Physics ELASTIC AND INELASTIC COLLISIONS, COLLISIONS IN ONE DIMENSION AND TWO DIMENSION
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CLASS 11 CHAPTER 6 - WORK, ENERGY AND POWER

ELASTIC AND INELASTIC COLLISIONS, COLLISIONS IN ONE DIMENSION AND TWO DIMENSION

Topic covered

`color{red} star` COLLISIONS
`color{red} star` ELASTIC AND INELASTIC COLLISIONS
`color{red} star` COLLISIONS IN ONE DIMENSION AND TWO DIMENSION

COLLISIONS

`➢` In this section we shall apply these laws to a commonly encountered phenomena, namely collisions.
`➢` We shall study the collision of two masses in an idealised form. Consider two masses `m_1` and `m_2`. The particle `m_1` is moving with speed `v_(1i)` , the subscript ‘i’ implying initial.
`➢` We can cosider `m_2` to be at rest. No loss of generality is involved in making such a selection. In this situation the mass `m_1` collides with the stationary mass ` m_2` and this is depicted in Fig. 6.10.

`➢` The masses `m_1` and `m_2` fly-off in different directions. We shall see that there are relationships, which connect the masses, the velocities and the angles.

Elastic and Inelastic Collisions

`➢` In all collisions the `"total linear momentum is conserved;"` the initial momentum of the system is equal to the final momentum of the system.
`➢` When two objects collide, the mutual impulsive forces acting over the collision time `Δt` cause a change in their respective momenta :

`color {blue}{Δ p_1=F_(12) Δt}`

`color {blue}{Δp_2=F_21 Δt}`



`=>` where `F_12` is the force exerted on the first particle by the second particle. `F_21` is likewise the force exerted on the second particle by the first particle.

Now from Newton’s Third Law, `F_(12) = − F_(21)`. This implies

`color {blue}{Δ p_1 + Δp_2=0}`



`➢` The above conclusion is true even though the forces vary in a complex fashion during the collision time `Δt.`

`➢` Since the third law is true at every instant, the total impulse on the first object is equal and opposite to that on the second.
`color{red}☞` On the other hand, the total kinetic energy of the system is not necessarily conserved.

`➢` The impact and deformation during collision may generate heat and sound. Part of the initial kinetic energy is transformed into other forms of energy.
`➢` A useful way to visualise the deformation during collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains its original shape without loss in energy, then the initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time `Δt` is not constant. Such a collision is called an elastic collision.

`➢` On the other hand the deformation may not be relieved and the two bodies could move together after the collision. A collision in which the two particles move together after the collision is called a completely inelastic collision. The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision.

Collisions in One Dimension

`➢` Consider first a completely inelastic collision in one dimension. Then, in Fig. 6.10,

`color {blue}{θ_ 1 = θ _2 = 0}`

`m_1v_(1i) = (m_1+m_2)v_f` (momentum conservation)

`color {blue}{v_f =(m_1)/(m_1+m_2) v_(1i)}`................(6.23)

`➢` The loss in kinetic energy on collision is

`color{orange} {Δ K = 1/2 m_1v_(1i)^2 -1/2 (m_1 +m_2)v_f^2}`

`=1/2 m_1v_(1i)^2 -1/2 (m_1^2)/(m_1+m_2) v_(1i)^2` [using Eq. (6.23)]

`=1/2 m_1v_(1i)^2 [ 1- (m_1)/(m_1+m_2)]`

`=1/2 (m_1m_2)/(m_1+m_2) v_(1i)^2`



which is a positive quantity as expected.

`➢` Consider next an elastic collision. Using the above nomenclature with `θ_1 = θ_2 = 0`, the momentum and kinetic energy conservation
equations are

`color {blue}{m_1v_(1i)= m_1v_(1f)+m_2v_(2f)}`...........(6.24)

`color {blue}{m_1v_(1i)^2 =m_1v_(1f)^2 +m_2v_(2f)^2}`.............(6.25)

From Eqs. (6.24) and (6.25) it follows that,

`m_1v_(1i)(v_(2f)-v_(1i))=(m_1v_(1f) (v_(2f) -v_(1f))`

or `v_(2f)(v_(1i) -v_(1f))(v_(1i)+v_(1f))`

Hence,

`color {blue}{:. v_(2f) = v_(1f) + v_(1f)}`

...............(6.26)

Substituting this in Eq. (6.24), we obtain

`color {blue}{v_(1f)=((m_1-m_2))/(m_1+m_2) v_(1i)}`.............(6.27)

and `color {blue}{v_(2f)=(2m_1v_(1i))/(m_1+m_2)}`.............(6.28)

`➢` Thus, the unknowns `{v_(1f), v_(2f)}` are obtained in terms of the knowns `{m_1, m_2, v_(1i)}`.


`color{red} ►` Special casesof our analysis are interesting.
`color {blue}{{"Case I"}` : If the two masses are equal

`v_(1f)=0`

`color {blue}{v_(2f)=v_(1i)}`

`➢` The first mass comes to rest and pushes off the second mass with its initial speed on collision.

