0
$\begingroup$

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?.

$\endgroup$
  • 1
    $\begingroup$ Hint: What is the trajectory of the center of mass? $\endgroup$ – amd Dec 8 '19 at 22:55
  • $\begingroup$ @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!. $\endgroup$ – Chris Steinbeck Bell Dec 9 '19 at 12:33
0
$\begingroup$

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 $x_0-120=240$.

$\endgroup$
0
$\begingroup$

We are apparently meant to assume that the two fragments of the projectile are the only significant pieces of mass that result from the "explosion". So the effect of the explosion is the same as if the two fragments pushed each other away; no external force is applied to the projectile-fragment system of masses, so the total momentum of that system is conserved.

Now you can proceed in various ways from there. Since you have posted another problem soon after this one whose easiest solution method is to track the center of mass of two objects, perhaps you are doing a sequence of problems based on center of mass of a system. In that case, since there is no external force, the center of mass continues on the same trajectory that the original projectile had, until one or both fragments hit the ground. You already have an answer explaining this method.

Alternatively, you can work out the velocity of the projectile at the instant before the explosion, find its momentum, and find an expression of the total momentum of the two fragments afterward. The only unknown in all of that is the velocity of the second fragment. Setting the two expressions for momentum equal, solve for the velocity of the second fragment, and plot where it lands given its initial position and velocity (a straightforward projectile question).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.