• Support PF! Buy your school textbooks, materials and every day products Here!

Need help showing work for these problems

  • #1
Question Details:
1) A skateboard track has the form of a circular are with a 4.00 m radius, extending to an angle of 90 degrees relative to the vertical on either side of the lowest point. A 57.0 kg skateboarder starts from rest at the top of the circular arc. What is the normal force exerted on the skateboarder at the bottom of the circular arc?

2) A 135-g rubber ball is thrown with a speed of 12m/s at an oncoming 1100-kg car which is approaching at 35 m/s, and undergoes a one-dimensional elastic collision with the car.
a) What is the speed of the ball after the collision
b) Show that kinetic energy of the ball is conserved
Answers: a)Vf, ball=70.16 m/s b) KEi=KEf=1.35x10^6J

3) a plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 245-g piece moves along the x-axis with a speed of 2.50 m/s, a 275-g piece moves along the y-axis with a speed of 1.50m/s. The third piece has a mass of 100g. What are the magnitude and direction of its velocity?
Answer: V3=7.384 m/s and Theta = 34 degrees

I do not know how to solve these equations can someone please show me the work so I can learn the appropriate equations to use and how to use them so I can solve the rest of my questions on my own thanks!
 

Answers and Replies

  • #2
209
1
You're supposed to do the work and show us where you get stuck and then we can give you a hint. This forum is not an answer machine.

To get you started:
1) Since it is a circular track, there is a centripetal force acting on the skateboarder, as well as a gravitational force.

2) Momentum is conserved. Use the equation for momentum and set the initial momentum equal to the final momentum of both objects.

3) Again, momentum is conserved.
 
  • #3
I have the answers already as you can see. What I don't understand is how to get the answers. and if I thought it was an answer machine then I would have posted all 25 questions. If you can't honor the request do not reply to my question Thanks. No time for smart remarks.
 
  • #4
209
1
I have the answers already as you can see.
Great..

What I don't understand is how to get the answers.
That's why I tried to get you started

and if I thought it was an answer machine then I would have posted all 25 questions.
Ok...

If you can't honor the request do not reply to my question. Thanks.
Look, I'm trying to help you. I'm trying to get you started and I'm saying that you shouldn't expect someone to solve the entire problems for you. If you start solving the problems and show us where you get stuck, or just tell us what you think you might have to do we can help you further. If you just throw a couple of questions at us you have a small chance someone will help you. It's ok with me, but I think that's not what you want.

No time for smart remarks.
...


Do you know the equation for centripetal force? Do you know the equation for weight? Do you understand what (conservation of) momentum is?
 
Last edited:
  • #5
Fine I'll bite: for the first problem here is what I got
Ui + Ki= Uf+Kf
mgh + 0 = 0 +1/2 mv^2---v^2= 2gh
sigma Fy=macp
N-mg=m v^2/r
N=mg+m 2gh/r= mg(1+2h/r)
(57kg)(9.8 m/s^2)(2(4m)/4m)--1.68 kN

second problem:Elastic collision means both momentum and total KE are conserved:

mv + MV = mv' + MV'

(1/2)mv^2 + (1/2)MV^2 = (1/2)mv'^2 + (1/2)MV'^2

Don't know how to solve to get answer.

third problem: Clueless even with ur hint
 
  • #6
209
1
Fine I'll bite: for the first problem here is what I got
Ui + Ki= Uf+Kf
mgh + 0 = 0 +1/2 mv^2---v^2= 2gh
sigma Fy=macp
N-mg=m v^2/r
N=mg+m 2gh/r= mg(1+2h/r)
(57kg)(9.8 m/s^2)(2(4m)/4m)--1.68 kN

second problem:Elastic collision means both momentum and total KE are conserved:

mv + MV = mv' + MV'

(1/2)mv^2 + (1/2)MV^2 = (1/2)mv'^2 + (1/2)MV'^2


Don't know how to solve to get answer.

third problem: Clueless even with ur hint
Very well! First answer is correct.

As for the second problem:
mv + MV = mv' + MV'
You know m, M, v and V. The two unknowns you're left with are v' and V'. You can express v in terms of V' (or the other way around). This can be plugged into:

mv^2 + MV^2 = mv'^2 + MV'^2
Leaving you with only one unknown.
 
Last edited:
  • #7
209
1
Last problem:

Momentum in the x direction is conserved, and in the y direction. Draw the momentum vectors. You know that the momentum of the third piece is equal and opposite to the other two, respectively along the x and y axis.

Does that help?
 
  • #8
I hope u are right about the first problem

The second problem although I can set it up I can't solve I am having difficulty plugging the numbers in I know m1 is mass of the ball m2 is mass of car v1 is speed of ball after collision and v2 is speed of car. But I still can't solve. algebra isn't my strongest subject

third problem
does this make sense? mvx + mvy +mv3=what
mv3=-mvx-mvy--v3=v(-x) + v(-y)
but again I am having difficulty plugging and chugging the numbers.
 
  • #9
209
1
I hope u are right about the first problem
Why do you think it would be wrong? I calculated it myself and got the same answer.


The second problem although I can set it up I can't solve I am having difficulty plugging the numbers in I know m1 is mass of the ball m2 is mass of car v1 is speed of ball after collision and v2 is speed of car. But I still can't solve. algebra isn't my strongest subject
Could you write it out?

third problem
does this make sense? mvx + mvy +mv3=what
mv3=-mvx-mvy--v3=v(-x) + v(-y)
but again I am having difficulty plugging and chugging the numbers.
I'm not sure what you're doing.

This is what you got along the x-axis:

Sum of [ px, i ] = Sum of [ px, f ] = 0
px,1= mv = .245 * 2.50
px,2= mv = ?
px,1 + px,2 = 0

So you know that px,2 = -.245 * 2.50. The minus sign tells you that the x component of the velocity is opposite to the object flying in the positive x-direction.

Do the same for y, draw it and then you should be able to figure out from trig how to find the angle.
 

Related Threads for: Need help showing work for these problems

Replies
1
Views
3K
Replies
5
Views
3K
Replies
5
Views
967
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
625
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
Top