# Help with mechanics as level exam question

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#1

At time t = 0, a small stone is thrown vertically upwards with speed 14.7ms−1 from a point A.
At time t = T seconds, the stone passes through A, moving downwards.
The stone is modelled as a particle moving freely under gravity throughout its motion.
Using the model,
(a) find the value of T,
(2)
(b) find the total distance travelled by the stone in the first 4 seconds of its motion.
(4)
(c) State one refinement that could be made to the model, apart from air resistance, that
would make the model more realistic.
(1)
Last edited by MedApplicantttt; 1 month ago
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#2
For part a I got 1.5s but the answer says 3s?
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1 month ago
#3
(Original post by MedApplicantttt)
For part a I got 1.5s but the answer says 3s?
Generally helps to see your working, but I agree with the answer. Im guessing you found the time to the peak height, whereas the question asks for the time to return to A.
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1 month ago
#4
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
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1 month ago
#5
(Original post by zoe2404)
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
Could argue conservation of energy. No change in GPE so no change in KE. Or using suvat
v^2 = u^2 + 2*g*0
as the displacement is zero so v^2=u^2 so v=-u. But the projectile motion is a quadratic which is symmetrical about the highest point, so the speeds will be the same.

The force on the body due to gravity (mg) does not change if youre going up or going down so its the same constant acceleration suvat scenario.
Last edited by mqb2766; 1 month ago
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1 month ago
#6
(Original post by zoe2404)
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.

For part b, collect the values again, and use the same equation.
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1 month ago
#7
(Original post by e^iπ + 1 = 0)
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.

For part b, collect the values again, and use the same equation.
If u is positive, a would have to be negative. Otherwise the displacement will always increase (upwards). But that is another common way to solve the problem.
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#8
Thanks so much.

I'm so confused about part b. Could someone please help. I started to work out the maximum height which I got to be 11m?
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1 month ago
#9
(Original post by MedApplicantttt)
Thanks so much.

I'm so confused about part b. Could someone please help. I started to work out the maximum height which I got to be 11m?
Agree with 11.025m above A. A sketch with questions such as these really helps, especially when its calculating total distance travelled as opposed to displacement

You could consider what happens in the next 2.5s from the peak point and then add the two distances together. Or calculate the (positive) displacement from A at 4s, then add the up and down again distances to the peak.
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#10
(Original post by e^iπ + 1 = 0)
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.

For part b, collect the values again, and use the same equation.
Hi, could you pls help with b
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1 month ago
#11
(Original post by MedApplicantttt)
Hi, could you pls help with b
What I did is collect all of the values you know, t = 4, u = 14.7 and a = -9.8.
Use a suvat equation with those terms to find the displacement below point A, so it will come out as a negative.
Then work out the distance from point A to the highest point; so v = 0. Find the displacement from that and double it (as it is the displacement for A to the highest point on one side) , and add on your previous displacement (not with the negative sign). Let me know if that matches the mark scheme.
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1 month ago
#12
I got an answer of -3, Did I do anything wrong?Btw do you have the mark scheme, I would love to check my answers!!Kind Regards
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1 month ago
#13
where did you find the mark scheme???
0
1 month ago
#14
(Original post by Spirereyli)
I got an answer of -3, Did I do anything wrong?Btw do you have the mark scheme, I would love to check my answers!!Kind Regards
Last edited by mqb2766; 1 month ago
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