# A level maths forces and motion questions

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#1
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#2
The answer is 2.22, but I got 1.93s
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3 years ago
#3
unfortunately the image link did not work
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#4
(Original post by the bear)
unfortunately the image link did not work

Does it work now?
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#5
Bump
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3 years ago
#6
(Original post by Janej77)
There's a lot to do in this question. Please post your working and show us everything you've tried.
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3 years ago
#7
(Original post by Janej77)
Does it work now?
yes...

the net force in the first case is T - weight of 560 kg

and since the acceleration is 0.75 ms-2 we can find the tension T.

we are told the tension is the same in the second case:

so the net force can be found: ( T - weight of 600 kg ) = 600 x accln.

using this new acceleration and a SUVAT equation you can find the velocity after 30 seconds.

after that the only force is gravity.
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#8
(Original post by Notnek)
There's a lot to do in this question. Please post your working and show us everything you've tried.

Can you tell me where I went wrong,please?
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3 years ago
#9
(Original post by Janej77)
Are any of these answers correct?
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#10
(Original post by the bear)
yes...

the net force in the first case is T - weight of 560 kg

and since the acceleration is 0.75 ms-2 we can find the tension T.

we are told the tension is the same in the second case:

so the net force can be found: ( T - weight of 600 kg ) = 600 x accln.

using this new acceleration and a SUVAT equation you can find the velocity after 30 seconds.

after that the only force is gravity.

Here is my working and I am still not getting the right answer .
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3 years ago
#11
(Original post by Janej77)

Here is my working and I am still not getting the right answer .
it's all good up until you put the gravity for freefall as positive 9.8... it should be negative.
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#12
(Original post by the bear)
it's all good up until you put the gravity for freefall as positive 9.8... it should be negative.
Isn’t downwards positive?
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3 years ago
#13
(Original post by Janej77)
Isn’t downwards positive?

it can be, in which case your value of u should be negative...
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#14
(Original post by the bear)
it can be, in which case your value of u should be negative...
So I can’t start the equation as a new journey in this case? I’ve been doing calculations in constant acceleration where I changed the positive direction for second journeys. So when do I know I can change the direction in maths?
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3 years ago
#15
(Original post by Janej77)
So I can’t start the equation as a new journey in this case? I’ve been doing calculations in constant acceleration where I changed the positive direction for second journeys. So when do I know I can change the direction in maths?
it is fine to restart with downwards as positive, but remember that the lift was moving upwards when the cable broke.

so you can either go with s = -21, u = 7/5, a = -9.8 OR

s = 21, u = -7/5, a = 9.8
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#16
(Original post by the bear)
it is fine to restart with downwards as positive, but remember that the lift was moving upwards when the cable broke.

so you can either go with s = -21, u = 7/5, a = -9.8 OR

s = 21, u = -7/5, a = 9.8
Thank you so much. I think I understand it now.
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3 years ago
#17
May I know from which board this is?
I solved it like this:
a = 9.8 m/s^2, u = 0 m/s, x = 21 m, t = ?
x = ut + 1/2at^2
21 = 0 + (1/2)(9.8)(t^2)
t = 2.1 s
Also, your equation is wrong since when taking v, x = vt - 1/2 at^2
And should you not have used a = 9.8 m/s^2 and not a = 7/150 m/s^2 while calculating v?
Correct me if I'm wrong, I just started m1 a few days ago
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#18
(Original post by Lury)
May I know from which board this is?
I solved it like this:
a = 9.8 m/s^2, u = 0 m/s, x = 21 m, t = ?
x = ut + 1/2at^2
21 = 0 + (1/2)(9.8)(t^2)
t = 2.1 s
Also, your equation is wrong since when taking v, x = vt - 1/2 at^2
And should you not have used a = 9.8 m/s^2 and not a = 7/150 m/s^2 while calculating v?
Correct me if I'm wrong, I just started m1 a few days ago
Edexcel
I haven’t used the wrong equation. I used s=ut + 1/2at^2
I didnt use 9.8 cos there’s tension so gravity is not the only force acting on it. Force=ma. The acceleration is upwards and I’ve worked out the velocity using the accerlation for the 30s one. This is because the velocity by the end of 30 s would be the starting velocity for when the lift comes down.
Btw the answer says 2.22 seconds to 2 dp
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