JEE Physics 2026 Question with Solution

Given: Total mass is $14 \, \text{kg}$ and the masses are in the ratio $2:2:3$. So the three fragment masses are $4 \, \text{kg}$, $4 \, \text{kg}$, and $6 \, \text{kg}$. The two equal fragments move perpendicular to each other with speed $18 \, \text{m/s}$ each.

Find: The velocity of the heavier fragment.

Since the body was initially at rest, total initial momentum is zero. Therefore, by conservation of momentum,

$$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$$

For the two equal fragments,

$$\vec{p}_1 = 4 \times 18 \, \hat{i} = 72\hat{i}$$ $$\vec{p}_2 = 4 \times 18 \, \hat{j} = 72\hat{j}$$

Hence,

$$\vec{p}_3 = -(72\hat{i} + 72\hat{j})$$

Its magnitude is

$$|\vec{p}_3| = \sqrt{72^2 + 72^2} = 72\sqrt{2}$$

Now for the heavier fragment of mass $6 \, \text{kg}$,

$$v_3 = \frac{|\vec{p}_3|}{6} = \frac{72\sqrt{2}}{6} = 12\sqrt{2} \, \text{m/s}$$

Therefore, the velocity of the heavier fragment is $12\sqrt{2} \, \text{m/s}$. The correct option is A.

Given: Two equal fragments each have momentum magnitude $4 \times 18 = 72$ and move perpendicular to each other.

Find: The speed of the heavier fragment.

The resultant of two perpendicular equal momentum vectors of magnitude $72$ is

$$72\sqrt{2}$$

Since total momentum must remain zero, the heavier fragment must have momentum of the same magnitude in the opposite direction.

$$p_3 = 72\sqrt{2}$$

Now divide by its mass $6 \, \text{kg}$:

$$v_3 = \frac{72\sqrt{2}}{6} = 12\sqrt{2} \, \text{m/s}$$

Therefore, the correct option is A.