Head on Collision : Velocities after Collision

Velocities after Collision

Two bodies of mass m1 and m2 moving along the same straight line collide such that their relative velocity is along the line of action.
Let the velocities before the collision be u1 and u2 and that just after the collision be v1 and v2 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,

$ \displaystyle m_1 \vec{u_1} + m_2 \vec{u_2} = m_1 \vec{v_1} + m_2 \vec{v_2}$

Since all the momenta are unidirectional

$ \displaystyle m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 $ ………(1)

Coefficient of Restitution

$\displaystyle e = -\frac{v_1 – v_2}{u_1 – u_2} $

$ \displaystyle v_1 – v_2 = e(u_2 – u_1 ) $ ……..(2)

Eliminating v2 from (1) and (2)

We obtain,

$ \displaystyle v_1 = \frac{(m_1 – e m_2)}{(m_1 + m_2)}u_1 + \frac{m_2(1 + e )}{(m_1 + m_2)}u_2 $

$ \displaystyle v_2 = \frac{m_1(1 + e )}{(m_1 + m_2)}u_1 + \frac{(m_2 – e m_1)}{(m_1 + m_2)}u_2 $

Exercise : An iron ball A collides with another stationary iron ball B. If the ratio of radii of the balls is n = 2, find the ratio of their speeds just after the collision (e = 0.5)
 Numerical

Impulse of Bodies :

The momentum delivered to the 2nd ball = Impulse of the 1st ball on the 2nd ball & vice versa.

Impulse = $ \displaystyle |\Delta P_1| = |\Delta P_2| $

$ \displaystyle |m_1 v_1 -m_1 u_1| = |m_2 v_2 – m_2 u_2| $

Now Putting the obtained value of v1 and v2 , we get

Impulse = $ \displaystyle \frac{m_1 m_2}{m_1 + m_2}(1+e)(u_1 – u_2) $

Therefore ,

Impulse = $ \displaystyle \mu (1 + e)u_{rel} $

Where $ \displaystyle \mu = Reduced \; mass = \frac{m_1 m_2}{m_1 + m_2} $

urel = Velocity of approach

e = coefficient of approach

Illustration : A ball of mass m collides with a stationary identical ball with a speed v0. Find the impulse of the colliding balls (e = 1/2).

Solution :
Impulse = $ \displaystyle \frac{m_1 m_2}{m_1 + m_2}(1 + e)(u_1 – u_2) $

m1 = m2

e = 1/2 , u1 = v0 , u2 = 0

We obtain ‘

Impulse = $ \displaystyle \frac{3}{4}m v_0 $

Impulse during Collision:

Since , $ \displaystyle e(numerically) = \frac{Velocity \; of \; separation}{Velocity \; of \; approach} $

$ \displaystyle e = \frac{v_2 -v_1}{u_2 -u_1} $

$ \displaystyle e = \frac{m(v_2 -v_1)}{m(u_2 -u_1)} $

$ \displaystyle e = \frac{Impulse \; of \; force \; of \; restitution}{Impulse \; of \; force \; of \; compression} $

Illustration : A block of mass m1 collides with the spring with a stationary block of mass m2 with a speed v0. Find the maximum impulse of the blocks during the impact.

 Numerical

Solution: During the impact, the spring force is an internal force to the system (m1 + m2) and no horizontal force acts on it. Therefore the horizontal momentum of the system remains constant.
The maximum impulse during the impact means maximum change in momentum during the impact. It happens when both the bodies move with same velocity at the end of period of deformation of the spring. At that time the compression of the spring is maximum.
Conservation of momentum in the horizontal direction between the initial position and maximum compressed position of the spring yields,

 Numerical

m1 v0 = m1 v + m2 v

$ \displaystyle v = \frac{m_1 v_0}{m_1 + m_2} $ …(i)

Therefore , change in momentum of m1 is

$ \displaystyle \Delta P_1 = |m_1 v – m_1 v_0| $

Impulse = $ \displaystyle \Delta P_1 = m_1 (v_0 – v ) $ …(ii)

Using (i) & (ii)

$ \displaystyle Impulse = \frac{m_1 m_2}{m_1 + m_2} v_0 $

Loss of Kinetic Energy in collision:

$ \displaystyle KE_{final} – KE_{initial} = -\frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(1-e^2)(u_1 -u_2)^2 $

$ \displaystyle \Delta KE = -\frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(1-e^2)(u_1 -u_2)^2 $

Since ΔKE is negative , the kinetic energy is lost as heat , light , sound etc.

