# How do I find the horizontal distance of a fragment in an explosion?

#### Chemist116

The problem is as follows:

A projectile is shot with an initial speed of $50\,\frac{m}{s}$ with an angle of $53^{\circ}$. On its highest point of its trajectory it explodes splitting in two fragments of the same mass, one of them inmediately after the collision has zero in speed and fells vertically. Find the horizontal distance in $m$ which the other fragment travels. You may use $g=10\,\frac{m}{s^2}$.

The alternatives are as follows:

$\begin{array}{ll} 1.&120\,m\\ 2.&180\,m\\ 3.&200\,m\\ 4.&240\,m\\ 5.&320\,m\\ \end{array}$

I'm not sure how to tackle this problem. I am assuming that it has to do with the conservation of momentum but I don't know how should I proceed with an explosion.

The only thing which I can say is that the components of the bullet will be:

$v_x=50\cos53^{\circ}= 50\times\frac{3}{5}=30$

$v_y=50\sin53^{\circ}= 50 \times \frac{4}{5}=40$

And i believe that these speeds might be used in the analysis for the explosion but I'm not sure what to do with those, am I right with my analysis?. Can someone help me with this thing please?.

#### topsquark

Math Team
Your $$\displaystyle v_x$$ and $$\displaystyle v_y$$ are your initial speeds. The explosion happens at maximum height. What are the velocity components at that point? The explosion won't conserve kinetic energy, but you won't need to use it. Conservation of momentum will work just fine.

-Dan

#### Chemist116

Your $$\displaystyle v_x$$ and $$\displaystyle v_y$$ are your initial speeds. The explosion happens at maximum height. What are the velocity components at that point? The explosion won't conserve kinetic energy, but you won't need to use it. Conservation of momentum will work just fine.

-Dan
At maximum height the only component will be $v_x$ and this moves with constant motion. No acceleration.

But how exactly should I use the conservation of momentum for the explosion?. Can you help me with this please?.

#### DarnItJimImAnEngineer

$\vec{p}_{before~explosion} = \vec{p}_{1,after~explosion} + \vec{p}_{2,after~explosion}$
All three momenta are in the same direction, so you can ignore the vectors, and one of them is zero, so the arithmetic is easy.