A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ____m/s.
Conservation of Momentum diagram for Q27 - JEE Main 2026 Evening
Vector diagram illustrating the trajectories of the explosive fragments.

Solution & Explanation

### Related Formula vecp_textinitial = vecp_textfinal M_1vecV_1 + M_2vecV_2 + M_3vecV_3 = 0 ### Core Logic
Conservation of Momentum diagram for Q27 - JEE Main 2026 Evening
Vector diagram illustrating the trajectories of the explosive fragments.
The total mass is 14 \, mathrmkg and breaks in ratio 2:2:3. Let the masses be M_1, M_2, and M_3. M_1 = frac27 times 14 = 4 \, mathrmkg M_2 = frac27 times 14 = 4 \, mathrmkg M_3 = frac37 times 14 = 6 \, mathrmkg ### Step 1: Setup Momentum Equations Since M_1 and M_2 fly off perpendicular to each other, let their velocity vectors be along the x and y axes. Because the total momentum must be zero, we align the fragments opposite to the final 3rd fragment's direction for simplicity, or just use standard axes. Let vecV_1 = -18hati and vecV_2 = -18hatj. Note that the solution simplifies the mass ratio directly to 2, 2, 3 as relative masses for the momentum equation: 2(-18hati) + 2(-18hatj) + 3vecV_3 = 0 ### Step 2: Solve for Velocity vecV_3 = frac36hati + 36hatj3 = 12hati + 12hatj Magnitude of vecV_3: |vecV_3| = sqrt12^2 + 12^2 = 12sqrt2 \, mathrmm/s ### Pattern Recognition In a 3-part explosion from rest, the momentum of the third piece must be equal and opposite to the vector sum of the other two pieces. You can use the ratio of masses directly in the momentum conservation equation instead of absolute masses to save time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Center of Mass and Collisions

Reference Study Guides

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