Q.

A projectile of mass 3 m is projected from ground with velocity 202m/s at 45°. At highest point it explodes into two pieces. One of mass 2 m and the other of mass m. Mass 2m falls at a distance of 100m from point of protection. Find the distance of second mass from point of projection where it strikes the ground.g=10ms-2

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answer is 40.

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Detailed Solution

Range of the projectile in the absence of explosion

R=u2sin2θg=(202)2sin90°10=80m

The path of center of mass of projectile will not change, i.e, xC.O.M is still 80m. Now, 
from the definition of center of mass.

Question Image

xCOM=m1x1+m2x2m1+m2

80=(m)x1+(2m)(100)m+2m

on solving this we get x1=40m

Therefore, the mass m will fall at a distance x1=40 cm from point of projection.

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A projectile of mass 3 m is projected from ground with velocity 202m/s at 45°. At highest point it explodes into two pieces. One of mass 2 m and the other of mass m. Mass 2m falls at a distance of 100m from point of protection. Find the distance of second mass from point of projection where it strikes the ground.g=10ms-2