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…

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 )

Aicharya · Accepted Answer

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

t = 1 s

,

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

∴ \vec{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

A particle of mass 2 m is projected at an angle of 45 ° with horizontal with a velocity of 20 2 m s - 1 .…

Solution