SHM and Projectile Motion questions (1 Viewer)

[acmilan]

Can someone help me out with a few questions.

1. A particle moves in a straight line so that its position x from a fixed point O at time t is given by x = 10 + 8 sin 2t + 6cos 2t. Prove that the motion is SH. Find the period and amplitude. (The amplitude is the part i cant do)

2. Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top of a wall 26.25 m high and 30 m away.

3. A particle projected from a point meets the horizontal plane through the point of prjection after travelling a horizontal distance a, and in the course of its trajectory attains a greatest height b above the point of projection. Find the horiztonal and vertical components of the velocity in terms of a and b. Show that when it has described a horizontal distance x, it has attained a height of [4bx(a-x)] / a^2

4. A stone is projected horizontally from the top of a building 30 metres high with a velocity of 15m/s. At the same instant another stone is projected from the foot of the building with a velocity of 30m/s at an angle of 60 degrees to the horizontal. The two stones move in the same vertical plane/ Prove that they collide and, if the coordinates of the foot of the building are (0,0), find the coordinates of the point of collision

 

[CM_Tutor]

Originally posted by acmilan1987
1. A particle moves in a straight line so that its position x from a fixed point O at time t is given by x = 10 + 8 sin 2t + 6cos 2t. Prove that the motion is SH. Find the period and amplitude. (The amplitude is the part i cant do)

Use the auxiliary angle method to rewrite 8sin 2t + 6cos 2t as Rsin(2t + α).

2. Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top of a wall 26.25 m high and 30 m away.

The equations of motion should be x = Vtcos α and y = -5t² + Vtsin α + 15, taking g = 10 ms^-2.
When x = 30, y = 26.25 and y' = 0. So you have three simultaneous equations:
30 = Vtcos α _____ (1)
26.25 = -5t² + Vtsin α + 15 _____ (2)
0 = -10t + Vsin α _____ (3)
in the three unknowns V, t and α.

You don't care about t, so make t the subject of (3), and back substitute into (1) and (2) to get two equations in the two unknowns you do care about, and then solve for V and α. I get V = 25 ms^-1 and α = tan^-1(3 / 4).

3. A particle projected from a point meets the horizontal plane through the point of prjection after travelling a horizontal distance a, and in the course of its trajectory attains a greatest height b above the point of projection. Find the horiztonal and vertical components of the velocity in terms of a and b. Show that when it has described a horizontal distance x, it has attained a height of [4bx(a-x)] / a^2

Taking O as the point of projection, the equations of motion should be x = Vtcos α and y = -gt² / 2 + Vtsin α.

The particle lands when y = 0, ie at t = (2V / g)sin α, and at this time x = a. So,
a = (2V² / g) * sin α * cos α _____ (1)

The particle reaches its maximum height when y' = 0, ie at t = (V / g)sin α, and at this time y = b:
b = (-g / 2) * (V / g)²sin²α + Vsin α(V / g)sin α = (V² / 2g)sin²α _____ (2)

Now, put t = x / Vcos α into the equation for y, and simplify using (1) and (2). You should get y = 4bx(a - x) / a², as required.

Note that there is an alternate approach to this question. You could eliminate t from the equations of motion to get the equation of path, which is a quadratic in x, and hence the path is parabolic. Then, note that we know two points on this path - (0, 0) and (a, 0). Since a parabolic path is symmetric about its axis, and it only attains one maximum height, this point must lie on the axis, and so the path also passes through (a / 2, b). Now, there is only one parabola that passes through these three points, and that is y = 4bx(a - x) / a², so this must be the equation of path.

4. A stone is projected horizontally from the top of a building 30 metres high with a velocity of 15m/s. At the same instant another stone is projected from the foot of the building with a velocity of 30m/s at an angle of 60 degrees to the horizontal. The two stones move in the same vertical plane/ Prove that they collide and, if the coordinates of the foot of the building are (0,0), find the coordinates of the point of collision

Top of building stone has equations of motion of: x = 15t and y = 30 - 5t², taking g = 10 ms^-2.

Foot of building stone has equations x = 15t and y = 15t * sqrt(3) - 5t², taking g = 10 ms^-2.

With the same x equation, the stones are clearly vertically above or below one another, or else they are colliding, which occurs when they are at the same height:

30 - 5t² = 15t * sqrt(3) - 5t²,
t = 2 * sqrt(3) / 3

So, they collide at time t = 2 * sqrt(3) / 3, at which time x = 10 * sqrt(3) and y = 70 / 3.

They collide at position (10 * sqrt(3), 70 / 3)

 

[acmilan]

thanks

 

[DcM]

how about Q17.

A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a 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.

 

[CM_Tutor]

Originally posted by DcM
how about Q17.

5 * (sqrt(2) + 1)

 

[nick1048]

yeah 17 is tricky... I dont get it... Yeah that's the answer but how

 

[Jumbo Cactuar]

By geometry;

(5t)2/h = ((Vh - 10) t + 5 t )2 / 2 h
2(5)2= (Vh -5)2
50 = 25 - 10 Vh + Vh2
Vh2 - 10 Vh -25 = 0

Vh = (10 +/- sqrt(100+100))/2
= 5 (1 +/- sqrt(2))
but Vh is larger than 10 so;
Vh = 5 (1 + sqrt(2))

 

[Pace_T]

I must be missing something there, can somoene please explain, lol.
What geometrical approach is being used in Jumbo Cactuar's post? Thanks!

 

[Jumbo Cactuar]

uh yeah, it is the geometry of a parabola;

x²/y = constant = (x₀)²/y₀

 

[Pace_T]

Cheers buddy.