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

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?.

• Hint: What is the trajectory of the center of mass? – amd Dec 8 '19 at 22:55
• @amd I don't know exactly how should I proceed with this problem. Perhaps do you mind to offer some answer or an extended hint!. – Chris Steinbeck Bell Dec 9 '19 at 12:33

The only external force on the projectile is due to gravity. Even when the projectile splits, the forces that cause this are internal. Thus after splitting, the centre of mass of the projectile continues free fall as before. Had it not split, the range of free fall for the projectile would be$$\frac{u^2\sin(2\theta)}g=240$$This means that when the two fragments land, the centre of mass of the system consisting of the two fragments is at $$x=240$$. The fragment with $$0$$ initial speed falls down at $$x=120$$. Let the other fragment land at $$x=x_0$$. Thus,$$\frac{\frac m2\cdot120+\frac m2\cdot x_0}{\frac m2+\frac m2}=240$$The horizontal distance travelled by this fragment alone is $$x_0-120=240$$.