Question

a ball of mass =5kg is thrown at an angle is 60 with the horizontal with an initial velocity u= 40mper s the change in its momentum during its flight in a time interval sec is ​

Answer 1

Explanation:

Change in momentum

( final momentum-inial momentum)

The object is projected at an angle θ

So, initial velocity be, u i = ucosθ i^+usinθ j^

The vertical component that lifting the body upward is usinθ j^

In projectile motion, horizontal component of velocity remains unchanged.

Therefore in total time of flight, ie; at the end of the motion,

velocity is, uf =cosθ i^-sinθ j^

Thus change in velocity in the motion, Δv=uf −ui

Δu=−2usinθ

So, change in momentum, ΔP=−2musinθ

This is the change in momentum in total time of flight.

In half of time of flight, change in momentum is= -musinθ

So the magnitude of change in momentum is in half of time of flight is musinθ