- #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.
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.
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