Question

A projectile of mass 50 gram is projected with velocity of 10 M per second from the ground at an angle 45 degree with the horizontal. Find the magnitude of the change in the momentum between 'living' and 'arriving' back to ground.

Answer 1

Answer:

- 0.35 kg. m s⁻¹ j

Explanation:

Initial Velocity = v cos 45 i + v sin 45 j

Final Velocity = v cos 45 i

Change in momentum = ( Final velocity - Initial Velocity ) m

Change in momentum = ( v cos 45 i -  v cos 45 i - v sin 45 j) 0.05 kg

Change in momentum = ( - v sin 45 j) 0.05

Change in momentum = ( - 10 x 0.707 j) 0.05

Change in momentum =  - 0.35 kg. m s⁻¹ j

Answer 2

A projectile of mass 50 gram is projected with velocity of 10 M per second from the ground at an angle 45 degree with the horizontal. Find the magnitude of the change in the momentum between 'living' and 'arriving' back to ground.

Initial Velocity = VCosθ + i Vsinθ

Final velocity = VCosθ - i V Sinθ

Chnage in Velocity = Final Velocity - Initial Velocity

= VCosθ - i V Sinθ - (VCosθ + i Vsinθ)

= - 2iVSinθ

Change in Momentum = m * Change in Velocity

= 50 * (- 2iVSinθ)

= -100 * i * 10 * Sin45°

= -1000i /√2

= 707i gm/s