Conservation of Momentum in Collisions

Linear Momentum

The total amount of motion possessed by a moving body is known as the momentum of the body.

It is the product of the mass and velocity of a body.
It is a vector quantity whose direction is along the instantaneous velocity.

Momentum, $p = m v$

$Momentum = Mass \times Velocity$

The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.

The direction of the momentum vector is the same as the direction of the velocity vector.
SI unit: kg-m/s, N-s ; Dimension: $[M L T^{- 1}]$

1.0Principle of Conservation of Momentum

Conservation of Linear Momentum

According to Newton’s second law,if net external force on the system is zero,then linear momentum of the system remains conserved. According to Newton’s second law.

$F_{n e t} = \frac{d p}{d t}$

If $F_{n e t} = 0 \Rightarrow \frac{d p}{d t} = 0 \Rightarrow p = constant \Rightarrow p_{in} = p_{Final}$

For two particle system $p_{1} + p_{2} = constant$

$Δ p_{1} + Δ p_{2} = 0$

$Δ p_{1} = - Δ p_{2}$

$(change in momentum of 1^{st} particle = - change in momentum of 2^{nd} particle)$

2.0Examples of the Law of Conservation of Momentum

The following are examples of the law of conservation of momentum:

Air-filled balloons
System of gun and bullet
Motion of rockets

Conservation of Linear Momentum of System And Impulse

As we have studied that

$F_{e x t} = \frac{d ( M v _{CM} )}{d t} = \frac{d ( p _{system} )}{d t}$

Now, If $F_{ext} = 0 Then, p_{system} = constant$

It means that the total linear momentum of a system of particles remains conserved in a time interval in which the impulse of external forces is zero. The total momentum of a system of particles cannot change under the action of internal forces, and if the net impulse of the external forces in a time interval is zero, the total momentum of the system in that time interval will remain conserved.

$p_{final} = p_{initial}$

The above statement is known as the principle of conservation of momentum.

Since force, impulse and momentum are vectors, the component of momentum of a system in a particular direction is conserved, if net impulse of all external forces in that direction vanishes.

$No external force \Rightarrow stationary mass relative to an inertial frame remains at rest$

Coefficient of Restitution

Usually, the force D applied by the bodies A and B on each other during the period of deformation differs from the force R applied by the bodies on each other during the period of restitution. Therefore, it is not necessary that the magnitude of the impulse due to deformation equals that of the impulse $\int R d t$ due to restitution.

The ratio of the magnitudes of impulse of restitution to that of deformation is called the coefficient of restitution and is denoted by e.

Coefficient of restitution depends on various factors such as elastic properties of materials forming the bodies, velocities of the contact points before impact, state of rotation of the bodies and temperature of the bodies.In general, its value ranges from zero to one but in collisions where additional kinetic energy is generated, its value may exceed one. Depending on the values of the coefficient of restitution, two particular cases are of special interest.

Perfectly Plastic or Inelastic Impact For these impacts e = 0, and bodies undergoing impact move with same velocity after the impact.

Perfectly Elastic Impact For these impacts e = 1.

Strategy to solve problems of head -on impact:

Write the momentum conservation equation;

$m_{A} v_{A} + m_{B} v_{B} = m_{A} u_{A} + m_{B} u_{B} \dots\dots\dots (A)$

Write the equation involving coefficient of restitution

$v_{B} - v_{A} = e (u_{A} - u_{B}) ........ (B)$ the

3.0Types of Collisions

Collisions

Collisions are broadly classified into three types:

Head on Collision

If the velocity vectors of the colliding bodies are directed along the line of impact, the impact is called a direct or Head-on impact.

Head on Elastic Collision

The head-on elastic collision is one in which the colliding bodies move along the same straight-line path before and after the collision.

Assuming initial direction of motion be positive and $u_{1} > u_{2}$ (so that collision may take place) and applying law of conservation of linear momentum

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \Rightarrow m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2}) \dots\dots\dots (1)$

For an elastic collision, kinetic energy before collision must be equal to kinetic energy after collision. i.e.,

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} \Rightarrow m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - u_{2}^{2}) \dots\dots .. (2)$

Dividing equation (2) by (1)

$u_{1} + v_{1} = v_{2} + u_{2} \Rightarrow (u_{1} - u_{2}) = (v_{2} - v_{1}) \dots\dots .. (3)$

In 1-D elastic collision 'velocity of approach' before collision is equal to the 'velocity of separation after collision, no matter what the masses of the colliding particles are. This law is called Newton's law for elastic collision.

