Question: An instrument-carrying rocket accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is L. The rocket breaks into two pieces that fly apart horizontally. The larger piece has three times the mass of the smaller piece. To the surprise of the scientist in charge, the smaller piece returns to Earth at the launching station. How far away does the larger piece land? Neglect air resistance and effects due to the Earth’s curvature.

My attempt:

Since there are no external forces in the horizontal direction - $ dp/dt_x=0 $

If the light object of mass $ m $ travels back to its starting location, it must take time sqrt(2h/g) where $ h $ is the height at which the explosion happens. Therefore, the horizontal velocity of the light block is $ -L /\sqrt{2h/g} $ and it follows that the velocity of the heavy mass is $ (X - L) / \sqrt{2h/g} $ where X is the distance from the launch point to where the heavier mass hits the ground.

The conservation of momentum equation is: $$ m*-L /\sqrt{2h/g} = 3m * (X - L) / \sqrt{2h/g} $$

Solving this equation gives $ X = 4/3 L $ but the actual answer is $ X = 8/3 L $. What is wrong with my solution?


1 Answer 1


The rocket had some horizontal momentum $4mv$ just before exploding. You have forgotten to consider this term in your equation for the conservation of momentum.

Assuming the rocket to be a simple projectile without any thrust, its horizontal velocity at the top of its trajectory can be given as $v = L/\sqrt{\frac{2h}{g}}$

The equation for the conservation of momentum can be given as

$$\frac{4mL}{\sqrt{\frac{2h}{g}}} = \frac{-mL}{\sqrt{\frac{2h}{g}}} + \frac{3m(X-L)}{\sqrt{\frac{2h}{g}}}$$

$$\implies X = \frac{8L}{3}$$

You can verify this intuitively. The center of mass of the system lies at a distance of $\frac{m(0) + 3m(\frac{8L}{3})}{m+3m} = 2L$, from the launch point, which is exactly how far the rocket would go if it didn't explode.

  • $\begingroup$ How come the center of mass continues on the same trajectory? $\endgroup$ Mar 7 at 6:07
  • $\begingroup$ @EmilSriram Because the explosion is an internal force. The only external force acting on the system is gravity in the downward direction. Hence, the CoM continues to move in a parabolic path. $\endgroup$
    – Amogh
    Mar 7 at 8:10

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