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Question

A projectile of mass 3m is projected from the ground with a velocity 30 m/s at 45 with the horizontal. At the highest point, it explodes into two pieces. One of pieces has mass 2m and the other has mass m. Both the pieces fly off after explosion. Mass 2m falls at a distance of 100 m from the point of projection. Find the distance of second mass from the point of projection, where it strikes the ground. (Take g=10 m/s2)
  1. 90 m
  2. 60 m
  3. 80 m
  4. 70 m


Solution

The correct option is D 70 m
Range of projectile in the absence of explosion :-
R=u2sin2θg
=(30)2×sin9010
=90 m


The path of the center of mass will not change. So xCOM will be 90 m.
Say, m1=m and m2=2m

Let x1 be the distance of m1 from the point of projection, where it strikes the ground.
Mass m2 falls at a distance x2=100 m from the projection point.
xCOM=m1x1+m2x2m1+m2
90=m(x1)+2m(100)m+2m
x1=70 m

So, the mass m will fall at a distance x1=70 m from the point of projection.

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