Back
to top
Newtons Laws
Watch this thread
Announcements
Page 1 of 1
Go to first unread
Skip to page:
ShaneK
Badges:
0
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
I'm learning physics on my own, without a tutor (I'm not at school). I've been managing fine, but I'm stuck on this question and I want to understand it.
It's part of "Homework Questions for New Higher Physics" but there are no answers at the back of the book and I can't find them anywhere online, so I can't even look ahead and try to figure it out by working backwards.
It's from the chapter "Newton 2".
Q8
The vehicle on the air track is pulled by a falling mass of 0.2kg. The mass of the air track vehicle is 1.1kg.
a) show, by calculation, that the air track vehicle and the falling mass both accelerate at 1.51ms-2.
I know that they accelerate together, since they're attached, so they'll have the same acceleration. They're on an air track, so I don't need to think about friction. I'm so lost, I have no idea what to do here, which makes me feel like I'm probably over-looking something simple, but it's been a couple of days now that I've been going back to it and I just can't work out what to do. *sigh*
I hope someone can help me out!
It's part of "Homework Questions for New Higher Physics" but there are no answers at the back of the book and I can't find them anywhere online, so I can't even look ahead and try to figure it out by working backwards.
It's from the chapter "Newton 2".
Q8
The vehicle on the air track is pulled by a falling mass of 0.2kg. The mass of the air track vehicle is 1.1kg.
a) show, by calculation, that the air track vehicle and the falling mass both accelerate at 1.51ms-2.
I know that they accelerate together, since they're attached, so they'll have the same acceleration. They're on an air track, so I don't need to think about friction. I'm so lost, I have no idea what to do here, which makes me feel like I'm probably over-looking something simple, but it's been a couple of days now that I've been going back to it and I just can't work out what to do. *sigh*
I hope someone can help me out!
0
reply
Muttley79
Badges:
20
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report
#2
So what equations of motion have you got?
1) for the falling particle
2) for the vehicle?
1) for the falling particle
2) for the vehicle?
0
reply
esi183
Badges:
0
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report
#3
You need to first work out the force on the falling mass, as this is the only one that has an external force acting on it causing acceleration (I.e gravity). To do this you use W = mg (which is the same as f = ma but uses the constant acceleration under gravity). This is the force that the mass exerts.
Then you need to rearrange F = ma to give the acceleration with the combined mass of the car and falling mass with the force calculated above. (a = f/m). This is because the pulling string couples the mass of both objects, meaning that the force acts on them as if they are one.
Then you need to rearrange F = ma to give the acceleration with the combined mass of the car and falling mass with the force calculated above. (a = f/m). This is because the pulling string couples the mass of both objects, meaning that the force acts on them as if they are one.
0
reply
Muttley79
Badges:
20
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report
#4
Sorry - we aren't encouraged to give solutions. Can't see where you've used the 'pulled' fact ....
0
reply
esi183
Badges:
0
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report
#5
The "pulled" fact is that their mass is coupled by the string, but the weight of only the falling mass is causing the acceleration. I was unaware that we are discouraged to give solutions, but my solution is correct. Imagine a car on a table, with a string attached, and a mass pulling it hanging down from the edge of the table on the string. If you can't see then ask, don't declare.
0
reply
esi183
Badges:
0
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report
#6
(Original post by Muttley79)
Sorry - we aren't encouraged to give solutions. Can't see where you've used the 'pulled' fact ....
Sorry - we aren't encouraged to give solutions. Can't see where you've used the 'pulled' fact ....
0
reply
Muttley79
Badges:
20
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
esi183
Badges:
0
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8
Report
#8
Tension is above the scope of this question; if it was not, the string would be described "light and inextensible" in the question. If the question asked for tension, you have to use simultaneous equations to work out tension.
Of course we are assuming the string is light and inextensible. To be pedantic, the solution formally requires us to state that the force acting on the car is T, and T = m(car) x a. Then, state the force acting on the falling mass as an equation: (m(mass) x g) - T = m(mass) x a. Then you can substitute T out by using the previous equation for the car, arriving in the same place as my solution above.
Note that even when using tension rather than intuition, if you require the acceleration what you do is consider the mass of both and the force experienced by the falling object by substituting tension out.
Of course we are assuming the string is light and inextensible. To be pedantic, the solution formally requires us to state that the force acting on the car is T, and T = m(car) x a. Then, state the force acting on the falling mass as an equation: (m(mass) x g) - T = m(mass) x a. Then you can substitute T out by using the previous equation for the car, arriving in the same place as my solution above.
Note that even when using tension rather than intuition, if you require the acceleration what you do is consider the mass of both and the force experienced by the falling object by substituting tension out.
0
reply
Stonebridge
Badges:
13
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9
Report
#9
(Original post by esi183)
I have also now changed my post so that it doesn't give the answer.
I have also now changed my post so that it doesn't give the answer.
http://www.thestudentroom.co.uk/wiki...ring_questions
0
reply
ShaneK
Badges:
0
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#10
Thank you! I'd done the first part and worked out the pulling force of the falling mass, but then got stuck.
I actually don't know how I hadn't been able to see how to do the second part. I think once I'd got confused I just couldn't see past that to the right answer, even though it makes complete sense now.
I actually don't know how I hadn't been able to see how to do the second part. I think once I'd got confused I just couldn't see past that to the right answer, even though it makes complete sense now.
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
