Dr. Wang forgot to charge his electric skateboard! Rather than push the skateboa

Question

Dr. Wang forgot to charge his electric skateboard! Rather than push the skateboard with his foot (he's far to lazy for that), he fills a box full of kittens and plans to get the skateboard rolling by throwing kittens behind him (he knows that kittens always land on their feet and. therefore, won't be harmed). Unfortunately, the mass of a kitten (0.5-kg) is very small compared to the mass of the skateboard (5-kg) and Dr. Wang (75-kg) (you can neglect the mass of the box used to carry the kittens). Because of the large difference in mass, he has to throw the kittens very fast (poor kittens!). If he starts with 10 kittens and throws each kitten with speed of 10 m/s relative to the skateboard, how fast will he be moving after all 10 kittens have been thrown? You can assume the skateboard is frictionless for this very disturbing analysis.

Explanation / Answer

Momentum of the each kitten after throwing :

Pk = -m*u = -0.5*10 = -5 kg.m/s <------ negative sign for opposite to the direction of Dr Wang.

So, by conservation of momentum, change in momentum of ( skateboard+Dr Wang) after each throw:

Pt = 5 kg.m/s

Now, after 1st throw,

the momentum of (skateboard +Dr Wang+9 kittens) = 5 kg.m/s

After 2nd throw, the momentum of (skateboard +Dr Wang+8 kittens) = 5 +5 = 10 kg.m/s

After 3rd, = 5 + 5 +5 = 15 kg.m/s

So, after 10 throws, the momentum of (skateboard +Dr Wang) = 10*5 = 50 kg.m/s

So, M*v = 50

So, (75+5)*v = 50

So, v = 50/80 = 0.625 m/s

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