# Doubt from Collisions

**Ishan_2020**#1

You can refer to some elementary level texts for getting to this problem, its a very basic one, just go through DCP, I think its given in that text, if not so, then ping me I will post the solution for you.

**Ishan_2020**#3

Sir I went through HC Verma and DCP LEVEL 1 before attempting this, similar question but it was elastic collision, this inelastic could you please help with the solution, another doubt can we use COME, because as far as I know in inelastic collisions we cannot use COME.

Yeah, obviously because in that case, there is loss of energy taking place due to the deformations that may have been produced in the body due to inelastic collision.

I am not completely sure of the solution, but after pondering over the question, I am posting the solution.

I'm open for discussions over the same, be it mentors or even students

**Part (a)**

So here, given that the velocity of m_1 initially, is v. When it hits the system of m_2 and m_3 it will get stuck to the former one. At that particular time instant, the latter block (m_3) will be at rest.

Apply COM for the other two blocks : mv=2mv_1 \implies v_1=\frac{1}{2}v (where v_1 is the speed at which the stuck combination of m_1 and m_2 will be moving).

So, for part (a), the final speeds of m_1 and m_2 will be \frac{1}{2}v each while the other block will be at rest itself.

**Part (b)**

So considering the movement, when the block hits the system of spring and blocks, at first contact, left block gains speed due to the contact and that will be its **maximum** attainable speed throughout the process. As it starts moving, it will compress the spring, which will exert an outward force on each blocks an hence the one at rest will start gaining speed and the one moving will start feeling decelerating force.

Thus, the speed of the m_3 block will be maximum when the spring is at its maximum possible extension. Also, at that time, the speeds of all the three blocks will be the same.

Applying COM between the initial and this state, taking v_2 as the common attained speed at maximum elongation, we get : 2m.\frac{v}{2}=3mv_2\implies v_3=\frac{1}{3}v

The kinetic energy of m_3 will be maximum at this point only and that will be K=\frac{1}{2}m\frac{v^2}{9}=\frac{mv^2}{18}.

**Part (c)**

Its a bit not nicely stated as to whether we have to obtain the minimum energy of the m_2 block in the whole process or is it after the impact?

For the former one, obviously the answer will be zero, which will be when no collision occurred.

For the latter one, as we explained earlier, the minimum velocity of (m_1+m_2) system will be at maximum extension and hence minimum kinetic energy of m_2 will be precisely equal to the maximum kinetic energy of m_3 which is \frac{mv^2}{18}.

**Part (d)**

Note that, in case of an inelastic collision, there is energy loss that occurs at the time of collision due to deformations, etc and hence we cannot equate the initial energy of \frac{mv^2}{2} with any of the subsequent energy values as some fraction of it will be lost after collision. But, if we consider the total e energy just after the collision, it will remain conserved as subsequently, no inelastic deformations or so are taking place.

Considering initial collision stage when m_3 was at rest and the stage of maximum compression of the spring and applying energy conservation between these two states, we get :

\implies \frac{1}{2}.2m.(\frac{v}{2})^2=\frac{1}{2}.3m.(\frac{v}{3})^2+\frac{1}{2}kx^2 (where k represents the spring constant of the spring)

\implies mv^2(\frac{1}{4}-\frac{1}{6})=\frac{1}{2}kx^2 \implies x_{max}=v\sqrt{\frac{m}{6k}}.

Do let me know if you have any doubts.

Sir in part (b) I have a doubt.

Initially after collision there is compression in the spring (relative velocities of blocks are towards the spring).

Then after attaining maximum compression (each will have a velocity of v/3)spring tries to come to its natural length.

So the force on m3 still acts in a fashion to increase its velocity.

So After attaining natural length only expansion of spring starts ,after which m3 starts decelerating .So won’t the m3 have max velocity when spring comes to its natural length state ?(after getting compressed and restored to natural length)

**Churchil_2018**#11

Okay @Ishan_2020 I will try to get back to you and @Bhuvanitha_2020 you too.

If you have any solution @Bhuvanitha_2020 do post it

**Bhuvanitha_2020**#12

Maximum kinetic energy of m3 occurs after the spring attains natural length(pls do see my before post).

Conserving momentum and energy and solving.

Consider when spring attains natural length the velocity of combined mass as V1 and velocity of m3 and V2.

Minimum kinetic energy of m2 also occurs here.

1/2*m*(v/6)^2=1/2mv^2/72

**Churchil_2018**#14

Awesome @Bhuvanitha_2020

Thanks for pointing out my glitch

And sorry @Ishan_2020 for not getting you the correct solution.