Edexcel Physics AS Physics on the Go Unit 1 RETAKE

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I got around 25N too, not seeing how the majority of people got 13/14N

Yeah because there was a scale factor of 2:1 and i measured the resultant as 12.5cm so the magnitude must have been 25N. I cant remember the angle but i just used the sine rule to work it out (sinA/a=sinB/b) because i was stumped at that point lol. Not sure how everyone is getting 13N newtons but me and you are right :wink:

Reply 101

idsjfsiojsdioj

Original post by jay_rob

This is not correct. If you calculate the horizontal components of the forces and add them and then take the vertical components of the forces and add them and then use Pythagoras you get 13.4N as the solution. Also If the diagram is drawn correctly you get this answer.

Reply 102

jay_rob

Original post by idsjfsiojsdioj

Wait, so how did you manage to draw a right angled triangle? Because my diagram did not have a pythagorean triangle. The question wanted you to draw a scaled vector diagram and if drawn correctly, it would never have a right angle in. The only reason it was scaled was so that you could measure the resultant with a ruler. No calculation regarding pythagorus should have been involved?!

Reply 103

idsjfsiojsdioj

Original post by jay_rob

I used Pythagoras to confirm my answer:

Reply 104

Ezio.

Original post by idsjfsiojsdioj

I used Pythagoras to confirm my answer:

You cannot use pythagroras on a non rightangled triangle.

Reply 105

idsjfsiojsdioj

Original post by Ezio.

You cannot use pythagroras on a non rightangled triangle.

If you derived the forces into their vertical and horizontal components you could.

Reply 106

jay_rob

Original post by idsjfsiojsdioj

I used Pythagoras to confirm my answer:

There is not way to calculate horizontal or vertical components from a triangle that does not have a right angle.

Reply 107

idsjfsiojsdioj

Original post by jay_rob

There is not way to calculate horizontal or vertical components from a triangle that does not have a right angle.

If you derived the forces into their vertical and horizontal components you could.

Reply 108

jay_rob

Original post by idsjfsiojsdioj

If you derived the forces into their vertical and horizontal components you could.

I can see where you're coming from but its not right. By splitting (was it 35 degrees? i cannot remember) the angle in 90 + 35 you possibly could but you had to use your graph to determine the magnitude. No pythag was needed for that

Reply 109

naybour2000

Yo so guys basically for the q where you had to find r of the raindrop, I accidentally put 4/3pir^2 instead of r^3?: How many marks is that lost? Obvs I'm hoping i can still get 2/3 but its probably more like 1/3. So annoying for such a small error :frown:

Reply 110

idsjfsiojsdioj

Original post by naybour2000

You probably get 1/3 (probably 1 mark for the answer and one for working out the volume of the raindrop) but don't worry its only 2 marks.

Reply 111

naybour2000

Original post by idsjfsiojsdioj

You probably get 1/3 (probably 1 mark for the answer and one for working out the volume of the raindrop) but don't worry its only 2 marks.

Yeah thought so aha yeah its only two marks but I was sitting there like YES I KNOW THIS but ce la vie sometimes life gives you lemons and those lemons give you 1/3 marks in your physics resit. Apart from that thought the rest of the paper was okay. :smile:

Reply 112

crimson1234

Got 9.8 from the vector diagram, dont know how people got 13.
Also for the errors in the viscosity exp, do you guys think human reaction time and parallax will be appropriate

Reply 113

Z1228

What did people write for the properties shown by the stress??

Reply 114

Z1228

Thought it was an easy paper that I will probably do **** on hahahaa

Reply 115

AdamThePhysicsMo

Does anyone have a list of the questions as I've already forgotten most of them lol

Reply 116

A.McCall

I got all the ones stated here write don't remember what I put for the projectiles question though. The paper was nicer than last year's that is for sure. Hopefully I have scored highly to couple with some good unit 4 and 5 hopefully :smile:

Reply 117

jtebbbs

For everyone saying work = f * s, this only works when the force is constant, so if the line was horizontal on a force-extension graph. The graph in the exam forms a triangle with the x-axis, so if you're multiplying force and extension at 6cm and getting around 130J then that's just a coincidence and not correct. Work done is technically the integral of force with respect to displacement, i.e. area under a graph, NOT just "force * distance" so you need to use AREA to get the marks

Reply 118

HoldtheDoor

What did you guys put for the free body diagram of the bridge at point A? What did you write for the reason in can handle larger masses?

Reply 119

Konanabanana

Original post by HoldtheDoor

What did you guys put for the free body diagram of the bridge at point A? What did you write for the reason in can handle larger masses?

i did mg downwards and x2 Tensions at angles as shown in the diagram then for the explanation i just said that the weight is split into 2 components and this component is lower than the vertical weight and so the force that the beams have to 2 carry is less so less chance they break....