Physics CENTRE OF MASS AND COLLISION OF BODIES

COLLISION

The event or the process, in which two bodies either coming in contact with each other or due to mutual interaction at distance apart, affect each other motion (velocity, momentum, energy or direction of motion) is defined as a collision.

`text(Changes during collision/Impact :)`

(a) During impact each of the colliding bodies experiences a strong force, hence during impact each of the bodies changes its momentum, but as a whole, the total momentum of the system does not change.

(b) Impact is practically followed by emission of light, sound, heat etc. Therefore, during an impact or collision, the mechanical energy of the system does not remain constant whereas the total energy of the system remains unchanged.

(c) In ideal case of collision such as collision between gas molecules, atoms, electrons etc., the K.E. of the system of colliding particles remains constant before and after the impact. This type of collision is known as perfectly elastic collision. Remember that, during collision, the K.E. of the particles changes due to large impulsive force.

(d) In some collisions, the K.E. of the system changes. This type of collision cannot be called as perfectly elastic.

(e) For perfectly elastic collision, the K.E. of the system at the beginning and at the end of the impact must remain constant, but not during the impact.

(f) If the colliding particles/bodies stick together, they move together with same velocity. This type of collision is known as perfectly inelastic collision. In this process some times K.E. is lost completely. For example, in dropping a stone into mud, the stone loses its total K.E. Some times, K.E. is fractionally lost, for example, the collision of a bullet with a hanging target.

(g) Relative velocity of the bodies just after the collision ( velocity of separation) may or may not be equal to that before the collision along the line of impact (velocity of approach)

`text(Insight into collision :)`

Consider two balls moving as shown in the fig(1.a). If `v_1 > v_2`, both the balls will eventually collide. Just at the instant of collision the point of contact A has a larger velocity ( `v_1`) compared to the contact point B . Let us assume that the two bodies interact (remain in contact) for a small time `Deltat` , for this period points A and B must move through equal distance. This is possible only if the contact point gets deformed. The deformation is maximum when the two balls have equal velocities. A fraction of total kinetic energy gets converted into elastic potential energy. If the balls are made of elastic material, they regain their original shape and gets separated with changed velocities. During the entire process both the ball keep on pushing each other, as a result of which, the velocity of first ball decreases from `v_1` and that of second ball increases from `v_2`. If the balls regain their original shape, the entire elastic potential energy is converted back to kinetic energy. Remember that we are neglecting energy loss on account of heat developed, noise produced etc. If the material of the bodies are perfectly inelastic, they continue to move with the deformed shape. In this case both the bodies have a common velocity (when their deformation is maximum) and they move together. Collision is said to be perfectly inelastic in this case. For partially inelastic material, the deformed body tries to regain its shape but is capable of achieving it partially. Some kinetic energy is still lost and remains in form of elastic potential energy.

Classification of Collisions :

(1) On the basis of direction :

(a) `text(One-dimensional collision, Head on collision or Direct collision :)`
Collision in which the particles move along the same straight line before and after the collision.

(b) `text(Two dimensional collision or Oblique collision :)`
Collision in which the particle move in the same plane at different angles before and after collision

(2) On the basis of conservation of kinetic energy

(a) `text(Elastic collision :)`
Collision in which total K.E. before and after then collision remains the same.

(b) `text(ln-elastic collision :)`
Collision in which total K.E. before and after collision does not remain the same.

(c) `text(Perfectly Inelastic collision :)`
Collision in which particles get stick together after the collision. Loss of energy is maximum.

Line of Impact

The line of impact is the line along which the impulsive force acts on the bodies or the impact takes place.

Locating line of impact :

Step 1: Let two bodies 1 and 2 collide. First of all find the point of contact just before the collision.

Step 2: Draw a straight line passing through the point of contact of the bodies along the surface of collision. This is known as the common tangent to the surfaces in contact.

Step 3: After drawing the tangent, draw a normal to the tangent at that point. This normal line is known as line of impact.

Newton's law of collision

Let `(v_12 )_n` = component of relative velocity just after collision along line of impact( called velocity of separation)

`(u_(12) )_n` = component of relative velocity just before collision along line of impact( called velocity of approach)

According to Newton's law of collision when two bodies collide, the relative velocity after the impact bears a constant ratio with the relative velocity before impact along the line of impact. This constant ratio is known as coefficient of restitution for the impact, denoted by `'e'`.

`(v_(12))_n = - e (u_(12)_n)` Where `e` = coefficient of restitution for collision.

