Question

A projectile moving vertically upwards with a velocity of200m/s breaks into two equal parts at aheight of 490m. One part starts moving vertically upward with a velocity of 400m/s .how much time after the break up will the other part hit the ground

Answer 1

as you know that there is only one force acting on body i.e gravity , thus the total acc. of the system should be 'g' only:

if we see closely at the time of breaking , the vel. of the system was 200m/s so , just after time dt it should be almost same as before i.e 200m/s....

let total mass be M

as one mass goes up with vel. 400 m/s,,,

what would happen to the second one?

as per conservation of momentum:-

vel. of com=[vel.(A)*mass(B)+vel(B)*mass(B)]/(total mass)

=200m/s=[400m/s*(M/2)+V*(M/2)]/M

=200m/s=400/2+V

thus v=0,

hence after that the second mass would free fall:

now using :

s=ut+1/2at^2

we get :

490=1/2(9.8)t^2

thus t^2=100s

t=10s