# Linearising equations question

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#1
I'm slightly stuck on these two questions, anyone got any hints?
Last edited by Slx.24; 4 weeks ago
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4 weeks ago
#2
wouldn't the first one have to be rearranged so that the s is the denominator and Ek the numerator and then substitute -0.98?
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4 weeks ago
#3
(Original post by Slx.24)
I'm slightly stuck on these two questions, anyone got any hints?
1) I recommend remembering the equation of a straight line which is y = mx+ c, where m is the gradient and c is the y intercept. It may also help if you see it as Ek = -Wsin(θ) s + E0.

In this case substitute the value for W in: Ek = -2.39 sin(θ) s + E.

Then set -2.39 sin(θ) = -0.98 as we can see that -Wsin(θ) is the gradient and we are given the value of the gradient. Remember it is -2.39 not 2.39 as it is -W.

This equation can be very simply solved (I hope!): -2.39 sin(θ) = -0.98 (solve for θ)

2) Again remember the equation of a straight line and we can see that the gradient which in this case is (e/h) = 2.41x10^14 and we know the value of e. It should be quite easy to find the value of h.

And the same can be done for the intercept to find the other value.
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#4
(Original post by Zain-)
1) I recommend remembering the equation of a straight line which is y = mx+ c, where m is the gradient and c is the y intercept. It may also help if you see it as Ek = -Wsin(θ) s + E0.

In this case substitute the value for W in: Ek = -2.39 sin(θ) s + E.

Then set -2.39 sin(θ) = -0.98 as we can see that -Wsin(θ) is the gradient and we are given the value of the gradient. Remember it is -2.39 not 2.39 as it is -W.

This equation can be very simply solved (I hope!): -2.39 sin(θ) = -0.98 (solve for θ)

2) Again remember the equation of a straight line and we can see that the gradient which in this case is (e/h) = 2.41x10^14 and we know the value of e. It should be quite easy to find the value of h.

And the same can be done for the intercept to find the other value.
(Original post by Zain-)

Thanks! I did like 90% of that but I got stuck because I thought there were two unknowns, one being theta (sorry on a phone rn) I knew what the gradient was and did equate the whole thing to -0.98 but just couldn't figure the whole thing out because I thought 0.98= E0 -2.39 sin(theta) BUT we don't know what E0 is as well as sin(theta) so I was completely stumped.
Are you saying that E0 is somehow not part of the gradient?

(Original post by englishhopeful98)
wouldn't the first one have to be rearranged so that the s is the denominator and Ek the numerator and then substitute -0.98?
Idts, the x and y values were in the right order so you dont need to rearrange further (for the first one anyways) s = x and Ek = y which means the things in the middle is the gradient 0
4 weeks ago
#5
(Original post by Slx.24)
(Original post by Zain-)

Thanks! I did like 90% of that but I got stuck because I thought there were two unknowns, one being theta (sorry on a phone rn) I knew what the gradient was and did equate the whole thing to -0.98 but just couldn't figure the whole thing out because I thought 0.98= E0 -2.39 sin(theta) BUT we don't know what E0 is as well as sin(theta) so I was completely stumped.
Are you saying that E0 is somehow not part of the gradient?

Idts, the x and y values were in the right order so you dont need to rearrange further (for the first one anyways) s = x and Ek = y which means the things in the middle is the gradient Yes, E0 is not part of the gradient. E0 is the y intercept or in this case the Ek intercept. I assume that E0 specifically is the kinetic energy when displacement is 0, you do not really need to understand that but just be able to recognise which part of the equation is the gradient and which part of the equation is the y intercept. Like I did above, it might help to rearrange the equation:

Ek = -Wsin(θ) s + E0.

This is in the form y = mx + c where m is the gradient, c is the y intercept, y and x are both axes.

Also to note, the gradient is always multiplying the x axis value (in this case s), as seen from the general equation above. We can see very clearly that E0 is not multiplying s, so therefore it is not part of the gradient. However, if the question had brackets like this: Ek = (E0 - Wsinθ) s, then here is an example of where E0 is part of the gradient but the question does not ask this.

Also, it is important to understand that the gradient is not "the things in the middle". I have explained what the gradient is directly above if you need to look back. The equation can be given in any order so it is important to understand what each part of the equation is as you will be required to rearrange it into the correct order - generalised by y = mx + c.
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