Question bank/JEE Main/Physics

A projectile of mass 3 m is projected from ground with velocity
$20\ \sqrt{2}$m/s at an angle of 45º. At highestpoint it explodes into
two pieces. One of mass 2 m and the other of mass m. Mass 2 m falls at a
distance of 100 m from point of projection towards right of point of
projection. If the distance(inm) of second mass from point of projection
where it strikes the ground is n × 10, then what is the value of n.(g=10

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Range of the projectile in the absence of explosion
The path of centre of mass of projectile will not change, i.e.,
x_COM is still 80 m, Now, from the definition of centre of mass
$x_{COM} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}$
or $80 = \frac{(m)\left( x_{1} \right) + (2m)(100)}{m + 2m}$
Solving this equation, we get
x_1 = 40m
Therefore, the mass m will fall at a distance x_1=40 cm from point of
projection.

Source: Grass-G / P-07 Physics paper 17 October.docx