`color {blue}{{"Case II"}` : If one mass dominates, e.g. `m_2 > > m_1`

`color {blue}{v_(1f) ≃ v_(1i) quad v_(2f) ≃ 0}`



`➢` The heavier mass is undisturbed while the lighter mass reverses its velocity



`
Q 3129380211

Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically `10^7 m s^(–1)`) must be slowed to `10^3 m s^(–1)` so that it can have a high probability of interacting with isotope `text()_92^(235) U` and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (`D_2O`) or graphite, is called a moderator.

Solution:

The initial kinetic energy of the neutron is

`color{lime} {K_(ii)= 1/2 m_1v_(1i)^2}`

while its final kinetic energy from Eq. (6.27)

`color{purple} {K_(1f) =1/2 m_1 v_(1f)^2= 1/2 m_1 ((m_1-m_2)/(m_1+m_2))^2 v_(1i)^2}`

The fractional kinetic energy lost is

`f_1=(K_(1f))/(K_(1i))=((m_1-m_2)/(m_1+m_2))^2`

while the fractional kinetic energy gained by the moderating nuclei `K_(2f) //K_(1i)` is

`f_2 = 1 − f_1` (elastic collision)

`=(4m_1m_2)/((m_1+m_2)^2)`



One can also verify this result by substituting from Eq. (6.28).

For deuterium `m_2 = 2m_1` and we obtain `f_1 = 1//9` while `f_2 = 8//9`. Almost 90% of the neutron’s energy is transferred to deuterium. For carbon `f_1 = 71.6%` and `f_2 = 28.4%`. In practice, however, this number is smaller since head-on collisions are rare.
If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one-dimensional collision, orhead-on collision. In the case of small spherical bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 which is at rest. In general, the collision is two dimensional, where the initial velocities and the final velocities lie in a plane.

Collisions in Two Dimensions

`➢` Fig. 6.10 also depicts the collision of a moving mass `m_1` with the stationary mass `m_2`. Linear momentum is conserved in such a collision.
`➢` Since momentum is a vector this implies three equations for the three directions `{x, y, z}`. Consider the plane determined by the final velocity directions of `m_1` and `m_2` and choose it to be the x-y plane.
`➢` The conservation of the z-component of the linear momentum implies that the entire collision is in the x-y plane. The x- and y-component equations are

`color {blue}{m_1v_(1i) = m_1v_(1f) cos θ_1 + m_2v_(2f) cos θ_2}`................ (6.29)

`color {blue}{0= m_1v_(1f) sin theta_1-m_2 v_(2f) sin theta_2}`...............(6.30)

`➢` One knows `{m_1, m_2, v_(1i)}` in most situations. There are thus four unknowns `{v_(1f) , v_(2f) , θ_1` and `θ_2`}, and only two equations. If `θ_ 1 = θ_2 = 0`, we regain Eq. (6.24) for one dimensional collision. If, further the collision is elastic,

`color {blue}{1/2 m_1v_(1i)^2=1/2 m_1v_(1f)^2+ 1/2 m_2v_(2f)^2}`................(6.31)

`➢` We obtain an additional equation. That still leaves us one equation short. At least one of the four unknowns, say `θ_ 1`, must be made known for the problem to be solvable. For example, `θ_1` can be determined by moving a detector in an angular fashion from the `x` to the `y` axis. Given `{m_1, m_2, v_(1i) , θ_1}` we can determine `{v_(1f) , v_(2f) , θ_2}` from Eqs. (6.29)-(6.31).

`➢` The matter simplifies greatly if we consider spherical masses with smooth surfaces, and assume that collision takes place only when the bodies touch each other. This is what happens in the games of marbles, carrom and billiards.

`➢` In our everyday world, collisions take place only when two bodies touch each other. But consider comet coming from far distances to the sun, or alpha particle coming towards a nucleus and going away in some direction.
`➢` Here we have to deal with forces involving action at a distance. Such an event is called scattering. The velocities and directions in which the two particles go away depend on their initial velocities as well as the type of interaction between them, their masses, shapes and sizes
Q 3280467317

Consider the collision depicted in Fig. 6.10 to be between two billiard balls with equal masses `m_1 = m_2`.

The first ball is called the cue while the second ball is called the target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle `θ_2 = 37°`. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain θ 1.

Solution:

From momentum conservation, since the masses are equal

`color{purple} {v_(1i) = v_(1f) + v_(2f)}`

or `v_(1i)^2 = (V_(1f) + v_(2f)) * (v_(1f) + v_(2f) )`

`=v_(1f)^2 + v_(2f)^2 + 2v_(1f) * v_(2f)`

`= { v_(1f)^2 + v_(2f)^2 + 2 v_(1f)v_(2f) cos ( theta_1 + 37^o)}` ............1

Since the collision is elastic and m1 = m2 it follows from conservation of kinetic energy that

`color{orange} {v_(1f)^2 = v_(1f^2) + v_(2f)^2}` ....................2

Comparing Eqs. (1) and (2), we get

`cos (θ_1 + 37°) = 0`

or `θ_1 + 37° = 90°`

Thus, `θ_1 = 53°`

This proves the following result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other.
 
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