Question

Two particles of masses m1, m2 move with initial
velocities u1, and u2. On collision, one of the
particles get excited to higher level, after absorbing
energy e. If final velocities of particles be v1 and v2,
then we must have
[AIPMT-2015]
!see the attached picture!​

Answer 1

Answer:

option 4 is correct... apply total conservation of energy...

Answer 2

Answer:

$\frac{1}{2}m1u1^{2} + \frac{1}{2}m2u2^{2} - e = \frac{1}{2} m1v1^{2} + \frac{1}{2} m2v2^{2}$

Explanation:

In the given question, we apply the concept of total energy conservation. From this concept the total initial energy must be equal to total final energy. We know that the kinetic energy is given 1/2 mv². Using this, we get;-

$\frac{1}{2} m1u1^{2} + \frac{1}{2} m2u2^{2} = \frac{1}{2} m1v1^{2} + \frac{1}{2} m2v2^{2} + e$

$\frac{1}{2} m1u1^{2} + \frac{1}{2} m2u2^{2} - e = \frac{1}{2} m1v1^{2} + \frac{1}{2} m2v2^{2}$

In the first question we added energy 'e' on the right side as it was added as it got excited to a higher level after collision.

In the second equation, we equated to match the option.

Moreover, option (a) and (c) are dimensionally wrong and option (b) has the square of mass which is impossible in kinetic energy. This way to after eliminating all the options, option (d) holds true.

Hope it helps! ;-))