A body A, of mass m = 0.1 kg has an initial velocity of 3 \widehat i ms-1 . It collides elastically with another body, B of the same mass which has an initial velocity of 5 \widehat j ms-1. After collision, A moves with a velocity \overrightarrow v = 4( \widehat i + \widehat j ) . The energy of B after collision is written as x \over 10 . The value of x is \underline\hspace2cm .

Center of Mass PYQ — JEE Main 2020 · 8 Jan · Shift 1 · Q51 with Solution

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Correct answer: 1

We need the kinetic energy of $B$ after collision.

Since masses are equal, we can write momentum conservation directly in velocity form:

\vec u_A + \vec u_B = \vec v_A + \vec v_B

Substitute the values:

3\hat i + 5\hat j = (4\hat i + 4\hat j) + \vec v_B

So,

\vec v_B = (3\hat i + 5\hat j) - (4\hat i + 4\hat j)

\vec v_B = -\hat i + \hat j

|\vec v_B|^2 = (-1)^2 + (1)^2 = 2

K_B = \frac12 m |\vec v_B|^2

K_B = \frac12 \times 0.1 \times 2 = 0.1\,\text{J}

Given that energy is written as

\frac{x}{10}

So,

\frac{x}{10} = 0.1 = \frac{1}{10}

Hence,

x = 1

Initial kinetic energy:

K_i = \frac12(0.1)(3^2) + \frac12(0.1)(5^2) = 0.05(9+25)=0.05\times 34=1.7\,\text{J}

Final kinetic energy of $A$ :

K_A' = \frac12(0.1)(4^2+4^2)=0.05(32)=1.6\,\text{J}

Final kinetic energy of $B$ :

K_B' = 0.1\,\text{J}

Total final kinetic energy:

1.6+0.1=1.7\,\text{J}

So the collision is indeed elastic.

Therefore, the answer is correct.