Question

A body of mass 2kg is projected from the ground with a velocity 20ms at an angle 30° with the
verticle. Ift. is the time in seconds at which the body is projected and t, is the time in seconds at
which it reaches the ground, the change in momentum in kgms during the time (t-t) is
1) 40
2) 40.53
3) 50.13
4) 60​

Answer 1

The given question

answer is 50.13

Answer 2

answer : 40√3 Kgm/s

explanation : let's first find out velocity of the body at t = 0 and when it reaches the ground.

the body of mass , m = 2kg is projected with speed 20m/s at an angle 30° with the vertical.

so, initial velocity of the body , u = $20sin30^{\circ}\hat{i}+20cos30^{\circ}\hat{j}$

= 10 i + 10√3 j

find time of flight , t = $\frac{2u_y}{g}$

where ux is the vertical component of initial velocity of the body.

so, t = $\frac{2ucos30^{\circ}}{g}$

= (2 × 20 × √3/2)/(10)

= 2√3 sec

now velocity of body after time t = 2√3sec , v = $20sin30^{\circ}\hat{i}+20cos30^{\circ}-g(2\sqrt{3})\hat{j}$

= 10i + (10√3 - 20√3)j

= 10i - 10√3j

now, change in linear momentum = mass of body × (final velocity - Initial velocity)

= m(v - u)

= 2kg × [(10i - 10√3j) - (10i + 10√3j)]

= 2(-20√3)j

= -40√3 j

magnitude of change in linear momentum = 40√3 Kgm/s