DO ME: SHM and Projectile Motion questions (1 Viewer)

[cl3nta]

Hey all,

I'm trying to come to grips with this whole topic. It doesn't help when your maths teacher is away and trials are in two weeks... I have a few questions. Thanks for any help

1. Show that d/dx(1/2v^2)=dv/dt

2. A particle moves in a straight line and its displacement x metres from the origin after t seconds is given by: x=cos^2 3t, t>0
i)when is the particle first at x=3/4 ?
ii)in what direction is the particle moving when it is first at x=3/4 ?
iii)express the acceleration of the particle in terms of x.
iv)Hence, or otherwise, show that the particle is undergoing simple harmonic motion.
v)What is the period of the motion?

3. An angler casts a fishing line so that the sinker is projected with a speed of V m/s from a point 5 metres above a flat sea. The angle of projection is theta. Assume the equations of motion are x_double_dot = 0 and y_double_dot=0
There was a picture provided which can be viewed below (excuse the very bad paint job)

i)let (x,y) be the position of the sinker at time t seconds after the cast, and before the sinker hits the water. It is known at x=Vtcos(theta). Show that y=Vtsin(theta)-5t^2+5
ii)Suppose the sinker hits the sea 60m away. Find V if theta=tan^-1(3/4)
iii) For the cast described in ii), find the maximum height above sea level that the sinker achieved.

4. A stone is thrown so that it will hit a bird at the top of a poel. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at the speed of 10ms^-1. The stone reaches a height double that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity of the stone.

It's a "few" questions - thanks again for any help

 

[cl3nta]

No one can help?? To help I got q1 done - required chain rule, I can post up answer if anyone else needs it The others... HELP!!

 

[kony]

i had a look at 2, and i think it's fairly hard - more of a challenge question. you need the sort of manipulation skills that are more expected of ext2.

i'll have a go at it tomorrow, but i just did a bit of it and it was a bit tough. (particularly part 2. weird chain rule)

 

[ssglain]

I think what kony is saying is that it might help with Q2 to write x = cos²(3t) as x = 1/2 + (1/2)[cos(6t)] - this move is not exclusive to the MX2 course.

i) 3/4 = 1/2 + (1/2)[cos(6t)] -> cos(6t) = 1/2
So 6t = pi/3 -> t = pi/18

ii) x' = -3sin(6t)
When t = pi/18, dx/dt < 0 -> moving left

iii) x" = -18cos(6t)
Since cos(6t) = 2(x - 1/2)
x" = -36(x - 1/2)

iv) x" = -6²X
Acceleration satisfies the form of x" = -n²X where n = 6 and X = (x - 1/2) -> SHM

v) period = 6/2pi = 3pi

 

[ssglain]

Q3 seems flawed. The vertical component of acceleration cannot = 0 because the downward acceleration due to gravity is -g = -10.

i) y" = -10
Integration gives y' = -10t + C
Now, t = 0 -> y' = Vsin@ (the vertical component of the inital velocity V)-> C = Vsin@
.: y' = -10t + Vsin@
Integration gives y = -5t² + Vtsin@ + C
Now, t = 0 -> y = 5 -> C = 5
.: y = -5t² + Vtsin@ + 5 as required

ii) Interpret the information provided - it means that y = 0 when x = 50. Also, @ = arctan(3/4) is given. Using these parameters you should be able to form two equations using x = Vtcos@ and y = -5t² + Vtsin@ + 5, which you can then solve simultaneously for V.

iii) Max height occurs when y' = 0. Solve this for time at which this occurs and substitute into y = -5t² + Vtsin@ + 5.

Someone else can have a go at explaining Q4.

 

[cl3nta]

you're right about Q3 ssglain. That was a typo on my part. Ofcourse y"=-10 (0r-9.8)