Note:
(a) `e` , a dimensionless quantity, is independent of shape and mass of object but depends on the material. It has constant value for a given pair of objects.

(b) `e = 1` : Collision is elastic `e = 0` : collision is perfectly inelastic 0 < e < 1 : collision is inelastic

(c) Definition of `e` can be applied only a long line of impacted.

`text(Velocity of approach and Velocity of separation In common situations :)`
Shown in Fig. (5.a), (5.b) & (5.c)

Analysis of Collision

(1) `text(Head on collision :)`

(a) Velocities after collision :
Two bodies of mass `m_1` and `m_2` moving uni-directionally collide such that their relative velocity is along the line of impact. Let the velocities before the collision be `vecu_1` and `vecu_2` and that just after the collision be `vecv_1`and `vecv_2` respectively. Assume that the direction of motion of the bodies remains the same just before and after the collision. Conserving the linear momentum along the line of impact we obtain

`m_1 vecu_1 + m_2 vecu_2 = m_1 vecv_1 + m_2 vecv_2`

Since all the momenta are unidirectional

`=> m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2`..........................(1)

Newton's experiment law yields

`e = - (v_1 - v_2)/(u_1 - u_2)`

`=> v_1 - v_2 = e(u_2 - u_1)` ................................(2)

Solving equations (1) and (2)

We obtain,

`v_1 = (m_1 - em_2)/(m_1 + m_2) u_1 + (m_2(1 + e))/(m_1 + m_2) u_2`

and

`v_2 = (m_1 (1 + e))/(m_1 + m_2) u_1 + (m_2 - em_1)/(m_1 + m_2) u_2`

Note : Take proper sign for `u_1` and `u_2`.

(b) Impulse of Bodies :
The linear momentum delivered to the `2^(nd)` ball= Impulse of the `1^(st)` ball on the `2^(nd)` ball & vice versa.

=> Impulse = `|triangle P_1| = |triangle P_2|`

= `|m_1v_1 - m_1u_1| = |m_2v_2 - m_2u_2|`

Putting the obtained values of `v_1` and `v_2` we obtain same result, numerically given as

= `(m_1m_2)/(m_1 + m_2) (1 +e)(u_1 - u_2)`

= `mu(1 + e) u_(rel)`

Where
`mu` = reduced mass = `m_1m_2//(m_1 + m_2)`

`u_(rel)` = Velocity of approach

`e` = coefficient of restitution for collision

(c) Loss of K.E. in collision :
Rewriting the momentum equation

`m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2`.............................(1)

From Newton's experimental law,

`v_2 - v_1 = e (u_1 - u_2)`.......................(2)

Square of (1) `+m_1m_2xx` square of (2) yields

`(m_1v_1 + m_2v_2)^2 + m_1m_2(v_2 - v_1)^2 = (m_1u_1 + m_2u_2)^2 + m_1m_2{e(u_1 - u_2)}^2`

`=>` `(m_1^2 + m_1m_2)v_1^2 + (m_2^2 + m_1m_2)v_2^2 = (m_1u_1 + m_2u_2)^2 + e^2 m_1m_2(u_1 - u_2)^2`

`=> (m_1 + m_2) (m_1v_1^2 + m_2v_2^2) = (m_1u_1 + m_2u_2)^2 + m_1m_2(u_1 - u_2)^2 {1 - (1 - e)^2}`

`=>` `(m_1 + m_2)(m_1v_1^2 + m_2v_2^2)=(m_1u_1 + m_2u_2)^2 + m_1m_2(u_1 - u_2)^2 - m_1m_2(1 - e)^2(u_1 - u_2)^2`

`=>` `(m_1 + m_2)(m_1v_1^2 + m_2v_2^2) = (m_1 + m_2)(m_1u_1^2 + m_2u_2^2) - m_1m_2(1 - e)^2(u_1 - u_2)^2`

Dividing both sides by `(m_1 + m_2)` and rearranging the terms, we obtain

`{1/2m_1v_1^2 + 1/2m_2v_2^2) - (1/2m_1u_1^2 + 1/2m_2u_2^2)} = -1/2 \ \(m_1m_2)/(m_1 + m_2)(1 - e^2)(u_1 - u_2)^2`

`=>` `KE_text(final) - KE_text(initial) = -1/2\ \(m_1m_2)/(m_1 + m_2)(1 - e^2)(u_1 - u_2)^2`

`=>` `DeltaKE = - 1/2 \ \ (m_1m_2)/(m_1 + m_2)(1 - e^2)(u_1 - u_2)^2`

`DeltaKE = -(mu u_(rel)^2(1 - e^2))/2`

Where `mu` = reduced mass and `u_(rel)` is velocity of approach
Since `DeltaKE` is negative the kinetic energy is lost as heat, light, sound etc.

