Calculating Velocity and Displacement in a Skateboarding Physics Experiment

AI Thread Summary
The discussion focuses on calculating the velocity and displacement of a boy and girl on skateboards after they push off each other with a force of 30N for 1.2 seconds. The boy's mass is 45kg, resulting in an acceleration of approximately 0.6666 m/s², while the girl's mass is 40kg, leading to an acceleration of about 0.75 m/s². The calculated velocities after pushing are 0.79 m/s for the boy and 0.9 m/s for the girl, with the girl's displacement during the push calculated at 0.54 meters. Participants suggest using the impulse/momentum theorem for a more straightforward calculation and emphasize considering the movement of their center of masses. The calculations and suggestions are generally affirmed as correct and helpful.
MetalCut
Messages
21
Reaction score
0
I just want to know if this is correct.
The question is:
A boy and a girl on a skateboard push each other away at 30N for 1.2 s
The boys' mass is 45kg and the girl's mass is 40kg.

Determine the velocity after they have stopped pushing each other?
How far did the girl move while their hands were touching?

BOY Girl
F=30N F=30N
m=45kg m=40kg
a=0.6666m/s/s a=0.75m/s/s

Now to get the velocity, do i use v=v0+at, I am not sure.
So for the boy then v=0+(0.66666)(1.2)
v=0.79m/s
And for the girl then v=0+(0.75)(1.2)
v=0.9m/s

And for the displacement of the girl: = v0t+(0.5)(0.75)(1.2)^2
= 0.54m
 
Physics news on Phys.org
MetalCut said:
I just want to know if this is correct.
The question is:
A boy and a girl on a skateboard push each other away at 30N for 1.2 s
The boys' mass is 45kg and the girl's mass is 40kg.

Determine the velocity after they have stopped pushing each other?
How far did the girl move while their hands were touching?

BOY Girl
F=30N F=30N
m=45kg m=40kg
a=0.6666m/s/s a=0.75m/s/s

Now to get the velocity, do i use v=v0+at, I am not sure.
So for the boy then v=0+(0.66666)(1.2)
v=0.79m/s
And for the girl then v=0+(0.75)(1.2)
v=0.9m/s

And for the displacement of the girl: = v0t+(0.5)(0.75)(1.2)^2
= 0.54m
Looks good to me:smile:
 
Thanx man.
I appreciate.
 
Looks OK to me. An easier way to find the velocities after they stop pushing is to use the impulse/momentum theorem. (Look it up.)

Also, realize that we are talking about the movement of their center of masses. (The boy and girl are deformable objects.)
 
Thanx man. I will do that.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top