Motion Q (1 Viewer)

[azureus88]

A smooth piece of ice is projected up a smooth inclined surface. Its distrance up the surface at time t seconds is x = 6t - t^2. At what angle @ should the surface be inclined to the horizontal to produce these equations.

the answer is 11 degrees and 47 minutes.

how would you approach this question?

 

[azureus88]

bump... anyone q

 

[Trebla]

Is that the whole question?

What does it mean by being "projected up a smooth inclined surface"? Is it going up an inclined plane with x being the distance it covers on the plane? or is it thrown into the air (as implied by the word "projected")?

 

[azureus88]

yeh thats all that was specified although it did come with a diagram of a small ball on the hypotenuse of a right angled triangle and and distance from that ball to the point on the hypotenuse that meets the horizontal was called x. it also gave a hint about resolution of forces whatever that means.

 

[Pwnage101]

yeh thats all that was specified although it did come with a diagram of a small ball on the hypotenuse of a right angled triangle and and distance from that ball to the point on the hypotenuse that meets the horizontal was called x. it also gave a hint about resolution of forces whatever that means.

are you sure this is MX1?

Maybe it's a mechanics question from MX2....

 

[azureus88]

it was from yr 12 cambridge 3u

 

[Pwnage101]

it was from yr 12 cambridge 3u

can u scan the page?

 

[azureus88]

can u scan the page?

sorry, dont have a scanner lol. but the question's on p93 q16 and it links to the diagram on p90 q7. maybe someone else who has the book can scan it...

 

[kaz1]

It doesn't ask for the angle of inclination in the textbook.

 

[azureus88]

it does...got to p93 q16

 

[kaz1]

it does...got to p93 q16

I think that's either 4unit or beyond as it's the last extension question.

 

[jpmeijer]

I think that's either 4unit or beyond as it's the last extension question.

Definitely not 3unit, it's actually a very simple 4unit mechanics question.

We are given that:
x = 6t - t^2

Therefore:
v = 6 - 2t
a = -2 which implies that the force along the direction that the particle is travelling is -2 (force = mass*acceleration, but let's assume mass = 1, so force = acceleration)

We know that the force acting downwards is gravity = -g = -9.8

So "resolving forces" into "orthogonal (right-angled) components" (see my awesome paint drawing) shows that:
sin@ = 2/9.8
@ = sin^-1(2/9.8)
@ = 11 degrees 47 minutes