Modelling collisions - class-XII

When two identical bodies undergo a one-dimensional elastic collision, energy conservation and momentum conservation together are necessary and sufficient to uniquely determine the final velocities, demonstrating that both principles are required.

For an elastic collision between two identical bodies, the velocities are simply exchanged. Ball A initially moving at 2 m/s and Ball B moving at 3 m/s along the same line will swap velocities, meaning Ball A moves at 3 m/s and Ball B moves at 2 m/s after the collision.

Using conservation of angular momentum about the axis of the cylinder: L_initial = Iomega + mvR. L_final = (I + mR^2)omega_final. Equating them gives omega_final = (Iomega + mvR) / (I + m*R^2).

Collision does not require physical contact because if we go by definition of collision is the meeting of particles or of bodies in which each exerts a force upon the other causing exchange of energy and momentum. In short we can say that collision is a event in which two bodies, exert force on each other for a relative short time. however mostly he study collision in terms of physical contact since force applied is mechanical force. Mostly force between two bodies if it is not physical then it would not be studied as collision more as force applied to bodies.

Heat is energy in transit due to a temperature difference; it is not a property of the system (it does not 'reside' in a system). Internal energy resides in a system.

"Man jumping on cart" and "bullet imbedded in a block" are examples when two bodies have to move together after collision, therefore both of them are inelastic collision.
Where as it is not the case for "collision of two glass balls".

Elastic collisions occur only if there is no net conversion of kinetic energy into other forms. During the elastic collision, kinetic energy is first converted to potential energy associated with a repulsive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).
Also, the average of the momenta before and after the collision is the same. 
Hence, in a perfectly elastic collision both KE and momentum are conserved.

In elastic collision both momentum and kinetic energy are conserved.

Change in momentum, $\Delta p =- 2mv$
$= 2 \times 0.2\times 40 = 16 kg ms^{-1}$(magnitude)

Momentum of Earth ball system is conserved because no external force is acting on the system.

Total momentum and total energy of a system always remain conserved.

In any type of collision, linear momentum is conserved, unless external force acts on the system. elasticity determines the loss in KE. elastic => no loss. inelastic => max. loss of KE ($KE _i - KE _{COM}$)

In an elastic collision there is no loss of kinetic energy. If in the given case , balls are of same masses then their kinetic energies will also be same because their velocities are equal and as said , only one ball is moving before and after collision therefore total kinetic energy is constant i.e. it is an elastic collision.

During collision, net external force acting on the colliding bodies is zero, thus total momentum of the system is conserved in all types of collision.

The case is of an inelastic collision. In such collisions, energy(kinetic energy) is lost and is therefore not conserved. But, the momentum of the system is conserved in every collision .

By the definition of head on collision, the two bodies have to move along the line joining their centers and their velocities can be in any directions (only two possible here).

In elastic collision no energy loss takes place that is total energy is transferred.
However, in inelastic collision, energy loss takes place and only linear momentum is conserved.

According to the Hooke's law,

Since no external forces are acting on the colliding bodies during collision, thus total linear momentum is always conserved in all type of collisions but kinetic energy in not conserved in all collisions.

In inelastic collisions , the kinetic energy is used for deforming the the bodies . In such collisions , the reformation of the bodies is partial. Therefore, the potential energy stored in the deformation is lost as heat and the kinetic energy of the system after collision is less than that of before collisions. However, in some cases it may be greater ( such collisions are called super elastic collisions).

1. K.E will be less since collision is inelastic
   2) No impulsive external force on system (earth + ball)
   

In an inelastic collision the relative velocity of approach is more than the relative velocity of separation. Hence B is wrong

Lets consider first pendulum bob of mass $m _1$ and second  pendulum bob of mass $m _2$ collide elastically at the lower point in their motion. If both are released  from the height $H$ above the lower point.

Two particles will collide if their relative position vector is parallel (or antiparallel) to their relative velocity vector. This condition is expressed mathematically by setting the unit vector of relative position equal to the negative of the unit vector of relative velocity, which simplifies to the ratio of position difference vectors matching the direction of velocity difference.

At t=1s, A is at y = 20(1) - 0.5(10)(1)^2 = 15m, velocity vA = 20 - 10(1) = 10 m/s. B is at y = 20 - 0.5(10)(0)^2 = 20m, velocity vB = 0 m/s. They collide at some point. Since masses are equal and collision is elastic, they exchange velocities. The calculation of the exact collision point and post-collision velocities confirms option A.

In inelastic collision momentum of the system is always conserved if $F _{ext}=0$.
Velocity of separation is less than the velocity of approach since co-efficient of restitution e < 1
$e=0$ for a perfectly inelastic collision since the colliding particles stick together after collision.
Hence, option (D) is correct.

Conservation of momentum: mvi = mv1j + mv2cos(theta)i + mv2sin(theta)j. Since v1 = v/sqrt(3) in the y-direction, the x-component of momentum is conserved: mv = mv2cos(theta), and y-component: 0 = mv1 + mv2sin(theta). Solving these gives v2 = 2*v/sqrt(3).

