A toy rocket traveling east explodes into 2 equal pieces the instant the rocket is at maximum height & 100 m east of the launch site. Both of the pieces continue to travel east. One lands 170 m...

Question

east of the launch site. Where does the other piece land?

Answer 1

Hi Sarah B.,

We can first conclude that if the toy rocket had not exploded, it would have landed 200 m east of the launch site, since at maximum height (half way), the toy rocket is 100 m east, so it will land at another 100 m east, or 200 m.

Next, the explosions force is assumed equal on the two equal pieces created, and both pieces continue to travel in an east direction. Since one of the toy rocket pieces landed 30 m short (200 m - 170 m) of where the toy rocket would have landed without the explosion, the force on this piece from the explosion was in the (-)east direction. Therefore the force of the explosion on the other toy rocket piece is in the (+)east direction, with equal force, thereby adds 30 m to where the toy rocket would have landed without the explosion.

So 200 m + 30 m = 230 m is where the other piece lands.

I hope this helps, Joe.

Answer 2

Hi Sarah B.,

Somewhere more than 100 m east of the launch site! You cannot say more, b/c you have 2 independent variables to solving the subsequent parabola problem: the instantaneous velocity of the rocket just before the explosion, and the height that it attains (more on this later). You could carry these variables forward into a formal expression for the momenta of the pieces (m(v+Δv) and m(v-Δv) respectively), and then their impact trajectories. (At least, you should be able to!).

But more to the point: something blew these pieces apart: it must have involved an expanding gas. Such a component is not guaranteed to eject with neutral east-west momentum, is it (even if the two solid pieces are equal mass)? So there's a wild card in your calculations.

Another wild card: the two fragments are not guaranteed to have negligible air resistance (call this the Challenger scenario, if you will). Even if they had identical air resistance coefficients, the effect on the faster-moving piece will be greater, hence the correction larger.

To see why you can't set any better limit, consider the two extremes: (1) the "nearer" piece moves negligibly east, and the intact rocket would have landed 135m east of the launch site; or, (2) the 170 m piece is the nearer piece, the intact rocket was moving at high velocity (there's no guarantee that the orbit up to that point was a parabola!), and the second piece is moving at about twice that velocity, hence lands far away. I think for toy-rocket-sized objects, the second piece is unlikely to attain escape velocity, given air resistance and all, but you never know!

-- Cheers, -- Mr. d.