Momentum and force combined question

[JayK]

Homework Statement

a. A skateboarder of mass 52kg is riding her 5.5kg skateboard at a speed of 4.3m/s. She jumps off her skateboard, sending the skateboard forward at a speed of 8.6m/s. what is the velocity of the skateboarder after she jumps off?
b. What is the net force if it takes 1.25 for her to come to a stop?
c. what is the coefficient of friction?

Homework Equations

m1v1+m2v2=m1v1'+m2v2'
Fnet=ma
Ffr=(coefficient of friction)(Fn)

The Attempt at a Solution

For a, what I did was, I used the first formula to figure it out, so..
(57.5)(4.3)+(5.5)(4.3)=(57.5)(v1')+(5.5)(8.6)
270.9=57.5v1'+47.3
v1'=3.9m/s
Is this right? for m1 I added 52kg and 5.5kg.

For b, is the initial velocity 4.3m/s and the final 3.9m/s??
I really don't get c.

 

[patzer234]

For your momentum equation there is initially only one velocity, 4.3 metres per second along the positive x-axis ( try drawing a sketch here as velocity is a vector and direction will affect the sign of the value of the velocity), so it should be written $\left(m_1+m_2\right) v_i$ on your left hand side (where the subscripts 1 and 2 refer to the skateboarder and skateboard respectively). Then it should be simple to rearrange and isolate the final speed of the skateboarder.
For b visualise what happens to the skateboarder in the 1.25 s it takes for her to come to a stop. She leaves the board with the speed calculated in the first part of the question then she has final velocity at the end of the 1.25 seconds. I'll let you figure what the final velocity is.
Then to find the net force you have to make a substitution. What is acceleration equal to?
For part c it should be simple to figure out the coefficient of friction if one knows the force calculated in part b).

 

[TomHart]

m1v1+m2v2=m1v1'+m2v2'

(57.5)(4.3)+(5.5)(4.3)=(57.5)(v1')+(5.5)(8.6)

For some reason, you plugged in the sum of the 2 masses (girl + skateboard) for m1.
For part b, you have 3 knowns for the girl after the moment she left the skateboard: Initial velocity, final velocity, and time. From that, you are able to find her acceleration and, if you wanted to, the distance it took her to stop. And from her acceleration, you should be able to find the force acting on her that caused her to stop. To clarify, she is sliding on the ground and the friction between her feet and the ground causes her to stop - that is, unless she fell and did a face plant into the ground. :) And, as patzer mentioned, based on that calculated force, you should be able to find the coefficient of friction. I'll give you a hint: Look at your relevant equations listed.

 

[haruspex]

for m1 I added 52kg and 5.5kg.

Why, when you have m2 for the mass of the board, with its own momentum term, in the same equation? Is the girl riding one board and carrying another?

What is the net force if it takes 1.25 for her to come to a stop?

It should ask for the average force to come to a stop in 1.25s.

what is the coefficient of friction?

There is no way to know from the information given.
If we are given that the girl slides to a stop in 1.25s then we can find the coefficient of kinetic friction.
But most people would run to a stop to avoid slipping, and thus avoiding falling over. In that case all we can calculate is the minimum possible coefficient of static friction.