This discussion is closed.
#1
I will just put all my mechanics question into this thread rather than creating mini ones, saves me a lot of time.

For now i just have i three questions i need help on.

Question 1

Spoiler:
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"The picture shows Romeo(use your imagination lol) trying to attract Juliet's attention without her nurse, who is in a downstairs room, noticing. He stands 10m from the house and lobs a small pebble at her bedroom window. Romeo throws the pebble from a height of 1m with a speed of 11.5 ms^-1 at an angle of 60 degrees to the horizontal.

(i) How long does the pebble take to the house? t = 1.74 secs(correct)
(ii)Does the pebble hit Juliet's window, the wall of the house or the downstairs room window? s = 3.50m, Juliet's window(correct)
(iii) What is the speed of the pebble when it hits the house?

This is where i get stuck, the answer is v = 9.12ms^-1 and tried splitting it up into components Vy and Vx, Vy = 11.5sin60 - 9.8*1.74(which is wrong) and Vx doesn't work, so what do i do here???

Question 2

Spoiler:
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Use g = 10 ms^-2......A firework is buried so that its top is at ground level and it projects sparks all at a spped of 8ms^-1. Air resistance may be neglected.

(i) Calculate the height reached by a spark projected vertically and explain why no spark can reach a height greater than this. v^ = u^2 + 2as....s = 3.27m
(ii) For a spark projected at 30 degrees to the horizontal over horizontal ground, show that its height in metres t seconds after projection is 4t-5t^2 and hence calculate the distance it lands from the firework.

s = ut+1/2at^2, 8cos60 = 4 and so on......when it lands y = 0, so solve for t, t = 4/5......s = ut(horizontal) so, 8sin60*4/5 = 5.5m(correct)

(iii) For what angle of projection will a spark reach a maximum height of 2 m??......i am not sure what to do here tbh, any hint would be useful

Question 3

Spoiler:
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A small stone is projected horizontally from a height of 19.6m over horizontal ground at a speed of 15ms^-1

(i) Calculate the time which elapses between projection and the stone hitting the ground and show that the horizontal distance travelled is 30m.....So = 19.6m, Sy = ut+1/2at^2 = -4.9t^2

-19.6/-4.9 = t^2 = t = 2, Sx = 15*2 = 30m

In a new situation two small stones are projected horizontally towards each other at the same instant over horizontal ground. Initially the stones are 19.6m above the ground and are 50m apart. The initial speeds are 15ms^-1 and v ms^-1.

(ii) Explain briefly why the stones will be at the same height as one another above the ground after the elapse of any time interval and hence much collide if they do not hit the ground first. Y1 = Y2 = 19.6-4.9t^2 which is the same for all time intervals (correct)

(iii) Find the value of v if the stones collide 4.9m above the ground.

i got no idea what i am supposed to be doing here???

0
8 years ago
#2
(Original post by boromir9111)
I will just put all my mechanics question into this thread rather than creating mini ones, saves me a lot of time.

For now i just have i three questions i need help on.

Question 1

Spoiler:
Show
"The picture shows Romeo(use your imagination lol) trying to attract Juliet's attention without her nurse, who is in a downstairs room, noticing. He stands 10m from the house and lobs a small pebble at her bedroom window. Romeo throws the pebble from a height of 1m with a speed of 11.5 ms^-1 at an angle of 60 degrees to the horizontal.

(i) How long does the pebble take to the house? t = 1.74 secs(correct)
(ii)Does the pebble hit Juliet's window, the wall of the house or the downstairs room window? s = 3.50m, Juliet's window(correct)
(iii) What is the speed of the pebble when it hits the house?

This is where i get stuck, the answer is v = 9.12ms^-1 and tried splitting it up into components Vy and Vx, Vy = 11.5sin60 - 9.8*1.74(which is wrong) and Vx doesn't work, so what do i do here???

Question 2

Spoiler:
Show
Use g = 10 ms^-2......A firework is buried so that its top is at ground level and it projects sparks all at a spped of 8ms^-1. Air resistance may be neglected.

