Two identical balls A and B having velocities of 0.5m/s−1 and −0.3m/s−1 ...
Q. Two
identical balls A and B having velocities of 0.5m/s−1 and −0.3m/s−1 respectively collide elastically in one
dimension. The velocities of B and A after the collision
respectively will be
Neet
2016 W P E Q 5 I told
you earlier in qus of year 2010 , 2011
& 2012. If you haven’t subscribed my channel then do it now and don’t miss
any important thing about Neet. In case
of collision the two bodies of masses m1 and m2 are moving along a line with
velocities u1 and u2 respectively, u1 should be greater then u2 for two bodies
to collide.
As the
velocity of m1 (u1) is greater than that of m2 (u2), there will be a
deformation in the bodies. This deformation continues till they acquire same
velocity. Now a question arises, that why will they acquire same velocity. The
velocity of the front body m2 (which was slower
initially) will gradually increase, and the velocity of the rear body m1 (which
was faster initially) will gradually decrease. By conservation momentum.
m1u1 + m2u2 = m1 v1+ m2 v2
During the whole process, the momentum of the two
-block system remains constant.
Before and after collision kinetic energy may be
same but not during the whole process.
Newton
This is an experimental law which relates the velocity of approach of the
bodies before collision and the velocity of separation after the impact.
When the collision is neither perfectly elastic nor perfectly inelastic, for
such cases, Newton did certain experiments and
gave another law which is now called as Newton
Relative
velocity of impact = e [Relative velocity of approach before impact]
Direct Approach : Remember that whenever two identical
bodies will collide elastically , they will exchange their component of velocities
along L.O.C. In case of one dimensional collision , we can say simply , they
will exchange their velocities.
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