a) it is possible for both of them to be at rest after the collision?
b) it is possible for one of them to be at rest after the collision?
I think no and yes, but i'm not sure.
2006-11-14 15:50:40 · 5 answers · asked by dreamclock 2 in Science & Mathematics ➔ Physics
Hi! You are right.
If they have equal masses and are elastic then b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :
Vf= final speed; Vcm=center of mass speed ; Vi=initial speed.
Thus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=
=m1*Vi/(m1+m2) => m1=m2.
" a)" isn't possible because the total momentum must be preserved.
Hope it helps!
2006-11-14 19:46:24 · answer #1 · answered by viktorpopescu 2 · 0⤊ 0⤋
a) it is impossible for both of them to be at rest. the initial and final momentum must be equal. the final momentum cannot be zero since at least one of the balls was moving in initial state. so NO it is impossible
b) most likely yes. and in order to achieve that either the one initialy at rest must be an impossible to move object, or after the collision the one that was moving becomes still and the other "adopts" the other momentum. so YES most likely
we are assuming perfect elastic collisions of course
2006-11-14 17:35:46 · answer #2 · answered by Emmanuel P 3 · 1⤊ 0⤋
In first case, if u take two bodies of same mass and project them towards each other with same velocity such that after collision the two mass stick together and move as combined mass, then the velocity of the composite mass is zero. Hence, the system is at rest.
In second case, if u take two bodies of same mass and put one at rest and project other with a certain velocity towards the first one then after collision the second body will come to rest and transfer all its energy to the first body. So after collision first body moves with the second body's initial velocity.
2006-11-15 00:55:55 · answer #3 · answered by Napster 2 · 0⤊ 0⤋
you may want to apply relative velocities, you may want to apply relative lots. i flow for the least complicated procedures, and for that reason that is relative lots. So, (80/60) = a million.334:a million ratio. it really is two.334 "parts" even as extra mutually. 12m/2.334 = 5.14. So, one may have travelled 5.14m, the different may have travelled (12 - 5.14) = 6.86m. Which one is which? The least mass may have done the most suitable distance, The more effective mass the least. So for that reason, the 60kg scholar might want to have moved 6.86m.
2016-11-29 03:52:48 · answer #4 · answered by ? 4 · 0⤊ 0⤋
Conservation of momentum is the approach to both questions.
You could also try to remember the sorts of things that happen in billiards.to answer the second question.
2006-11-14 17:13:00 · answer #5 · answered by arbiter007 6 · 0⤊ 0⤋