HSC Physics Marathon 2013-2015 Archive (1 Viewer)

InteGrand · Dec 13, 2014

re: HSC Physics Marathon Archive

Fizzy_Cyst said:
Question:

3 moons, X, Y and Z are in orbit around the same planet. The moons have identical orbital speeds, but masses of M, 9M and 25M respectively. Determine the ratio of their orbital radii and justify your answer.

The force of gravity between the planet and a moon of mass M is $F_G=\frac{GmM_P}{r^2}$ , where the planet's mass is M_P. This provides the centripetal force for the moon, which is equal to mv²/r, where v is equal for all three planets. So $\frac{mv^2}{r}=\frac{GmM_P}{r^2}$ , from which we see that the m's cancel out and we get that $r=\frac{GM_P}{v^2}$ (constant), so orbital radii are in ratio 1:1:1 (same).

InteGrand · Dec 13, 2014

re: HSC Physics Marathon Archive

Bert dropped a basketball from a window and recorded the time it took to reach the ground. He then climbed the stairs and dropped the basketball from a window 4.0 m higher up and recorded a time 3.0 s longer than the first time.

What was the first height from which Bert dropped the ball?

Fizzy_Cyst · Dec 13, 2014

re: HSC Physics Marathon Archive

InteGrand said:
The force of gravity between the planet and a moon of mass M is $F_G=\frac{GmM_P}{r^2}$ , where the planet's mass is M_P. This provides the centripetal force for the moon, which is equal to mv²/r, where v is equal for all three planets. So $\frac{mv^2}{r}=\frac{GmM_P}{r^2}$ , from which we see that the m's cancel out and we get that $r=\frac{GM_P}{v^2}$ (constant), so orbital radii are in ratio 1:1:1 (same).

Perfecto!

Nice question too

jyu · Dec 14, 2014

re: HSC Physics Marathon Archive

A 5 kg mass slides at a constant speed of 2.0 m/s down a plane inclined at 30 deg to the horizontal. Determine the reaction force of the plane on the mass.

Fizzy_Cyst · Dec 14, 2014

re: HSC Physics Marathon Archive

Please answer the current question before posting others, it is too confusing for me.

Current Question:

InteGrand said:
Bert dropped a basketball from a window and recorded the time it took to reach the ground. He then climbed the stairs and dropped the basketball from a window 4.0 m higher up and recorded a time 3.0 s longer than the first time.

What was the first height from which Bert dropped the ball?

If not answered in next ~48 hours, I will answer and respond with another question

PhysicsMaths · Jan 1, 2015

re: HSC Physics Marathon Archive

Let's restart this marathon:
An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.

Calculate how much more work needs to be done to move the object to a stationary point 80 000 km from the centre of the planet

InteGrand · Jan 1, 2015

re: HSC Physics Marathon Archive

PhysicsMaths said:
Let's restart this marathon:
An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.

Calculate how much more work needs to be done to move the object to a stationary point 80 000 km from the centre of the planet

Gravitational potential energy is inversely proportional to distance from the centre of the planet: $U(x)=-\frac{k}{x}$ , where U(x) is the gravitational potential energy at a distance x from the centre of the planet, and k is a positive constant (equal to GmM, where M is the planet's mass, m is the object's mass).

Work done to move from distance x₁ to x₂, where x₂ > x₁, is given by $W=U(x_2)-U(x_1)$

Since it is given that $1.0\times 10^6$ J of work is done to move from 10⁷ m to $2\times10^7$ m,

$10^6 \text{ J}=-\frac{k}{2\times 10^7 \text{ m}}-\left ( -\frac{k}{10^7\text{ m}} \right )$

Solving for k gives $k=2\times 10^{13}\text{ Jm}$

Now, the required extra work to move to 80000 km, W, is given by

$W=\left( -\frac{2\times 10^{13}}{8\times 10^7}-\left ( -\frac{2\times10^{13}}{2\times 10^7} \right )\right)\text{ J}$ , i.e. W = 7.5 × 10⁵ J.