(i) Calculate the height reached by a spark projected vertically and explain why no spark can reach a height greater than this. v^ = u^2 + 2as....s = 3.27m
(ii) For a spark projected at 30 degrees to the horizontal over horizontal ground, show that its height in metres t seconds after projection is 4t-5t^2 and hence calculate the distance it lands from the firework.

s = ut+1/2at^2, 8cos60 = 4 and so on......when it lands y = 0, so solve for t, t = 4/5......s = ut(horizontal) so, 8sin60*4/5 = 5.5m(correct)

(iii) For what angle of projection will a spark reach a maximum height of 2 m??......i am not sure what to do here tbh, any hint would be useful

Question 3

Spoiler:
Show
A small stone is projected horizontally from a height of 19.6m over horizontal ground at a speed of 15ms^-1

(i) Calculate the time which elapses between projection and the stone hitting the ground and show that the horizontal distance travelled is 30m.....So = 19.6m, Sy = ut+1/2at^2 = -4.9t^2

-19.6/-4.9 = t^2 = t = 2, Sx = 15*2 = 30m

In a new situation two small stones are projected horizontally towards each other at the same instant over horizontal ground. Initially the stones are 19.6m above the ground and are 50m apart. The initial speeds are 15ms^-1 and v ms^-1.

(ii) Explain briefly why the stones will be at the same height as one another above the ground after the elapse of any time interval and hence much collide if they do not hit the ground first. Y1 = Y2 = 19.6-4.9t^2 which is the same for all time intervals (correct)

(iii) Find the value of v if the stones collide 4.9m above the ground.

i got no idea what i am supposed to be doing here???

u is initial velocity, v is final v_x represent x component etc.
u_x= u cos60= v_x
10/ ucos60 = T

v_y= v_y - gT

(v_y^2 + v_x^2)^0.5= final speed

that looks like what you have, and if its wrong, I think possible the answer itself is wrong.

for q2 you know the absolute speed of firing out. from the first part, you just do the same:

max height h, projection speed u, final speed=0, projection angle Q measured from horizontal
using...
V^2= U^2+2as

0= u^2 sin^2(Q) -2gh
solve for Q

for q3 you can form equations for its vertical and horizontal position. becuase you know:

vertical displacement of Stone 1= vertical dis of stone 2
horizontal displacement of stone 1= hor. dis. stone 2

and the time at which this occurs has to be the same.

Therefore you could form 4 equations for each stones displacement as a fucntion of time. Then set the hor. equal and vert. equal, solving for t
0
#3
(Original post by hai2410)
u is initial velocity, v is final v_x represent x component etc.
u_x= u cos60= v_x
10/ ucos60 = T

v_y= v_y - gT

(v_y^2 + v_x^2)^0.5= final speed

that looks like what you have, and if its wrong, I think possible the answer itself is wrong.

for q2 you know the absolute speed of firing out. from the first part, you just do the same:

max height h, projection speed u, final speed=0, projection angle Q measured from horizontal
using...
V^2= U^2+2as

0= u^2 sin^2(Q) -2gh
solve for Q

for q3 you can form equations for its vertical and horizontal position. becuase you know:

vertical displacement of Stone 1= vertical dis of stone 2
horizontal displacement of stone 1= hor. dis. stone 2

and the time at which this occurs has to be the same.

Therefore you could form 4 equations for each stones displacement as a fucntion of time. Then set the hor. equal and vert. equal, solving for t
I've already sent you message but shall reply on here as well....for part (ii) v^2 = u^2sin^2(q) + 2as

s = 5.5m v = 0

2*-10*5.5 = -110-----> 110 = 64sin^2(q)
sin^2(q) = 1.71875
sin(q) = root(1.7185)....... and that doesn't work....what mistake am i making here?
0
8 years ago
#4
(Original post by boromir9111)
I've already sent you message but shall reply on here as well....for part (ii) v^2 = u^2sin^2(q) + 2as

s = 5.5m v = 0

2*-10*5.5 = -110-----> 110 = 64sin^2(q)
sin^2(q) = 1.71875
sin(q) = root(1.7185)....... and that doesn't work....what mistake am i making here?
sorry can't really understand what your doing but i'll try clarify.

you know the magnitude of a 2-d vector is just given by the pythagorean root of the sum of the square of the sides) like c^2=a^2+b^2 etc.

