At what velocity does the skateboard shoot off?

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SUMMARY

The skateboard's velocity after a person jumps off can be calculated using the conservation of momentum formula, m1v1 = m2v2. Given a person with a mass of 50 kg jumping off at a velocity of 3 m/s, the skateboard, which has a mass of 3.5 kg, would shoot off at a velocity of approximately 42.857 m/s. However, this calculation assumes the 3 m/s is relative to the skateboard, not the ground, which is a critical distinction. Realistically, achieving such a velocity in a short time frame is highly improbable due to the required acceleration and force.

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cody7688
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A person stands still on skateboard. He then jumps off at a velocity of 3 m/s. His mass is 50 kg and the skateboard has a mass of 3.5 kg. At what velocity does the skateboard shoot off?2. Is my answer correct?
If not what did I do wrong?3. m1v1=m2v2
V= 42.857 m/s
 
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cody7688 said:
A person stands still on skateboard. He then jumps off at a velocity of 3 m/s. His mass is 50 kg and the skateboard has a mass of 3.5 kg. At what velocity does the skateboard shoot off?2. Is my answer correct?
If not what did I do wrong?
3. m1v1=m2v2
V= 42.857 m/s
Unfortunately the question does not make clear whether the 3m/s is relative to the ground (as you have assumed) or relative to the skateboard.
(In fact, it is not even clear whether the jump is in the fore-aft direction of the skateboard!)
 
Hello Cody, welcome to PF :smile: !

I see your post was edited by a moderator; I wonder what was changed ?!

If you make the assumption Haru mentions, you end up with the result you obtained. It is worth checking if that is realistic (even if exercise composers and reallity are rather often at odds...)

3 m/s is 10.8 km/h, so modest running speed. It's difficult to give yourself that speed in a fraction of a second even when on solid ground, let alone when on a skateboard !
Assume you have 0.1 sec to push off (by then the skateboard is well under way - as we will see).
That requires an acceleration of 430 m/s2 and a force of 430 * 3.5 = 1500 N. Three times body weight !
And in 0.1 s, the board is ½ * 430 * 0.12 = 2.14 m away. More than 2 leg lengths. So even the 0.1 is too long !

Shorter legs ? 0.03 s pushing time (to get the 2.14 m down to 0.64 m) requires 5000 N, ten times body weight. No way !

(All these calculations are rough: assuming constant acceleration gives reasonable estimates, though)
--

My conclusion: the 3 m/s is wrt the board. Less sensational, I agree.
I do get reasonable numbers.
Anyone beg to differ ?

--
 

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