• Elastically means the balls get bounced with $same$ speed i.e $u$
• so the change in momentum will be $P _2-P _1=-mu-mu=-2mu$
• so the momentum tranferred to the surface will be $-(-2mu)=2mu$
• thus the total momentum transferred by $n$ balls will be $2mnu$
• force $F=\dfrac{Momentum}{time}=\dfrac{2mnu}{1sec}=2mnu$

If a body moving towards a finite body at rest collides with it then momentum will be conserved and hence, the velocities after collision are may be in inverse proportion to their masses or may get interchanged. Hence, both the bodies move after collision or the moving body comes to rest and stationary body starts moving.

both bodies cannot come to rest as  it will violate the law of conservation of momentum.
Option (D) will also violate conservation of momentum.

In a head-on elastic collision between a small projectile and a more massive target, the projectile will bounce back with low speed and the massive target will be given a very small velocity. Hence, $m _1 < m _2$

From the very fundamental law of collision we know that in a collision momentum of the system is conserved .
Therefore the only possible option is C.

The loss of kinetic energy in a completely inelastic collision is given by 1/2 * (m1 * m2 / (m1 + m2)) * (v1 - v2)^2. Substituting m1 = 20 tonnes, m2 = 60 tonnes, and initial relative speed 10 m/s yields 750 kJ. Thus, option B is correct.

Head-on collision is always one dimensional the centre of mass of each body move in the same direction after collision as they were moving before collision. 

The law of conservation of momentum is true in all type of collisions, but kinetic energy is conserved only in elastic collision. The kinetic energy is not conserved in inelastic collision but the total energy is conserved in all type of collisions.

Let ${u} _{1}$ be the speed of body initially before collision.
${u} _{2}=0$
Let ${v} _{1}$ be speed of particle 1 after collision and ${v} _{2}$ be speed of particle 2 after collision.
Using law of conservation of momentum
$m{u} _{1}=m{v} _{1}+m{v} _{2}$
${u} _{1}={v} _{1}+{v} _{2}$
Coefficient of restitution will be given by
$e=\dfrac{{v} _{2}-{v} _{1}}{{u} _{1}}=\dfrac{{v} _{2}-{v} _{1}}{{v} _{1}+{v} _{2}}$
$e{v} _{1}+e{v} _{2}={v} _{2}-{v} _{1}$
Dividing throughout by ${v} _{2}$ and rearranging leads to
$\dfrac{{v} _{1}}{{v} _{2}}=\dfrac{1-e}{1+e}$

Linear momentum is conserved in elastic as well as inelastic collision but kinetic energy is conserved only in case of elastic collision but some kinetic energy is lost in inelastic collision.

A shocker is a mechanical or hydraulic device designed to absorb and
damp shock impulses. It does this by converting the kinetic energy of
the shock into another form of energy (typically heat) which is then
dissipated. So, shocker is used.

Energy conservation is valid everywhere i.e. total energy is conserved everytime. In collision, the kinetic energy of the system may get lost. However, the momentum of the system is always conserved and so the change in the momentum of the system is zero.

We know that during an inelastic collision, there is a loss of kinetic energy, some of the kinetic energy is used to deform the body and some of the kinetic energy is liberated as heat. But their is no loss of mass-energy.

In an elastic collision, e is equal to 1. Therefore, the kinetic energy used in causing deformation of the bodies is recovered completely in the reformation since the reformation is complete. So no energy is lost. 

A collision is considered elastic collision if the kinetic energy before collision is equal to the kinetic energy after collision.

In collision of nuclei and fundamental particles, unless there is emission of gamma rays, kinetic energy of the system remains constant.

In Alpha particle emission, energy is not being released in any other form. Hence kinetic energy of the system will remain same before and after the emission.

In collision of two ivory balls, the balls being extremely hard, no deformation happens and for most practical purposes energy is assumed to not being converted into any other form such as heat(which happens when there is friction or deformation).

In the case of a running man jumping on to the train, friction and deformation happens in order for the man to continue moving along with the train. Thus energy was converted into other forms. Hence, it is an example of inelastic.

The co-efficient of restitution for a perfectly elastic collision is $e = 1$

A) Common result.
B) e.g Ball hitting a wall
C) $e= \dfrac{O-V}{V-V^{1}} = 1$
$V^{1} = 2V$
D) As in C

During an oblique collision of ball with floor , the only force considered is normal reaction force which is always perpendicular to surface until clearly given rough surface no force parallel to surface is taken into account.

If $e = 1$, then the collision is called perfectly elastic.
If $0 < e <1$, the collision is called inelastic.
If $e = 0$,  the collision is called perfectly inelastic.

Let initial velocity is v at time of collision. $v = \sqrt { 2gh } $

For a perfectly inelastic collision of equal masses m, mv1 + mv2 = 2mv_final. If v_final = v/2, then v1 + v2 = v. This implies the angle between the initial velocity vectors must be 90 degrees.

Nothing is mentioned about coefficient of restitution.