Two particles of equal mass m have respective initial velocities u\widehat i and u( \widehat i + \widehat j \over 2 ) . They collide completely inelastically. The energy lost in the process is :

Center of Mass PYQ — JEE Main 2020 · 9 Jan · Shift 1 · Q55 with Solution

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Correct answer: B

Given velocities

For two particles of equal mass $m$ :

Particle 1: $\vec v_1 = u\hat i$
Particle 2: $\vec v_2 = u\left(\dfrac{\hat i + \hat j}{2}\right)=\dfrac{u}{2}\hat i+\dfrac{u}{2}\hat j$

They collide completely inelastically, so after collision they stick together.

Use conservation of momentum to find common final velocity

Initial total momentum:

\vec P_i = m\vec v_1 + m\vec v_2

\vec P_i = m\left(u\hat i\right) + m\left(\frac{u}{2}\hat i+\frac{u}{2}\hat j\right)

\vec P_i = mu\left(\frac{3}{2}\hat i+\frac{1}{2}\hat j\right)

Since total mass after sticking is $2m$ , final velocity $\vec V$ is

\vec V = \frac{\vec P_i}{2m} = \frac{u}{2}\left(\frac{3}{2}\hat i+\frac{1}{2}\hat j\right)

\vec V = \frac{3u}{4}\hat i + \frac{u}{4}\hat j

Initial kinetic energy

For particle 1:

K_1 = \frac{1}{2}m u^2

For particle 2, speed squared is

\left|u\left(\frac{\hat i+\hat j}{2}\right)\right|^2 = u^2\cdot \frac{1^2+1^2}{4} = \frac{u^2}{2}

So,

K_2 = \frac{1}{2}m\cdot \frac{u^2}{2} = \frac{1}{4}mu^2

Hence total initial kinetic energy:

K_i = \frac{1}{2}mu^2 + \frac{1}{4}mu^2 = \frac{3}{4}mu^2

Final kinetic energy

Magnitude squared of final velocity:

V^2 = \left(\frac{3u}{4}\right)^2 + \left(\frac{u}{4}\right)^2

V^2 = \frac{9u^2}{16} + \frac{u^2}{16} = \frac{10u^2}{16} = \frac{5u^2}{8}

Final kinetic energy of combined mass $2m$ :

K_f = \frac{1}{2}(2m)V^2 = mV^2 = m\cdot \frac{5u^2}{8} = \frac{5}{8}mu^2

Energy lost

\Delta K = K_i - K_f = \frac{3}{4}mu^2 - \frac{5}{8}mu^2

\Delta K = \frac{6}{8}mu^2 - \frac{5}{8}mu^2 = \frac{1}{8}mu^2

Check options

A: $\frac{1}{3}mu^2$ ❌
B: $\frac{1}{8}mu^2$ ✅
C: $\frac{3}{4}mu^2$ ❌
D: $\sqrt{\frac{2}{3}}mu^2$ ❌

Therefore, the correct option is B.

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View written solutionFree

Correct answer: B

Given velocities

For two particles of equal mass $m$ :

Particle 1: $\vec v_1 = u\hat i$
Particle 2: $\vec v_2 = u\left(\dfrac{\hat i + \hat j}{2}\right)=\dfrac{u}{2}\hat i+\dfrac{u}{2}\hat j$

They collide completely inelastically, so after collision they stick together.

Use conservation of momentum to find common final velocity

Initial total momentum:

\vec P_i = m\vec v_1 + m\vec v_2

\vec P_i = m\left(u\hat i\right) + m\left(\frac{u}{2}\hat i+\frac{u}{2}\hat j\right)

\vec P_i = mu\left(\frac{3}{2}\hat i+\frac{1}{2}\hat j\right)

Since total mass after sticking is $2m$ , final velocity $\vec V$ is

\vec V = \frac{\vec P_i}{2m} = \frac{u}{2}\left(\frac{3}{2}\hat i+\frac{1}{2}\hat j\right)

\vec V = \frac{3u}{4}\hat i + \frac{u}{4}\hat j

Initial kinetic energy

For particle 1:

K_1 = \frac{1}{2}m u^2

For particle 2, speed squared is

\left|u\left(\frac{\hat i+\hat j}{2}\right)\right|^2 = u^2\cdot \frac{1^2+1^2}{4} = \frac{u^2}{2}

So,

K_2 = \frac{1}{2}m\cdot \frac{u^2}{2} = \frac{1}{4}mu^2

Hence total initial kinetic energy:

K_i = \frac{1}{2}mu^2 + \frac{1}{4}mu^2 = \frac{3}{4}mu^2

Final kinetic energy

Magnitude squared of final velocity:

V^2 = \left(\frac{3u}{4}\right)^2 + \left(\frac{u}{4}\right)^2

V^2 = \frac{9u^2}{16} + \frac{u^2}{16} = \frac{10u^2}{16} = \frac{5u^2}{8}

Final kinetic energy of combined mass $2m$ :

K_f = \frac{1}{2}(2m)V^2 = mV^2 = m\cdot \frac{5u^2}{8} = \frac{5}{8}mu^2

Energy lost

\Delta K = K_i - K_f = \frac{3}{4}mu^2 - \frac{5}{8}mu^2

\Delta K = \frac{6}{8}mu^2 - \frac{5}{8}mu^2 = \frac{1}{8}mu^2

Check options

A: $\frac{1}{3}mu^2$ ❌
B: $\frac{1}{8}mu^2$ ✅
C: $\frac{3}{4}mu^2$ ❌
D: $\sqrt{\frac{2}{3}}mu^2$ ❌

Therefore, the correct option is B.

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