(d) Perfectly elastic collision :

Putting e = 1, we obtain,

`v_1 = ((m_1 - m_2)/(m_1 + m_2)) u_1 + ((2m_2)/(m_1 + m_2)) u_2`

and `v_2 =((2m_1)/(m_1 + m_2)) u_1 + ((m_2 - m_1)/(m_1 + m_2)) u_2 `

`=>` Impulse = `((2m_1m_2)/(m_1 + m_2)) (u_1 - u_2) ` and `DeltaKE = 0`

Cases:
(a) When one of the bodies is at rest, say `m_2` is at rest putting `u_2=0`, we obtain

`v_1 = ((m_1 - m_2)/(m_1 + m_2)) u_1` & `v_2 =((2m_1)/(m_1 + m_2)) u_1`

(b) If the bodies are identical ( same mass ) putting `m_1 = m_2`:
we obtain `v_1 = u_2 & v_2 = u_1`. That means the bodies exchange their momenta. If one of them, say `m_2` is at rest, just after collision., will move with same velocity as that of `m_1 & m_2` will be at rest.

(c) If one of bodies, say `m_2` is very massive, putting `m_1/m_2 approx 0`
we obtain `v_1 = -u_1 + 2u_2`
`v_2 = u_2`
(i) If `u_2 = 0` then `v_1 = -u_1` and `u_2 = 0`. That means the lighter mass gets reflected back with approximately same speed.

(e) Kinetic energy delivered by an Incident particle to a stationary particle in elastic collision :

The K.E. lost by the incident particle is equal to the K.E. delivered by it to the second particle or K.E. gained by the `2^(nd)` particle.

`=>` K.E. delivered = K.E. received by the `2^(nd)` particle from the `1^(st)` particle.

`=>` `(DeltaKE) = 1/2\ \ m_2v_2^2`

putting `v_2 = (2m_1)/(m_1 + m_2) u_1 (therefore u_2 = 0)` we obtain

`(DeltaKE)_1 = 1/2\ \ (4m_1^2m_2)/(m_1 + m_2)^2 u_1^2 => (triangle KE_1)/(KE_1) = (4m_1m_2)/(m_2 + m_2)^2`

`=>` Fraction of energy lost by the particle `1 = eta = (4m_1m_2)/(m_1 + m_2)^2`

Note: `eta` is maximum when `m_1 = m_2` `eta` is minimum when `m_2 rightarrow infty`

In perfectly elastic collision the (total) change of KE of the system `= 0 =>` `triangle KE` = 0

(f) Loss of Kinetic energy in perfectly inelastic collision:

putting `e= 0`, we obtain

`v_1 = m_1/(m_1 + m_2) u_1 + m_2/(m_1 + m_2) u_2`

and `v_2 = m_1/(m_1 + m_2) u_1 + m_2/(m_1 + m_2) u_2`

`=>` impulse = `(m_1m_2)/(m_1 + m_2) (u_1 - u_2)`

Since `v_1 = v_2` the particles move together with the same velocity; one sticks to the other.

When the second particle `(m_2)` is stationary `v_1 = m_1/(m_1 + m_2) u_1 = v_2`

Loss of K.E of the system, `DeltaK.E = 1/2m_1 u_1^2 - 1/2(m_1 + m_2)v^2` where `v = v_1 = v_2`

= `(m_1u_1^2//2){m_2//(m_1 + m_2)}`

Fractional loss in K.E. of the system = `{m_2//(m_1 + m_2)}`

(2) Oblique Collision :

In this type of collision the relative velocity of approach of the bodies doesn't coincide with the line of impact.
Let `u_1, u_2` are velocities before collision and `v_1 , v_2` are velocities after collision.

Conserving the linear momentum of the system along and perpendicular to the line of impact (due to absence of any other external impulsive force) we obtain, from shown in figure(7.a)

` m_1u_1 cos theta_1 + m_2u_2 cos theta_2 = m_1v_1 cos beta_1 + m_2v_2 cos beta_2`............(3)

& `m_1u_1 sin theta_1 + m_2u_2 sin theta_2 = m_1v_1 sin beta_1 + m_2v_2 sin beta_2`

Since no force is acting on `m_1 & m_2` along the tangent, the individual momentum of `m_1`, and `m_2` remains conserved

`=>` `m_1u_1 sin theta_1 = m_1v_1 sin beta_1`.......................(4)

& `m_2u_2 sin theta_2 = m_2v_2 sin beta_2`....................(5)

Newton's experimental Law : Along the line of impact

`e= -(v_1 cos beta_1 - v_2 cos beta_2)/(u_1 cos theta_1 - u_2 cos theta_2)`................(6)

Now we have four equations and four unknowns `v_1, v_2, beta_1` and `beta_2`.

