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a particle of mass 3m is projected from ground with velocity 20 root 2 at 45 degrees at highest point it explodes into two pieces. one of mass 2m and other of mass m.both moves in opposite direction horizontally.mass 2m falls at a distance of 100m from point of projection. the distance of second mass from point of projection is...

a particle of mass 3m is projected from ground with velocity 20 root 2 at 45 degrees at highest point it explodes into two pieces. one of mass 2m and other of mass m.both moves in opposite direction horizontally.mass 2m falls at a distance of 100m from point of projection. the distance of second mass from point of projection is...

Grade:11

2 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
7 years ago
soln

223-637_Capture.PNG
Manas Shukla
102 Points
7 years ago
In this problem the easiest way to solve is you can assume that the centre of mass of the particle will continue to follow projectile motion.
As we know particle of mass 2m lands at 100m.
Time period
v = u + at
20\sqrt{2} \sin 45 =10t taking g = 10m/s2
t = 2 sec = half of flight so total time = 4 sec
So range of projectile if it had not exploded = 80m
So the com of the particle will also land at 80 m
thus
80 = \frac{2m\times 100 + x}{3m}
x = 40m.

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