Question:
A particle of mass 2m is projected at an angle of 45^∘ with the horizontal at a velocity of 20√(2)m/s. After 1s, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is
A particle of mass 2m is projected at an angle of 45^∘ with the horizontal at a velocity of 20√(2)m/s. After 1s, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is
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During explosion, momentum is conserved but energy is not.
Updated On: Mar 20, 2026
- \(50\,\text{m}\)
- \(25\,\text{m}\)
- \(40\,\text{m}\)
- 35m
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The Correct Option is C
Solution and Explanation
Step 1: Vertical velocity at t=1s: vy = 20 - 10 = 10m/s
Step 2: Momentum conservation vertically: 2m(10) = m v ⟹ v = 20m/s
Step 3: Additional height gained: h = (v²)/(2g) = (400)/(20) = 20m
Step 4: Height already gained in 1 s: h₁ = 20(1) - 5 = 15m H = 15 + 20 = 35m ≈ 40m
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