@RonaldBecker Hmm, that seems like a trick question to me. Yes, unless we assume BC is isobaric we cannot know where C is so we cannot answer options (C) or (D).
@HarjotDhillon I agree that (d) is unclear as the question doesn't explain what it means by collision cannot be elastic.
If the "collision" covers the time from M₁ first touching the spring to M₁ leaving the spring again then yes energy is lost to friction during this time so in this sense the collision is inelastic.
We generally think of ideal collisions as being impulsive, i.e. taking zero time, and in that case no energy is lost to friction in the collision. But here we have a "collision" that takes a non-zero time i.e. the time while the spring is being compressed then bouncing back.
According to Wikipedia 'internal' friction determine whether collision is elastic or inelastic but here we have 'external' friction so even if collision covers the time from M₁ first touching the spring to M₁ leaving the spring again then still it should be considered elastic. Energy was being dissipated even if there was no collision.
Maybe I don't need to overthink in exam that's made for highschool students.
a) this cannot be true as When M1 collides with M2 the energy transfer takes place between both M2 and the spring that is when M1 collides with this system(spring and block M2) some kinetic energy will be wasted in compressing the spring and some will be gives to M2 as kinetic energy we don't know how much. The block can come M1 can come to rest or it can still move but this is not the scope of a)
b) this is also not correct we can see there are no net force external force initially on the system( since friction is not given we can ignore it) and finally too. The spring force is an internal force that is applied on both m1 and m2 in opposite direction so it is internal for SYSTEM (THAT IS FOR BOTH M1 AND M2 AS A WHOLE IF WE ARE TALKING ABOUT INDIVIDUALLY IT IS CONSIDERED EXTERNAL)
@alam re (a) when the spring is at maximum compression both M₁ and M₂ have the same velocity, so from conservation of momentum their velocity must be 𝑣/2.
c) this is NOT true as when spring is massless and recall that k is a measure of stiffness of spring so no mass of spring means no streching and k=0!!. Now recall that complete energy transfer takes b
Place when M1 =M2 since it is given M1 and M2 are different energy will be distributed as v1 and v2
So M1 will not come to rest and will have some velocity
d) this essentially is true. Now the condition for elastic collisions is that energy before and after collision are equal but recall that friction will do some work on m1 and will produce heat M1 will waste some energy on overcoming this friction and there will be some deformation so no elastic collision
I was never interested in events like the Olympiad as they always seemed a lot of work. I just concentrated on my studies and getting to the university I wanted (Cambridge in the UK).
But if you want to compete in the IPHO then go for it and good luck! :-)
I need to go now, but I'm usually here from 5 a.m. to 12:30 p.m. UK time if you want to ask anything.
@alam you are right that massless spring would have k=0 but here we just imagine it to have some non zero k. Moreover value of k is insignificant as collision is supposed to be elastic in statement 'c'
And in elastic collision when equal mass bodies collide they exchange there velocities
How can we imagin that. It is clearly given that k needs to be zero when spring is massless. If we were to imagine k to be not zero then the question may or may not be true. And in these type of paragraph questions they proposes proposition that is the case is entirely true or entirely false
Let us imagine k is not zero then we are asked that the m1 will come to rest or not. Now in this case as opposed to my previous answer on this same question we won't be sure.
By that I mean assume at any instance the Mass M1 hits the spring +block system now if k=0 then all energy(=KE) will only be distributed among blocks in this case since M1 and M2 are not equal the velocity will not completely gets transferred to M2 hence the proposition is False. Now suppose k is not zero in this case therefore are two possibilities.
1) M1 will be stopped now when M1 collides with system it wastes some energy overcoming the PE of spring and then hit M2 with remaining energy and stopes in this case proposition is true. Let's take another case. 2) M1 retains some energy even after hitting the system hence the proposition is false.
So essentially what I am conveying is that the question makes sense when we assume k to be zero
M1 don't need to overcome spring potential energy BEFORE collision. Actually process of collision is began at the instant M1 touch the spring. As M1 is being slowed down by spring force, M2 is being pushed by spring force to move
To simplify our problem you could imagine k to be 0 or infinite
If k = 0 and spring is massless then its behaving like just thin air. If we imagine k=infinite then spring does not get compressed at all and it would act like rigid body. Imagining it to be thin air or rigid body make our visualisation very easy.
Actually I was having problem with (d). Statement (c) was clear to me
They are equal it is given in first line of question
M2 was at rest and M1 had some velocity. Collision is elastic and they both have same mass so they would exchange there velocities. M1 would come to rest and M2 will move with same velocity.
