Solving Momentum Questions: 4 Problems Explained

[JayDub]

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

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

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

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

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.

 

[andrewchang]

to me, the first question looks like an energy problem...

 

[Eus]

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.

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

 

[Eus]

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 F_{average} of the braking force depicted in the graph.
It is easy to obtain it as follows.
F_{average}(total~distance~travelled)=total~area~under~the~graph
F_{average}=\frac{160(15)+\frac{1}{2}(160)(25-15)}{(25)}
F_{average}=128
Therefore, the quoted text below

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

should be as follows.
F_{net}=m(a)
-128=m(-2)
m=64~kg

 

[Hootenanny]

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

 

[daniel_i_l]

For your third question first find the work done, with that find the kinetic energy and then the speed.