A particle of mass 2 m is projected at an angle of 45 ° with horizontal with a velocity of 20 2 m s - 1 . After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. The maximum height attained by the other part is (Take g = 10 m s - 2 )

Question

A particle of mass 2 m is projected at an angle of 45 ° with horizontal with a velocity of 20 2 m s - 1 . After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. The maximum height attained by the other part is (Take g = 10 m s - 2 ), 35 m, 40 m, 15 m, 20 m

MARKS App · Accepted Answer

Applying conservation of linear momentum at the time of the collision, or at t = 1 s , m v → + m 0 = 2 m 2 0 i ^ + 1 0 j ^ ∴ v → = 4 0 i ^ + 2 0 j ^ At 1 sec, masses will be at height : h 1 = u y t + 1 2 v y t 2 = 2 0 1 + 1 2 - 1 0 1 2 = 15 m After explosion other mass will further rise to a height : h 2 = u y 2 2 g = 2 0 2 2 × 1 0 = 2 0 m : u y = 20 m/s just after the collision. ∴ Total height h = h 1 + h 2 = 35 m

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