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Mathematics: Post your doubts here!

I

iYuuki

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Hi, can anyone help me with this question? Any help would be appreciated :)
 

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Ashjay

Ashjay

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iYuuki said:
Hi, can anyone help me with this question? Any help would be appreciated :)
Click to expand...

i) In the first part you have the show the points of intersection of the two functions, cosec x = 1/sin x. So the graph of sin x can be used to draw its graph.
iii b) The symmetry of the graph is used to calculate the value of B. The distance of B from the line of symmetry is the same as its distance from A.
 

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Haya Ahmed

Haya Ahmed

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udiTWG1.png

Mechanics Question
part (iii) please someone help me in sketching this ! .. I'm so confused because of the marking scheme answer ..
this is May/June 2002 P4 9709
 
Ashjay

Ashjay

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Haya Ahmed said:
udiTWG1.png

Mechanics Question
part (iii) please someone help me in sketching this ! .. I'm so confused because of the marking scheme answer ..
this is May/June 2002 P4 9709
Click to expand...
motion.PNG From 0-2 seconds, both particles are travelling with the same acceleration in opposite directions. At 2 seconds the blue particle hits the ground and comes to rest, while the red particle's velocity begins to decrease due to its weight because the string has become slack. At 2.5 seconds its velocity reaches 0 and it begins to fall downwards. At 3 seconds the string gets taut again as calculated in (ii) so the blue particle feels an upward tension and begins to move again. From this point onwards we are not required to show the subsequent motion in the graph.
 
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Haya Ahmed

Haya Ahmed

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Ashjay said:
View attachment 50580 From 0-2 seconds, both particles are travelling with the same acceleration in opposite directions. At 2 seconds the blue particles hits the ground and comes to rest, while the red particle's velocity begins to decrease due to its weight because the string has become slack. At 2.5 seconds its velocity reaches 0 and it begins to fall downwards. At 3 seconds the string gets taut again as calculated in (ii) so the blue particle feels an upward tension and begins to move again. From this point onwards we are not required to show the subsequent motion in the graph.
Click to expand...
Thanks so much you're awesome :p :)
 
slisjunknown

slisjunknown

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Can anyone explain to me where I should use permutaions or combinations?:((n)
I would also appreciate if there is a shortcut to answering those type of questions...:cry:
 
Ashjay

Ashjay

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Haya Ahmed said:
I still didn't get it can you please explain in a more detailed way ? :( .. or another way
Click to expand...
Okay hmmm lets see... As we know that the system is in equilibrium, this means that the resultant of all of the forces = zero. We also know that the force exerted on the pulley by the string is 3*3^(1/2). This means that the force exerted on the string by the pulley is also 3*3^(1/2) ( because the system is in equilibrium! the two forces are equal and opposite). Now this force is the resultant of the two tensions in the string which are coloured blue in my diagram. Adding the two Tensions by the vector triangle and using the cosine rule enables us to find the Tension. If you still dont understand i will be happy to make another diagram! Hope you get it :)
 
Haya Ahmed

Haya Ahmed

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Ashjay said:
Okay hmmm lets see... As we know that the system is in equilibrium, this means that the resultant of all of the forces = zero. We also know that the force exerted on the pulley by the string is 3*3^(1/2). This means that the force exerted on the string by the pulley is also 3*3^(1/2) ( because the system is in equilibrium! the two forces are equal and opposite). Now this force is the resultant of the two tensions in the string which are coloured blue in my diagram. Adding the two Tensions by the vector triangle and using the cosine rule enables us to find the Tension. If you still dont understand i will be happy to make another diagram! Hope you get it :)
Click to expand...
I'm sorry for wasting your time, well I got everything but when it comes to the diagram I didn't get it :(
 
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