Solving Momentum Questions: 4 Problems Explained

  • Thread starter JayDub
  • Start date
  • Tags
    Momentum
In summary: For the final question, use the same approach as question 37, splitting the motion into x and y directions and conserving momentum. However, you will also need to use the coefficient of restitution to find the final velocities of both pucks.
  • #1
JayDub
30
0
Hey there I am having some trouble with some questions here in fisix so here I am asking for some help.

So the first question:
http://www.elarune.net/admins/josh/question7.jpg [Broken]

Now I am not sure if this is how you would go about solving it.

Fnet = ma.

Since the only force would be the breaking force and since it goes from 160 to 0 the F would be- 160 N. Next I find the acceleration with Kinematics, so:

Vf^2 = Vo^2 + 2ad
0^2 = 10^2 + 2(a)(25)
a = -2 m/s^2

so going back to the question I would go

Fnet = ma
-160 N = m(-2 m/s^2)
m = 80kg?

That I am not sure on.


Ok, the next question:
http://www.elarune.net/admins/josh/question37.jpg [Broken]

Now the only way I can think of going about this is first finding the V2' of the small ball by using it's angle and find the Vx. Then I could use this formula

m1v1 + m2v2 = m1v1' + m2v2'

This would give me v1' which is the Vx of the big ball but I do not have any angle or another side to find the V or the direction.


Question 3:
http://www.elarune.net/admins/josh/question41.jpg [Broken]

Again I am not sure how I could find this one out. I could find the area under the curve to the 200m which is equal to work. Then since the work is equal to the kenetic energy and I have the mass I could find the velocity. Again I am not sure if that correct?


Final Question
http://www.elarune.net/admins/josh/question42.jpg [Broken]

So this is the same as two questions ago. I could find the Vx of the bigger puck, use the same formula to find the Vx of the smaller one, but again I do not have another side or an angle to find the angle and V.

Thank you.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
to me, the first question looks like an energy problem...
 
  • #3
Answer to 'question7.jpg'

Using law of conservation of enery, treating the cyclist, the cycle and the ground as an isolated system, and assuming that the ground is flat and the brake transfers all of the kinetic energy into thermal energy, we can solve the problem presented in 'question7.jpg' easily as follows.

[tex]W_{system}=\Delta E_{system}=\Delta E_{mechanic}+\Delta E_{thermal}[/tex].
There is no external force applied to the system. Hence, [tex]W_{system}=\Delta E_{system}=0[/tex].
[tex]0=\Delta E_{mechanic}+\Delta E_{thermal}[/tex].
We know that [tex]\Delta E_{mechanic}=\Delta E_{kinetic}+\Delta E_{potential~gravity}[/tex]
[tex]0=\Delta E_{kinetic}+\Delta E_{potential~gravity}+\Delta E_{thermal}[/tex]
The ground is flat. Therefore, [tex]\Delta E_{potential~gravity}=0[/tex].
[tex]0=\Delta E_{kinetic}+0+\Delta E_{thermal}[/tex]
The initial speed of the cycle is [tex]10\frac{m}{s}[/tex] and the graph provides the fact that [tex]W=\Delta E_{thermal}=\int F~dx[/tex]. Thus,
[tex]0=\frac{1}{2}m(v^{2}_{final}-v^{2}_{initial})+\int F~dx[/tex]
Solving the above equation for m yields a final equation as follows.
[tex]m=-2\frac{\int F~dx}{(v^{2}_{final}-v^{2}_{initial})}[/tex]
Substituting the numbers will yield a result as follows.
[tex]m=-2\frac{160(15)+\frac{1}{2}(160)(25-15)}{(0^{2}-10^{2})}[/tex]
[tex]m=\frac{6400}{100}[/tex]
[tex]m=64~kg[/tex]
 
Last edited:
  • #4
My Comment Regarding Your Answer to 'question7.jpg'

Actually, you can correctly solve your problem that way. Unfortunately, you made a mistake in the quoted text below.
Since the only force would be the breaking force and since it goes from 160 to 0 the F would be- 160 N.
Well, you have to obtain the [tex]F_{average}[/tex] of the braking force depicted in the graph.
It is easy to obtain it as follows.
[tex]F_{average}(total~distance~travelled)=total~area~under~the~graph[/tex]
[tex]F_{average}=\frac{160(15)+\frac{1}{2}(160)(25-15)}{(25)}[/tex]
[tex]F_{average}=128[/tex]
Therefore, the quoted text below
Fnet = ma
-160 N = m(-2 m/s^2)
m = 80kg?
should be as follows.
[tex]F_{net}=m(a)[/tex]
[tex]-128=m(-2)[/tex]
[tex]m=64~kg[/tex]
 
Last edited:
  • #5
Eus, Please do not provide full solutions. This is not the purpose of PF, please consult the posting guidelines.

For question 37, split the motion into x and y directions. Remember that momentum must be conserved in both directions. In this case, intially there is no momentum in the y direction, therefore after the collision, the net momentum must be zero.

-Hoot
 
Last edited:
  • #6
For your third question first find the work done, with that find the kinetic energy and then the speed.
 

What is momentum?

Momentum is a measure of an object's motion, calculated by multiplying its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

How do you solve momentum problems?

To solve momentum problems, you can use the equation p = mv, where p is momentum, m is mass, and v is velocity. You can also use the law of conservation of momentum, which states that the total momentum of a system remains constant in the absence of external forces.

What are the units of momentum?

The SI unit of momentum is kilogram meters per second (kg·m/s). In other systems, momentum can be expressed in gram centimeters per second (g·cm/s) or pound feet per second (lb·ft/s).

What is an elastic collision?

An elastic collision is a type of collision where both kinetic energy and momentum are conserved. In other words, the total kinetic energy before and after the collision is the same, and the total momentum of the system is also the same.

How do you determine the direction of momentum?

The direction of momentum is determined by the direction of the velocity vector. If the velocity is positive, the direction of momentum is in the same direction. If the velocity is negative, the direction of momentum is in the opposite direction.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
867
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
923
  • Introductory Physics Homework Help
Replies
7
Views
774
  • Introductory Physics Homework Help
Replies
6
Views
886
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
976
  • Introductory Physics Homework Help
Replies
7
Views
822
  • Introductory Physics Homework Help
Replies
11
Views
627
Back
Top