Exercise : Find the loss of KE of the two particles of mass m1 = 3 Kg & m2 = 6 Kg, moving towards each other with speed u1 = 5 m/sec, u2 = 10 m/sec respectively. The coefficient of restitution of collision of the particles is e = 0.5 ;

Exercise : Find the loss of kinetic energy of the system during collision if the particles are moving undirectionally.

Perfectly Elastic Collision :

Putting e = 1 , We obtain

Velocity of body after collision ,

$ \displaystyle v_1 = \frac{(m_1 – m_2)}{(m_1 + m_2)}u_1 + \frac{2 m_2}{(m_1 + m_2)}u_2 $

$ \displaystyle v_2 = \frac{2 m_1}{(m_1 + m_2)}u_1 + \frac{(m_2 – m_1)}{(m_1 + m_2)}u_2 $

Impulse = $ \displaystyle \frac{2 m_1 m_2}{m_1 + m_2}(u_1 – u_2) $

and , ΔKE = 0

Cases :

(a) When one of the bodies is at rest, say m2 is at rest putting u2 = 0, we obtain

$ \displaystyle v_1 = \frac{(m_1 – m_2)}{(m_1 + m_2)}u_1 $

$ \displaystyle v_2 = \frac{2 m_1}{(m_1 + m_2)}u_1 $

(b) If the bodies are identical putting m1 = m2 we obtain
v1 = u2 & v2 = u1
That means the bodies exchange their momenta. If one of them, say m2, is at rest, just after collision m2 will move with same velocity as that of m1 & m1 will be at rest.

(c)If one of bodies, say m2, is very massive, putting

$ \displaystyle \frac{m_1}{m_2} \approx 0 $

we obtain ,

u1 + 2u2 & v2 = u2

(i) If u2 = 0 then

v1 = – u1 & v2 = 0

That means the lighter mass gets reflected back with approximately same speed. Example- Elastic collision of ball and earth. The ball recoils back with same speed where as the motion of earth remains unchanged. Therefore recoiling of earth is hardly experienced with collision of any object with it.

(ii) If u1 = 0, the lighter particle is at rest and the heavier particle moves with speed v2 , then putting u1 = 0
we obtain, v1 = 2u2 , v2 = u2
That means, the lighter particle moves along the direction of motion of the massive particle with double the speed of the incident (massive) particle, where as the velocity of the massive particle remains unchanged.

K.E. 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 2nd particle.

K.E. delivered = K.E. received by the 2nd particle from the Ist particle

$ \displaystyle \Delta KE = \frac{1}{2}m_2 v_2 ^2 $

Putting $ \displaystyle v_2 = \frac{2 m_1}{m_1 + m_2}u_1 $ (since u2 = 0)

We obtain ,

$ \displaystyle (\Delta KE)_1 = \frac{1}{2}\frac{4 m_1^2 m_2}{(m_1 + m_2)^2} u_1 ^2 $

$ \displaystyle \frac{\Delta KE_1}{KE_1} = \frac{1}{2}\frac{4 m_1 m_2}{(m_1 + m_2)^2} $

Fraction of energy lost by the particle 1 is

$ \displaystyle \eta = \frac{1}{2}\frac{4 m_1 m_2}{(m_1 + m_2)^2} $

Note:

η is maximum when m1 = m2

η is minimum when m2 → ∞

In perfectly elastic collision the (total) change of KE of the system = 0 ⇒ Δ KE = 0.

Also Read :

Impulse & Impulsive Force
Conservation of linear momentum during impact
Solved Examples : Impulse
Analysis of Collision & Coefficient of restitution
Line of impact during collision
oblique impact in Collision

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