# A projectile of mass M is projected so that its horizontal range is 4 km. At the highest point, the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part starts falling down vertically with zero initial speed? [cont.]

## The horizontal range of the lighter mass is?

Mar 3, 2017

It is not clearly stated however, assumed that the projectile is fired from the ground as shown in the figure below

Maximum height is reached because of $\sin \theta$ component of the velocity where it becomes $= 0$. From symmetry we see that this occurs at a distance $= \frac{r}{2}$ from the origin and time taken is half the time if flight $\frac{t}{2}$. We have ignored the air resistance.
At this point the un-split projectile has only $x$-component of velocity $= u \cos \theta$

As the projectile explodes in two pieces of mass $\frac{M}{4} \mathmr{and} \frac{3 M}{4}$ respectively.

Considering momenta of both the pieces in the $x$-direction.
Let part of mass $\frac{M}{4}$ has velocity ${v}_{\frac{M}{4}}$. Using Law of conservation of momentum, we get
$\text{Initial Momentum"="Final Momentum}$
$M u \cos \theta = \frac{3 M}{4} \times 0 + \frac{M}{4} \times {v}_{\frac{M}{4}}$
${v}_{\frac{M}{4}} = 4 u \cos \theta$

In the vertical direction both pieces fall freely under the action of gravity. As such there is no change in the time of flight which is independent of mass.

Considering smaller piece before explosion,
velocity $= u \cos \theta$, time $= \frac{t}{2}$, horizontal distance traveled $\frac{r}{2} = 2 k m$
$\implies u \cos \theta \times \frac{t}{2} = 2 k m$ ......(1)

Let smaller piece travel a further distance $d$ after explosion,
velocity $= 4 u \cos \theta$, time $= \frac{t}{2}$
Distance traveled $d = 4 u \cos \theta \times \frac{t}{2}$
Using (1) we get
$d = 4 \times 2 = 8 k m$

Horizontal range of lighter part is given as
${r}_{\frac{M}{4}} = \frac{r}{2} + d = 2 + 8 = 10 k m$