A particle of mass 2m 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 2m 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 ) (A) 35m (B) 40 m (C) 15 m (D) 20 m

Tardigrade Admin · Accepted Answer

Applying conservation of linear momentum at the time of the collision, or
at

t = 1 s

m v + m (0) = 2 m (20 \hat{i} + 10 \hat{j})

∴ v = 40 \hat{i} + 20 \hat{j}

At 1 sec, masses will be at height:

h_{1} = u_{y} t + \frac{1}{2} v_{y} t^{2} = (20) (1) + \frac{1}{2} (- 10) (1)^{2} = 15 m

After explosion other mass will further rise to a height:

h_{2} = \frac{u _{y}^{2}}{2 g} = \frac{( 20 ) ^{2}}{2 \times 10} = 20 m : u_{y} = 20 m / s

just after the collision.

∴

Total height

h = h_{1} + h_{2} = 35 m

$35 m$

$40 m$

$15 m$

$20 m$

35m

40m

15m

20m

$35 m$

$40 m$

$15 m$

$20 m$