Inelastic Collision Formula: Derivation and Sample Questions

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Inelastic collision is a type of collision in which the momentum is conserved but kinetic energy does not remain conserved as a bit of it is lost in the collision of two bodies. The loss of the kinetic energy is in the form of some other type of energy such as thermal energy or sound energy. For example, the car crashing against the tree and the car stops completely. There are two types of inelastic collision which are named as perfectly inelastic collision and partially inelastic collision.

Also Read: Inelastic collision

Table of Content

Types of inelastic collision
1. Perfectly inelastic collision
2. Partially inelastic collision
Formula of inelastic collision
Things to Remember
Sample Questions

Key Terms: Perfectly inelastic collision, elastic and inelastic collision, inelastic collision examples, inelastic collision formula, Mass, Initial Velocity

Types of Inelastic Collision

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There are mainly two types of inelastic collision:

Perfectly Inelastic Collision

When two bodies collide and one sticks to the other completely, then the collision is called a perfectly inelastic collision. In this, the maximum amount of kinetic energy is lost due to the energy loss in bonding of two bodies.

Momentum Transfer

Momentum Transfer

Partially Inelastic Collision

Partially inelastic collisions are the most common form of collisions in the real world. In this type of collision, the objects involved in the collisions do not stick, but some kinetic energy is still lost. Friction, sound and heat are some ways the kinetic energy can be lost through partial inelastic collisions.

Also Read: Elastic collision

Formula of Inelastic Collision

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Let us consider two bodies A and B with masses m₁ and m₂ respectively. Suppose that initially A moves with the velocity u₁ and B is moving with the velocity u₂.

So, the perfectly inelastic collision happens between both bodies and they stick together and move with the velocity v.

Law of conservation of momentum states that for a collision occurring between object A and object B in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

So, by the law of conservation of momentum

Momentum of object A + Momentum of object B = Momentum of the collided body

Here, momentum of object A = m₁ × u₁

momentum of object B = m₂ × u₂

momentum of collided body = (m₁ + m₁) × v

effects of collison

Putting the value gives,

m₁ u₁ + m₂ u₂ = (m₁ + m2) v

v = m₁U₁+m₂U₂m₁+m₂

This formula is used to calculate the final velocity of the objects after the collision in one dimension if the collision is a perfectly inelastic collision.

Things to Remember

In all types of collision, momentum remains conserved.
In a perfectly inelastic collision, the two collided bodies stick together and move with the same velocity v.

v = m₁U₁+m₂U₂m₁+m₂

Coefficient of restitution generally lies between 0 and 1.
Coefficient of restitution is 0 for the perfectly inelastic collision.
Most collisions in nature are inelastic collisions.
Elastic and inelastic collision differs because of conservation of kinetic energy.

Also Read: Elastic and Inelastic collision

Sample Questions

Question 1: What is the difference between elastic and inelastic collision? (2 Marks)

Answer: Kinetic energy is the physical quality which differentiate the types of collision. In elastic collision, the kinetic energy remains conserved along with the momentum but there is some energy loss in the inelastic collision. This loss is in some form of energy such as thermal or sound energy.

Question 2: What is an elastic collision? (3 Marks)

(a)An elastic collision is one in which the objects after impact are deformed permanently.
(b)An elastic collision is one in which the objects after impact lose some of their internal kinetic energy.
(c)An elastic collision is one in which the objects after impact do not lose any of their internal kinetic energy.
(d)An elastic collision is one in which the objects after impact become stuck together and move with a common velocity.

Answer: (c)

Explanation: An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms such as heat, noise, or potential energy.

Question 3: What is coefficient of restitution for an inelastic collision? (2 Marks)

Answer: The coefficient of restitution (COR, also denoted by e), is the ratio of the final to initial relative speed between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic collision. A perfectly inelastic collision has a coefficient of 0, but a 0 value does not have to be perfectly inelastic.

Question 4: What is the equation for conservation of momentum for two objects in a one-dimensional collision? (3 Marks)
p1 + p1′ = p2 + p2′
p1 + p2 = p1′ + p2′
p1 − p2 = p1′ − p2′
p1 + p2 + p1′ + p2′ = 0
where p1 and p1′ is the initial and final momentum of object A and p2 and p2′ is the initial and final momentum of object B.

Answer: (b)

Explanation: Law of conservation of momentum states that for a collision occurring between object A and object B in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Initial Momentum of object A + Initial Momentum of object B = Final Momentum of the object A + Final Momentum of object B.