Solving four equations for four unknown we obtain

`v_1cos beta_1 = ((m_1 - em_2)u_1 cos theta_1 + m_2 (1 + e)u_2 cos theta_2)/(m_1 + m_2)`...............(v)

& `v_2 cos beta_2 = (m_1(1 + e)u_1 cos theta_1 + (m_2 - em_1)u_2 cos theta_2)/(m_1 + m_2)`

`therefore v_1 sqrt((v_1 sin beta_1)^2 + ( v_1 cos beta_1)^2)`

& tan `beta_1 = (v_1 sinbeta_1)/(v_1 cos beta_1) => beta_1 = tan^-1(v_1sin beta_1)/(v_1cos beta_1)`

[Put `v_1sin beta` from (i i) and `v_1 cos beta_1` from (v)]

Similarly `v_2` and `beta_2` can be calculated.

Impulse = `mu(1 + e)u_(rel) = (m_1m_2)/(m_1 + m_2)(1 + e^2)(u_1 cos theta_1 - u_2 cos theta_2)`

Energy loss = `mu(1 - e^2)u_(rel^2)//2 = mu(1 + e)u_(rel)^2//2 = (m_1m_2)/(m_1 + m_2)(1 - e^2)(u_1 cos theta_1 - u_2 cos theta_2)^2`

Here `u_(rel)` = velocity of approach along the line of impact :

(3) Oblique collision on a fixed horizontal plane :

Let a small ball collides with a smooth horizontal floor with a speed u at an angle `theta` vertical as shown in the figure(7.b). Just after the collision let the balI leave the floor with a speed v at an angle `beta` to vertical. It is quite clear that line of impact is perpendicular to the floor. Therefore, the impact takes place along the (normal) vertical. Now we can use Newton's experimental law as

`e = text(velocity of separation)/text(velocity approach)`

`=>` e = [velocity approach ] = velocity of separation

`=>` `e(ucos theta)(-hatj) = -(v cos beta)(+ hatj)`

`v cos beta = e ucos theta`...............(3.a)

Since impulsive force acts on the body along the normal we can conserve the horizontal momentum of the body.

(a) Velocity after collision.

`=>` `(P_x)_(body) = Constant => (P_x)_(initial) = (P_x)_(final)`

`=>` `m usin theta = mv sin beta => vsin beta = u sin theta`.....................(3.b)

`a^2 + (b^2) => v^2 cos^2 beta + v^2 sin^2 beta = e^2u^2 cos^2 theta + u^2 sin^2 theta`

`=>` `v^2 = u^2 [e^2 cos^2 theta + sin^2 theta] => v = u sqrt(sin^2 theta + e^2 cos^2 theta)`

(3.a) `div` (3.b) = `(vcos beta)/(u sin beta) = (eu cos theta)/(u sin theta)`

`=>` `cot beta = e cot theta => beta = cot^-1 (e cot theta)`

(b) Impulse

Impulse of the collision = change of momentum of the body

`{mv sin beta hati + (mv cos beta)hatj} - (mu sin theta hati - mu cos theta hatj)`

`=>` Impulse = `m(v sin beta - u sin theta)hati + m(v cos beta + u cos theta)hatj`
Since `vsin beta = usin theta`

`=>` Impulse = `m(v cos beta + u cos theta)hatj`
Putting `vcos beta = eucos theta` from eq.(3.a) we obtain

Impulse = `m(1 + e)u cos theta hatj`

`therefore` Magnitude of the impulse = `m(1 + e)u cos theta`

(c) Loss of kinetic energy

`triangleKE = (1/2) mv^2 - (1/2)m u^2`

Putting the value of v we obtain

=`1/2\ \m[[sqrt(u(sin^2 theta + e^2 cos^2 theta))]^2 - u^2]`

= `1/2 \ \ mu^2[sin^2 theta + e^2 cos^2 theta - 1] = -1/2\ \ mu^2[cos^2theta - e^2 cos^2 theta]`

`triangle K.E =1/2(1 - e^2) mu^2 cos^2 theta` -ve sign indicates the loss of K.E.


 
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