Newton's Laws Question: Skateboard Speeds After Release | F=ma Explained

[beancurd]

Jack and Jill are on skateboards. Jack weighs 3 times heavier than Jill, and is pushing Jill(a horizontal force on Jill's back). Immediately after Jack let's go, what are the speeds they're moving at? (No need to take friction into consideration)

F=ma
Okay, I do understand that they will be moving away from each other, and that Jill's acceleration will be 3 times faster than Jack. But that does not mean that Jill's speed is three times faster than Jack IMMEDIATELY after he let's go, right?I think they will have equal speed... but I'm not sure. Can someone answer this question?

 

[Filip Larsen]

Welcome to PF!

What principle do you think apply here? Can you write up an equation modelling that principle for Jack and Jills speed and mass?

 

[beancurd]

Newton's third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear

F= ma for Jill and
F = 3ma for Jack

Since force is the same, I know that the magnitude of acceleration of Jill will be three times greater than Jack's. Generally at any given time then, Jill's speed will be three times greater than Jack's. But at t=0, the initial speed? That, I'm not too sure. My guess is that the initial speed of both Jack and Jill should be the same...

Yeah.. that's all I know.

 

[Filip Larsen]

You are correct about the accelerations of Jack and Jill, and I agree that it makes most sense to assume their relative speed is zero to start with.

I assumed before that you had heard of momentum and the principles associated with it, but if not, you can assume that F is constant over a small time interval t and use that to prove that their speeds also are opposite and a factor three different in magnitude (assuming you know how to find the speed increase from an object under constant acceleration as a function of time).

 

[beancurd]

Ah, I see. Thank you for your help.