# How can the bodies attain same velocities during the max. deformation during collision?

My book writes:

Collision between two bodies is almost similar to the case where there is a spring between the two colliding bodies. At the instant they come into contact , the rear ball has a larger velocity and the front ball has a smaller velocity. During the impact, the balls deform. The deformed balls push each other and the velocities of the two balls change. The total kinetic energy decreases as some of the energy is converted to elastic potential energy of the deformed balls. The deformation is maximum when the two balls attain equal velocities.

Now, my query is: If the kinetic energy of both the ball decreases, how can their velocities be equal?? The front one,from the beginning, had low KE; if it decreases during the deformation, how can its velocity be equal to the velocity of the rear ball??? Please help.

If the kinetic energy of both the ball decreases, how can their velocities be equal?? The front one ( B ),from the beginning, had low KE; if it decreases during the deformation, how can its velocity be equal to the velocity of the rear ball ( A )?

In order to get a clear picture, let's consider the extreme case when the velocity of B = 0

Let's make a concrete example with numbers $m_A = 1, m_B = 2, M = 3$:

Suppose that:

$v_a = 6m/s$ and $v_b, p, E_k = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p = 1 * 6 = 6, v_{cm} = p/M = 2$ Kinetic energy and momentum are conserved only in elastic collisions, but if the bodies stick together the collision is inelastic an only momentum is conserved:

After the collision velocity of A would be anyway lower as KE should be distributed among more mass, but some KE is lost in the crash. How much?

Momentum is conserved: $p_{ab} = 6$ , from this datum you can calculate its velocity which now coincides with the velocity of center of mass: $$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_{AB} = 0.5 * 2^2 *3 = 6 \rightarrow E_A = 2 + E_B = 4$.

Some energy has been transferred to B (4 J), but two thirds of the kinetic energy(12 J) have been changed to other forms of energy. The general law of 'conservation of energy' has not been, anyway, violated

Velocity of center of mass is the same, although KE has changed. Note that momentum is conserved because we are assuming that on the surface of contact there is no friction.

I hope your main question has got an answer by now, velocities can and must be equal because AB is now one single body: the rear ball has decreased and the front ball has increased its v and the two values level out.

(This is not due to the loss of KE, even if it had been conserved the two bodies would have levelled their v to 3.464, but this would violate the principle of conservation of momentum that would have increased to 10.4)

As to the queries in your comments: when the bodies have reached the maximum deformation they will move at final and same v. It is impossible to determine how much of the amount of KE lost and transformed will be absorbed by each body, as this depends on the material they are made of: the more a body is deformable the more energy it will absorb

.. But what about the case when the front ball is moving?

It makes no difference! Just think of communicating vessels, once two bodies are joined and become a single body... energy, velocity and momentum level out and are unified.

Think of it this way:

• Ball A is moving at 10 m/s
• Ball B is moving at 3 m/s
• Both balls have the same mass

Here we have Ball A collides with Ball B, transferring energy. During this transfer (ignore the deformation for now, since that doesn't seem to be an issue at the moment, so we've got an elastic collision), consider what happens in the exchange of velocities:

After some $\Delta t$ we have

• Ball A is moving at 9.7 m/s
• Ball B is moving at 3.3 m/s

After some more $\Delta t$, we have

• Ball A is moving at 8.4 m/s
• Ball B is moving at 4.6 m/s

And we can keep doing this (infinitesimal) transfer of velocity, but you should see that Ball A is losing speed while Ball B is gaining speed. At some point during this transfer, both balls will have the same velocity.

Note that the statement is the velocities of the two balls decreases. This is not the same as the first one losing velocity, it's really an awkward way of saying that energy is not conserved.

• +1 . Sorry, Sir , I mistook in quoting the book's words. Now I have edited it. I also thought like that; it would happen only when the kinetic energy lost by the rear ball is gained by the front ball. But this doesn't happen really... – user36790 Nov 27 '14 at 4:52
• ....During the collision, part of their KE gets converted to elastic PE. So, if this happens, the KE lost by the rear ball willn't be gained completely by the front ball as part of it is converted to elastic PE . – user36790 Nov 27 '14 at 4:54
• ... And, Sir , which ball will lose more KE? The rear one having more velocity or the front one having smaller velocity?? The book says "... kinetic energy of the two balls decreases." Now, if both decreases their KE, how can the front ball equate its velocity to that of the rear?? – user36790 Nov 27 '14 at 5:08

by conservation of linear momentum the center of mass of the system will keep moving, thus the two ball are together they will keep moving in the same direction than the faster ball, at the same speed than the center of mass, that is, slower than the initial speed of the faster ball. Assuming a totally elastic collision, after the balls push back each other, the center of mass will still move at the same speed.