A projectile ,of mass 50g is projected with a velocity of 10 m/s from the ground at an angle of 45* with the horizontal. Find the magnitude of the change in its momentum between 'leaving' and 'arriving back' to ground.
Answers
The block arrives at the ground with the same velocity at which it is projected
Projected velocity in horizontal direction= 10cos45= arriving velocity in horizontal direction
Projected velocity in vertical direction=10sin45= arriving velocity in vertical direction
While the block arrives the direction of horizontal component of velocity will be same as that of horizontal component of projected one.
While the velocity of vertical component reverse it's direction
Change in momentum= Final momentum-Intial momemtum
momentum= mass×velocity
While finding the change in momentum the horizontal component of velocity gets cancelled out, while the vertical component gets added up and double
There change in momentum= 2×m×v×sin 45
=2×50×10÷√2
= 500√2 gm/s