The value of coefficient of restitution for elastic collision ; $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = 1$

If we multiply equation(3) by m_2 and subtract it from (1)

$(m_{1} - m_{2}) u_{1} + 2 m_{2} u_{2} = (m_{1} + m_{2}) v_{1}$

$v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2}$

Similarly multiplying equation(3) by m_1 and subtract it from (1)

$2 m_{1} u_{1} + (m_{2} - m_{1}) u_{2} = (m_{2} + m_{1}) v_{2} \dots\dots\dots . (4)$

$v_{2} = \frac{m _{2} - m _{1}}{m _{1} + m _{2}} u_{2} + \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} \dots\dots\dots (5)$

Special cases of head on elastic collision

Case-1: If two bodies are of equal masses: $m_{1} = m_{2} = m$ then $v_{1} = u_{2}$ and $v_{2} = u_{1}$ . Thus if two bodies of equal masses undergo an elastic collision in one dimension, then the bodies exchange their velocities after the collision.

Case-2: If two bodies are of equal masses and the second body is at rest

$m_{1} = m_{2}$ and the initial velocity of the second body $u_{2} = 0$ , then $v_{1} = 0$ and $v_{2} = u_{1}$ . When body A collides against body B of equal mass at rest, then body A comes to rest and body B moves on with the velocity of body A. In this case transfer of energy is hundred percent e.g. Billiard's Ball, Nuclear moderation.

Case-3: If the mass of one body is negligible as compared to other

If $m_{1} ≫ m_{2}$ and $u_{2} = 0$ then $v_{1} = u_{1}, v_{2} = 2 u_{1}$

When a heavy body A collides against a light body B at rest, then body A should keep on moving with same velocity whereas body B moves with velocity double that of A.

If $m_{2} ≫ m_{1}$ and $u_{2} = 0$ then $v_{2} = 0, v_{1} = - u_{1}$

When a light body A collides against a heavy body B at rest, body A starts moving with the same speed just in the opposite direction while body B practically remains at rest.

(b) Inelastic Collisions

Head on Inelastic Collision

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \dots\dots .. (1)$

By definition of coefficient of restitution

$= (v_{2} - v_{1}) = e (u_{1} - u_{2}) .... (2)$

$v_{1} = (\frac{m _{1} - e m _{2}}{m _{1} + m _{2}}) u_{1} + (\frac{( 1 + e ) m _{2}}{m _{1} + m _{2}}) u_{2}$

$v_{2} = (\frac{m _{2} - e m _{1}}{m _{1} + m _{2}}) u_{2} + (\frac{( 1 + e ) m _{1}}{m _{1} + m _{2}}) u_{1}$

Can also be expressed as

$v_{1} = \frac{p _{i} + m _{2} e ( u _{2} - u _{1} )}{m _{1} + m _{2}} v_{2} = \frac{p _{i} + m _{1} e ( u _{1} - u _{2} )}{m _{1} + m _{2}} (p_{i} = m_{1} u_{1} + m_{1} u_{2})$

Loss in Kinetic Energy of particles $= \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (1 - e^{2}) (u_{1} - u_{2})^{2}$

Value of the coefficient of restitution for an inelastic collision is 0<e<1

(c) Perfectly Inelastic Collisions

Head on Perfectly Inelastic Collision

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

Where final velocity $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$

Loss in Kinetic Energy of particles $= \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}$

Value of the coefficient of restitution for an inelastic collision is e=0

Oblique Collision

A collision in which the particles move in the same plane at different angles before and after the collision is called an oblique collision.

BY COLM along x-axis,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} cos θ + m_{2} v_{2} cos ϕ$

BY COLM along y-axis,

$0 + 0 = m_{1} v_{1} sin θ - m_{2} v_{2} sin ϕ$

By conservation of kinetic energy,

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ Its

4.0Impulse and Its Relation to Momentum

Impulse

Impulse of a force $F$ acting on a body for the time interval $t = t_{1}$ to $t = t_{2}$ is defined as

$I = \int_{t_{1}}^{t_{2}} F d t \Rightarrow I = \int F d t = \int m \frac{d v}{d t} d t = \int m d v$

$I = m (v_{2} - v_{1}) = Δ P$ { change in momentum due to force }

Also, $I_{Res} = \int_{t_{1}}^{t_{2}} F_{Res} d t = Δ P (impulse - momentum theorem)$

Note: Impulse applied to an object in a given time interval can also be calculated from the area under force time (F-t) graph in the same time interval