So, p1 + p2 = p1′ + p2′

Question 5: Give an example of a perfectly inelastic collision? (2 Marks)

Answer: In a perfectly inelastic collision, the collided bodies stick together. The example of the perfectly inelastic collision is when a ball made of clay is dropped on the ground from some height. It does not bounce back but sticks to the floor. The loss of kinetic energy is in the form of the deformation of the ball shape.

Question 6: Why is energy loss in an inelastic collision? (2 Marks)

Answer: The most energy in an inelastic collision is lost if the two colliding objects stick together and act as one object with a mass that is the sum of the two masses. Such a collision deforms the objects, which causes the loss in kinetic energy and an equal increase in heat or sound.

Question 7: Two football players collide head-on in mid air while trying to catch a thrown football. The first player is 95 kg and has an initial velocity of 6.00 m/s, while the second player is 115 kg and has an initial velocity of 3.50 m/s. What is their velocity just after impact if they cling together? (3 Marks)

Answer: We know that, v = m₁U₁+m₂U₂m₁+m₂

Here, m₁ = 95 kg, m₂ = 115 kg

And u₁= 6.00 m/s, u₂ = 3.50 m/s

So, putting value in the equation

v = 95 ×6+115 × 3.5095+115 m/s

v = 570 + 402.50210 m/s

v = 972.5210 m/s

v = 4.63 m/s

Hence, the velocity just after the impact if they cling together is 4.63 m/s.

Question 8: What is the speed of a garbage truck that is 1.20 × 104 kg and is initially moving at 25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at rest? (3 Marks)

Answer: Let us consider garbage truck as object A and trash can as object B.

So, m₁ = 1.20 × 104 kg, m₂ = 80 kg

And u₁= 25.00 m/s, u₂ = 0 m/s (as trash can is at rest so u₂ = 0 m/s)

We know that, v = m₁U₁+m₂U₂m₁+m₂

Putting the value in equation,

v = 12000 × 25+80 × 012000+80 m/s

v = 30000012080 m/s

v = 24.83 m/s

Hence, the speed of the truck after colliding with the trash can is 24.83 m/s.

Question 9: How change of kinetic energy is calculated in a perfectly inelastic collision? (2 Marks)

Answer: Two objects A and B with masses m₁ and m₂ respectively moving with the velocity of u₁ and u₂. The initial kinetic energy of the objects is:

KE₁ = 12 m₁ u₁² + 12 m₂ u₂²

After they collide, the kinetic energy is

KE₂ = 12 (m₁ + m₂) v₂

Where v is the final velocity when they stick together.

So, the change in kinetic energy KE is

KE = KE₁ – KE₂

KE = 12 m₁ u₁² + 12 m₂ u₂² - 12 (m₁ + m₂) v₂

Question 10: During an ice show, a 60 kg skater leaps into the air and is caught by an initially stationary 75 kg skater. What is their final velocity assuming negligible friction and that the 60 kg skater’s original horizontal velocity is 4.00 m/s? Also, calculate the change in kinetic energy? (3 Marks)

Answer: We know that, v = m₁U₁+m₂U₂m₁+m₂

Here, m₁ = 60 kg, m₂ = 75 kg

And u₁= 4.00 m/s, u₂ = 0 m/s

So, putting value in the equation

v = 60 ×4+75 × 060+75 m/s

v = 240135 m/s

v = 1.78 m/s

Hence, the final velocity is 1.78 m/s.

Also, the initial kinetic energy is

KE₁ = 12 m₁ u₁² + 12 m₂ u₂²

KE₁ = 12( 60 )× 4² + 12(75) × 0²

KE₁ = 12 (60) × 16 + 0

KE₁ = 480 J

The final kinetic energy is

KE₂ = 12 (m₁ + m₂) v₂

KE₂ = 12 (60 + 75) 1.782

KE₂ = 12 (135) 3.1684

KE₂ = 213.867 J

Hence, change in kinetic energy is

KE = KE₁ – KE₂

KE = 480 – 213.867 J

KE = 266.133 J

Also Read:

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Inertia	Unit of Acceleration	Types of Springs
Definition of Micrometer	What is Density	Types of Radiation
Pulley	Unit of Power	What is Pulley?
Types of Energy	Elastic Collision	Scalar and Vector Quantities

Inelastic Collision Formula: Derivation and Sample Questions

Types of Inelastic Collision

Perfectly Inelastic Collision

Partially Inelastic Collision

Formula of Inelastic Collision

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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Inelastic Collision Formula: Derivation and Sample Questions

Types of Inelastic Collision

Perfectly Inelastic Collision

Partially Inelastic Collision

Formula of Inelastic Collision

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

2.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

5.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: