1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic equation -- 2 masses attracted in space and collision...

  1. Oct 16, 2015 #1

    I like to ponder basic physics puzzles to pass time in boring meetings on metro etc...but I have one that I don't understand and I am obsessing over it (to detriment of working out etc.) it should be easy but I don't understand.

    Problem: Imagine you have two blocks (Mass of Block 1 is 1Kg...Mass of Block is 10Kg). They are free floating in space (no friction/air/gravity). They are 10meters apart from each other. If they both feel a force of 50 Newtons in opposite directions (towards each other). And then hit each other in an inelastic collision...what is the velocity of the final composite mass?

    My thought is ... 50 Newtons = 1kg * (acceleration)...so acceleration of block 1 is 50(m/S^2). I don't know how to determine the time the block is accelerating but I am thinking it is the same time as block 2 so I just make it accelerate for 1 second. So final velocity is 1 second * 50m/s^2 so...total energy at collision is .5*1kg * 2500= joules of energy for block 1. Then I do the second block and the total energy is 1 second * 10Kg* 5m/s^2 so total energy at collision is 125 Joules. Then I would take 2500 Joules and subtract 125 joules and then my net total energy is 2375 Joules going in the direction of block 1 so...the total velocity would be ((2375*2)/11Kg)^.5 = 20.7m/s going in the direction that block 1 was moving.

    But...the actual answer is 0 velocity with each block cancelling each other out...am I wrong? or is the answer wrong?
  2. jcsd
  3. Oct 16, 2015 #2


    User Avatar
    Gold Member

    Assuming the distance D is known (10 m). Suddenly the blocks are both affected by a constant F (50 N) towards each other till they hit. In this situation to calculate the time till the collision happens:

    D = (a1 + a2) ⋅ t2 / 2
    a1 = F / m1
    a2 = F / m2

    Then you can calculate the velocities at the collision with the time t and use the conservation of momentum for the final velocity of masses stuck together.

    EDIT: As the collision is inelastic you must not forget to take into account the energy transformed into heat due to friction, permanent deformation and other irreversible phenomena (and generally it's not possible to calculate with energy like you tried).
    Last edited: Oct 16, 2015
  4. Oct 16, 2015 #3
    The acceleration of the 1 kg block is 50 m/s2, while the acceleration of the 10 kg block is 1/10 as great, at 5 m/s2. So, the 1 kg block is always traveling 10 x as fast as the 10 kg block. Therefore, the momentum of the two blocks at any given time is exactly the same, but in opposite directions. So, if they stick together, their final momentum is zero.

  5. Oct 16, 2015 #4
    Choosing the time arbitrary to 1s has no relevance for the final results. You need the time at which the collision takes place.
    But it's no point to do it anyway. Conservation of momentum tells you that the final momentum should be equal to the initial one, which is zero.
  6. Oct 16, 2015 #5
    Thank you all so much! Now I am seeing my confusion. But one more question- how does the kinetic energy equation of motion (.5*m*v^2) comport with the conservation of momentum (m*v)? It seems that kinetic energy is 'lost' when object 1 (10kg * 1m/s=5 Joules) collides directly/inellastically with object 2(1kg * 10m/s=50 Joule). Whew! Thanks again.
  7. Oct 16, 2015 #6
    okay...you all have made Beethoven's 9th symphony happen in my mind...I understand now. Thank you! ahhh
  8. Oct 16, 2015 #7
    There are a couple of questions that I have for clarity.

    1. Why did you subtract the kinetic energies to get the "net energy", and what does net energy mean?
    2. What do you mean by the "total velocity"?
    3. Given the initial conditions, you can exactly calculate the time, position and velocities at collision
  9. Oct 16, 2015 #8
    You're too late. The OP's issues have all been resolved by the posts of the previous responders.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook