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The question (from Sparknotes SAT Physics):

An athlete of mass 70.0 kg applies a force of 500 N to a 30.0 kg luge, which is initially at rest, over a period of 5.00 s before jumping onto the luge. Assuming there is no friction between the luge and the track on which it runs, what is its velocity after the athlete jumps on?

Here is what I believe is the correct solution:

After 5 seconds of application of 500 N to the luge, its momentum will be 2500 N.s, so its velocity will be $\frac{250}{3}$ m/s. As the athlete is at rest with respect to the luge, when he jumps onto the luge, it can be assumed that mass (and therefore, energy) is being added to the system containing the luge. Hence, there should be no changed in the final velocity of the luge-athlete system since its combined energy is equal to the sum of the energies of the systems containing the luge and the athlete individually.

The solution given is:

The athlete imparts a certain impulse to the luge over the 5-s period that is equal to $F.∆t$. This impulse tells us the change in momentum for the luge. Since the luge starts from rest, this change in momentum gives us the total momentum of the luge. The total momentum of the luge when the athlete jumps on is 2500 kg · m/s. Momentum is the product of mass and velocity, so we can solve for velocity by dividing momentum by the combined mass of the athlete and the luge, and get 25 m/s as the final velocity.

I have a gut feeling that my reasoning is indeed faulty, but I'm unable to figure out why.

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  • $\begingroup$ I agree with your solution. Your solution is akin to someone running and pushing (of course the athlete has to run to accelerate himself as well, or else he won't be able to apply the force for so long!) while their solution is akin to, say, a rollercoaster accelerating (500N for 5s) and the athlete standing on a platform idly grabbing on and being dragged along. $\endgroup$
    – user12029
    Aug 15, 2015 at 18:15
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    $\begingroup$ This is a good example of a problem that is poorly worded. $\endgroup$
    – David White
    Aug 15, 2015 at 19:01

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As the athlete pushes off the ground, she and the luge would both be accelerating relative to the ground. This force of 500N for 5 seconds would result in a final momentum of 2500Kgm/s. If the system has 100kg total mass then you simply divide the momentum of 2500kgm/s by the mass 100kg to find the velocity which I believe is 25m/s not 250m/s. Also note that 250m/s is an unreasonable velocity for a luge and that 25m/s makes a bit more sense.

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  • $\begingroup$ So basically, I should assume the force to be actually external to the athlete-luge system? Also, thanks for the point about the unreasonable velocity, never thought of that. $\endgroup$
    – duci9y
    Aug 15, 2015 at 18:04
  • $\begingroup$ The ground is pushing on them, yes. $\endgroup$
    – Alex
    Aug 15, 2015 at 18:05
  • $\begingroup$ @Alex: Why do you think the force exerted on the athlete by the Earth is the same as that exerted by the athlete on the luge? $\endgroup$
    – akhmeteli
    Aug 15, 2015 at 18:14
  • $\begingroup$ The athlete and the luge are a single object connected and traveling at the same rate in this simplified homework problem. $\endgroup$
    – Alex
    Aug 15, 2015 at 18:18
  • $\begingroup$ @Alex: This is not at all obvious: the athlete's hands can move faster than the athlete's center of mass for 5 s. $\endgroup$
    – akhmeteli
    Aug 15, 2015 at 19:17
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I have a gut feeling that my reasoning is indeed faulty, but I'm unable to figure out why.

If a 500N (net) force acts on a 30kg object, the acceleration of the object is

$$a = \frac{500}{30} \mathrm {\frac{m}{s^2}} \approx 1.7g$$

which gives a 0 to 100 km/h time of 1.67 seconds thus beating all of the quickest supercars.

In other words, there's something not quite right with this problem as written. So, it must be that 500N is the net force accelerating the combination athlete/luge producing an acceleration of

$$a = \frac{500}{100} \mathrm {\frac{m}{s^2}} \approx 0.51g$$

which, after 5 seconds, produces a change in velocity of

$$\Delta v = at = 25\mathrm{\frac{m}{s}}$$

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In general, I might prefer your solution to the official one, the more so as their numbers are wrong, as Alex noted. There is, however, a problem with your solution as well: the athlete could not achieve the speed of 250/3 m/s (I guess it's 300 km/hour), even if (s)he did not have to push the luge. So I guess before s(he) jumped onto the luge her (his) speed should have been much less than that of the luge.

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