InteGrand · Jan 1, 2015

re: HSC Physics Marathon Archive

Next question:

InteGrand said:
Bert dropped a basketball from a window and recorded the time it took to reach the ground. He then climbed the stairs and dropped the basketball from a window 4.0 m higher up and recorded a time 3.0 s longer than the first time.

What was the first height from which Bert dropped the ball?

PhysicsMaths · Jan 1, 2015

re: HSC Physics Marathon Archive

InteGrand said:
Next question:

Any hints please?

InteGrand · Jan 1, 2015

re: HSC Physics Marathon Archive

PhysicsMaths said:
Any hints please?

There's two unknowns in the question: the initial height that the ball was dropped from, and the initial time it took to fall.

So you'll need to form two equations to find these two unknowns, then solve to find the initial height.

One equation is formed by using the fact that $t_2 = t_1 + 3 \text{ s}$ , where $t_1$ and $t_2$ are the times of the first and second drops respectively.

So form equations for $t_1$ and $t_2$ in terms of the initial height h, and use that $t_2 = t_1 + 3 \text{ s}$ to solve for h.

Kaido · Jan 2, 2015

re: HSC Physics Marathon Archive

9.1m & 13.1m?
(from observation, on phone Cbb working out)

InteGrand · Jan 2, 2015

re: HSC Physics Marathon Archive

Kaido said:
9.1m & 13.1m?
(from observation, on phone Cbb working out)

That's right.

How did you do it just by observation?

InteGrand · Jan 2, 2015

re: HSC Physics Marathon Archive

This question is still unanswered:

jyu said:
A 5 kg mass slides at a constant speed of 2.0 m/s down a plane inclined at 30 deg to the horizontal. Determine the reaction force of the plane on the mass.

InteGrand · Jan 2, 2015

re: HSC Physics Marathon Archive

It should be assumed that the plane is frictionless.

Edit: Not frictionless, didn't read the problem statement thoroughly before.

Fizzy_Cyst · Jan 2, 2015

re: HSC Physics Marathon Archive

InteGrand said:
That's right.

How did you do it just by observation?

Having a look at the question, where it says it takes 3s to go a further 4m, that impossible assuming g =-9.8ms^-2.

I get no real solutions for g =-9.8! Show me working!

I am able to get the 9.1m, but that involves an algebraic mistake.

Are you sure it is not 0.3s or something?

InteGrand · Jan 2, 2015

re: HSC Physics Marathon Archive

Fizzy_Cyst said:
Having a look at the question, where it says it takes 3s to go a further 4m, that impossible assuming g =-9.8ms^-2.

I get no real solutions for g =-9.8! Show me working!

I am able to get the 9.1m, but that involves an algebraic mistake.

Are you sure it is not 0.3s or something?

You're right, sorry, I didn't realise I needed the condition that $2\delta _h-g {\delta _t}^2 >0$ to have real solutions when setting the problem, where $\delta _h$ is the height difference and $\delta _t$ is the time difference.

I originally did this problem in the general case where the heights and times were just pronumerals, so I forgot to make sure when giving numerical values to them that they needed to satisfy that inequality, which comes about when solving the general case.

Kaido · Jan 2, 2015

re: HSC Physics Marathon Archive

Well, by observation I meant sitting down at a restaurant waiting for meal to be served and doing a bit of mental maths

Kaido · Jan 2, 2015

re: HSC Physics Marathon Archive

InteGrand said:
This question is still unanswered:

Not sure what you're looking for.
Frictional force, gravitational force, Perp. force, Parallel force ???

InteGrand · Jan 2, 2015

re: HSC Physics Marathon Archive

Kaido said:
Not sure what you're looking for.
Frictional force, gravitational force, Perp. force, Parallel force ???

I didn't make that question, but I think he was looking for the normal force of the plane on the object.

Kaido · Jan 2, 2015

re: HSC Physics Marathon Archive

Logically makes no sense. You can't have a frictionless plane and expect to achieve constant velocity.

Oh and what is the 'normal force'?

(Hella bad at english as u can tell)

HSC Physics Marathon 2013-2015 Archive (1 Viewer)

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