A body is projected from the ground for horizontal
range 100 m. At the highest point of its path it
breaks apart into two identical part P and Q. If P
comes to rest, then the horizontal distance of the
landing point of part Q, from the point of projection
is
(1) 200 m
(2) 150 m
(2) 250 m
(4) 50 m
Answers
hey mate...
here is your answer...
Answer : (2) 150 m.
SOLUTION :
Velocity of Centre of mass is constant, since internal forces don't chage the motion of Centre Of Mass.
The object breaks into 2 equal masses
The COM will trace the path of a projectile, since it's motion remains unchanged.
X coordinate of COM will be 100, i.e. given Range.
Let M be the total mass of the object
Let m be mass of each half
2m=M
X coordinate of COM = [mx + m(50)]/m +m
= m(x +50)/2m
100 = (x + 50)/2 (m gets cancelled)
200= x + 50
x = 150
hope it helps...
please mark as the brainliest...
Answer:
x = 150m
Explanation:
As there is no external force acting on the system
If the body wouldn't break into 2 pieces it would land at a distance of 100 m
Then its center of mass would be at 100 m
Even if the body breaks the center of mass of the system should remain at 100 m
As one of the pieces falls down vertically it reaches 50 m
2m(100)=m(50)+mx
x=150m