Question

A particle of mass 2 kg is projected at an angle of 60° above the horizontal, with a speed 20 m/s. The magnitude of angular momentum of the particle about point of projection, when it is at maximum height will be​

Answer 1

Answer: 300kgm^2/s

Explanation: taking Mass = 2 kg and speed (v) = 20m/s

This is your answer...(FTS test 3 physics)

Answer 2

Given ,

Mass of the particle = 2 kg

Angle of the projection of the  particle = 60°

speed (v)  of the projection of the particle

v = 20 m/s

To find :

The magnitude of angular momentum of the particle about point of projection when it is at its highest point=?

Solution :

∴We know that the angular momentum of a particle about any point  is the cross product of the distance of the particle from that point and linear momentum of the particle at that instant.

• So here 1st we find the velocity of the particle at the highest point which is = v cos 60° m/s

           = $20 \times \frac{1}{2} = 10$

           =10 m/s :

• Now the maximum height of the particle

    H = $V^{2} sin^{2} (60)\frac{1}{2g}$

               =$\frac{20^{2}\times3 }{2\times10\times4} =15$

                ⇒ H=15 m/s     :

                                                                         

• Now the linear momentum of the particle at the highest point:

    P=$m\times v$

      =$2 \times 10$

      = 20 kg m/s :

•   Finally the angular momentum L of the particle will be:

        $L=r\times p\\ =H\times p\\ =15\times 20\\ =300$