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HSC Tips - Mechanics (1 Viewer)

ezzy85

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with that q, it would say the force of something with UNIT mass if kv^2. therefore yes, the mass is there. however if it said the force is kv^2 and the mass was 20kg, youll have some problems because your m's wont cancel. F = ma, therefore you have to have tge mass.
 

Fosweb

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So what do you do when the question says "negligable mass"? Assume that mkv^2 = kv^2 ?
 

freaking_out

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Originally posted by Fosweb
So what do you do when the question says "negligable mass"? Assume that mkv^2 = kv^2 ?
no, since m is constant, so it doesn't matter if u put kv or mkv, in the end there is a constant (i.e the coefficient of v)..thats what i was taught.:confused:
 

ezzy85

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escape velocity is v when x approaches infinity and terminal is when t appraoches infinity. alternatively, you can make accelareration = 0
 
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wogboy said:
I think an important point for a conical pendulum is that a centripetal force is NOT a "causal" force. It is not a force that just exists due to nature, like gravity & tension do. Rather the centripetal force is caused by gravitation (weight or mg) and the tension in the string, both of which are "causal" forces.
So in other words Centripetal force is a resultant caused by gravitational force and tension of the string? Cool, that clears things up, I was just wondering about that. Thanks. :)
 

Stefano

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ezzy85 said:
escape velocity is v when x approaches infinity and terminal is when t appraoches infinity. alternatively, you can make accelareration = 0
"accelareration" ? I'm afraid I'm not familiar with that topic. Care to enlighten us ?
 

eccentricity

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In relation to your question,
F(Net)=mg-mv
But F(Net)=m*dv/dt
Therefore, dv/dt=g-v

It is important to realise that since mg is positive, you are treating the release point as the origin in your acceleration expression.

Therefore, from the chain rule,
dt/dv=1/(g-v)
so t= integral of [1/{g-v}]- limits {0 to H}

But you could also consider the case that the release point as having (H) as displacement and the ground as 0.
In this case,
F(net)=-mg-mv (as resistance is always negative)- but now gravity is opposing the motion.
 

Maths Champion

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Tip page 6? 7? Oh well, one of them anyway ...

Tips for Mechanics:
- DRAW a diagram of the situation.
- Try not to let the physics you have learn't for the HSC influence you, this is maths, they do things differently.


I can't think of any more tips (pathetic really) so please someone else post some more ...
The beauty of doing 4 unit maths is doing physics as well, as this may help you understand the PHYSICAL concepts in mechanics. For projectile motion, definitely know how to derive the equations for horizontal and vertical displacement. Know how to derive equation for max height where v=0, know how to derive the range of the projectile where y=0, or find the time it takes to reach max height and double it, HOWEVER this only works when projected from ground level. Know acceleration is the rate change of velocity: a=dv/dt. For circular motion physics also really helps. Angular velocity, rate change of angular displacement dθ/dt. Know the centripetal force is what keeps the object in uniform circular motion directed towards the centre. When object is on table, there is the normal force pushing it up, with gravity acting down and the tension in the string. Finally know about retarding forces, which means the decelerating forces of an aircraft for instance, with braking forces and reverse thrusters. Also with resisted motion, when the mass is projected upwards take up as positive and down as negative, know there is force of gravity acting downwards constantly and the air resistance opposing the direction of motion. When coming back down, take down as positive. That pretty much sums it up.
 
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