V in mutually opposite directions. They collide and stick together, Find the resultant velocity of the system?

Sol.

Two bodies moving with the same velocity in the opposite direction

Two bodies moving having mass m and moving with same velocity v, after collision two bodies stick together with common velocity $v_{f}$ . By using Conservation of linear momentum,

$p_{i} = p_{f}$

$m v + (- v) = 2 m v_{f}$

$0 = 2 m v_{f}$

$\Rightarrow v_{f} = 0$

Q.6 A bullet having mass m and speed v is fired into a large block of wood with mass M. Determine the final velocity of the combined system.

Sol. By using Conservation of Linear Momentum

$p_{i} = p_{f}$

$m v + M (O) = (m + M) V_{system}$

$V_{system} = (\frac{m}{m + M}) v$

Q.7 A 50-gram bullet moving with a velocity of 10m/s gets embedded into a 950-gm stationary body. Find the percentage loss in Kinetic energy of the system?

Sol.

$Δ K = \frac{1}{2} μ u_{re l}^{2}$

Percentage Loss in K. E $= \frac{Change in K . E}{Initial K . E} \times 100$

$\frac{Δ K . E}{K \cdot E _{i}} \times 100 = \frac{\frac{1}{2} \frac{50 \times 950}{1000} \times 100}{\frac{1}{2} 50 \times 100} \times 100 = 95%$

Q-8.Find speed just after collision?

Solution:

$e v = \frac{1}{2} \times 2 = 1 m / s$

Q-9.Find final velocities of balls:

(a) e=1

(b) e=0

Solution:

(a)

$V_{CM} = \frac{m ( v _{0} ) + m ( 0 )}{2 m} = \frac{v _{0}}{2}$

(b)

$v = V_{COM}$

(C )

Two uneven mass moving in the same direction with unequal velocity

$\frac{1}{2} = \frac{v _{2} - v _{1}}{v _{0}}$

$v_{0} = 2 v_{2} - 2 v_{1}$

$m v_{0} = (\frac{m v _{0}}{2} + \frac{m v _{0}}{2})$

$m v_{0} = m (v_{1} + v_{2})$

$v_{1} + v_{2} = v_{0}$

$v_{2} = \frac{3 v _{0}}{4}$ , $v_{1} = \frac{v _{0}}{4}$

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