Instantaneous Impulse :

There are many cases when a force acts for such a short time that the effect is instantaneous, e.g., a bat striking a ball. In such cases, although the magnitude of the force and the time for which it acts may each be unknown, the value of their product (i.e., impulse) can be known by measuring the initial and final momenta. Thus, we can write

$I = \int F d t = Δ P = P_{f} - P_{i}$

Important Points :

It is a vector quantity.
Dimensions = $[M L T^{- 1}]$
SI unit = kg m/s.
Direction is along change in momentum.
Magnitude is equal to area under the F–t graph.
$I = \int F d t = F_{a v} \int d t = F_{a v} Δ t$
It is not a property of a particle, but it is a measure of the degree to which an external force changes the momentum of the particle.

5.0Real-Life Applications of Collisions

Vehicle Safety: Airbags and crumple zones reduce force by increasing collision time.
Sports: Billiards, snooker, and cricket balls involve momentum transfer.
Space Physics: Elastic collisions of particles in cosmic phenomena.
Engineering: Particle accelerators, Newton’s cradle, and collision tests.

These applications show the practical relevance of momentum conservation principles in real-world scenarios.

6.0Solved Problems on Momentum in Collisions

Illustration-1: A ball of mass 2 kg moving with a speed of 5 m/s collides directly with another ball of mass 3 kg moving in the same direction with a speed of 4 m/s. The coefficient of restitution is $\frac{2}{3}$ . Find the velocities after the collision.

Solution:

Denoting the first ball by A and the second ball by B, velocities immediately before and after the impact are shown in the figure.

By COLM

$2 (5) + 3 (4) = 2 v_{1} + \Rightarrow 2 v_{1} + 3 v_{2} = 22 \dots\dots . (1)$

By definition of e

$e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} \Rightarrow \frac{2}{3} = e = \frac{v _{2} - v _{1}}{5 - 4} \Rightarrow 3 v_{2} - 3 v_{1} = 2........ (2)$

By solving equation(1) and (2)

$v_{1} = 4 m / s and v_{2} = 4.67 m / s$

Illustration-2.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 collision both bodies stick together and move with a common velocity. Find the velocity in m/s.

Solution:

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

Illustration-3.A body strikes obliquely with another identical stationary rest body elastically.Prove that they will move perpendicular to each other after collision.

Solution:

Conservation of linear momentum in x-direction gives

$m u_{1} = m v_{1} cos θ_{1} + m v_{2} cos θ_{2} \Rightarrow u_{1} = v_{1} cos θ_{1} + v_{2} cos θ_{2} \dots\dots (1)$

Conservation of linear momentum in y-direction gives

$0 = m v_{2} sin θ_{2} - m v_{1} sin θ_{1}$

$0 = v_{2} sin θ_{2} - v_{1} sin θ_{1} \dots\dots .... (2)$

By conservation of kinetic energy,

$\frac{1}{2} m u_{1}^{2} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} \Rightarrow u_{1}^{2} = v_{1}^{2} + v_{2}^{2} \dots\dots (3)$

By $(1)^{2} + (2)^{2}$

$u_{1}^{2} + 0 = v_{1}^{2} cos^{2} θ_{1} + v_{2}^{2} cos^{2} θ_{2} + 2 v_{1} v_{2} cos θ_{1} cos θ_{2} + v_{1}^{2} sin^{2} θ_{1} + v_{2}^{2} sin^{2} θ_{2} - 2 v_{1} v_{2} sin θ_{1} sin θ_{2}$

$u_{1}^{2} = v_{1}^{2} (cos^{2} θ_{1} + sin^{2} θ_{1}) + v_{2}^{2} (cos^{2} θ_{2} + sin^{2} θ_{2}) + 2 v_{1} v_{2} (cos θ_{1} cos θ_{2} - sin θ_{1} sin θ_{2})$

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} cos (θ_{1} + θ_{2})$

$cos (θ_{1} + θ_{2}) = 0 \Rightarrow θ_{1} + θ_{2} = 9 0^{\circ}$

Conservation of Momentum in Collisions

1.0Principle of Conservation of Momentum

2.0Examples of the Law of Conservation of Momentum

Coefficient of Restitution

3.0Types of Collisions

Collisions

(b) Inelastic Collisions

(c) Perfectly Inelastic Collisions

Oblique Collision

4.0Impulse and Its Relation to Momentum

5.0Real-Life Applications of Collisions

6.0Solved Problems on Momentum in Collisions

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