So if you work out the horizontal and vertical components then do Vx^2 +Vy^2 =Vtotal^2 to get Vtotal

I think you already knew that, but anyway.

Here's my working:

Vertical component (changes with time)
USING V=U+AT
11.5sin(60)- Tg= Vy where T is 1.74 since thats the time it takes to hit the window

Horizontal component (constant)
=11.5cos(60)

if you work out the magnitude now:

V^2=Vy^2 +Vx^2= (11.5sin(60)-1.74*9.81)^2+(11.5cos(60)^2

Gives V=9.14 ms^-1

which is probably slightly different due to the rounding of 1.74
0
8 years ago
#5
In fact, looking again at your OP, thats the exact same method you used. So if your wrong i'm wrong (very possible)
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#6
(Original post by hai2410)
sorry can't really understand what your doing but i'll try clarify.

you know the magnitude of a 2-d vector is just given by the pythagorean root of the sum of the square of the sides) like c^2=a^2+b^2 etc.

So if you work out the horizontal and vertical components then do Vx^2 +Vy^2 =Vtotal^2 to get Vtotal

I think you already knew that, but anyway.

Here's my working:

Vertical component (changes with time)
USING V=U+AT
11.5sin(60)- Tg= Vy where T is 1.74 since thats the time it takes to hit the window

Horizontal component (constant)
=11.5cos(60)

if you work out the magnitude now:

V^2=Vy^2 +Vx^2= (11.5sin(60)-1.74*9.81)^2+(11.5cos(60)^2

Gives V=9.14 ms^-1

which is probably slightly different due to the rounding of 1.74

that method i get now mate, i didn't make my post clear lol, i was talking about the second question....finding the angle for maximum height, 2m and i showed working for that???
0
8 years ago
#7
(Original post by boromir9111)
I've already sent you message but shall reply on here as well....for part (ii) v^2 = u^2sin^2(q) + 2as

s = 5.5m v = 0

2*-10*5.5 = -110-----> 110 = 64sin^2(q)
sin^2(q) = 1.71875
sin(q) = root(1.7185)....... and that doesn't work....what mistake am i making here?
oh right

considering vertical velocities only:
V^2=U^2+2AS

you know v=0 at the top of a parabolic path

0= (8sin(theta))^2 -2*10*2

solve for theta=52.2
0
8 years ago
#8
(Original post by hai2410)
oh right

considering vertical velocities only:
V^2=U^2+2AS

you know v=0 at the top of a parabolic path

0= (8sin(theta))^2 -2*10*2

solve for theta=52.2
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#9
(Original post by hai2410)
oh right

considering vertical velocities only:
V^2=U^2+2AS

you know v=0 at the top of a parabolic path

0= (8sin(theta))^2 -2*10*2

solve for theta=52.2
Thanks for that
0
8 years ago
#10
(Original post by someperson)
and yours is a mind****!

(Original post by boromir9111)
Thanks for that
I don't want to read through the whole thread but have all of the questions been solved?
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#11
(Original post by Clarity Incognito)
and yours is a mind****!

I don't want to read through the whole thread but have all of the questions been solved?
just got question 3 left..... when it says it collides 4.9m above the ground, do i take that as S?
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#12
Sy = 4.9m, 4.9m = 19.6 -4.9t^2

4.9t^2 = 14.7..... t = 1.73secs....on the right track?
0
8 years ago
#13
(Original post by boromir9111)
Sy = 4.9m, 4.9m = 19.6 -4.9t^2

4.9t^2 = 14.7..... t = 1.73secs....on the right track?
Yeah, you can indeed go about it that way. Find out how far particle on the left has travelled horizontally, which will allow you to find how far the particle on the right has travelled horizontally to the left.
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#14
(Original post by Clarity Incognito)
Yeah, you can indeed go about it that way. Find out how far particle on the left has travelled horizontally, which will allow you to find how far the particle on the right has travelled horizontally to the left.
Well, i think i have the answer. U = 13.9ms^-1???......

On the left, S = 25.95......so to get how far away the other stone is, i did 50-that and so, from there i get 13.9....is that the right method?
0
8 years ago
#15
(Original post by boromir9111)
Well, i think i have the answer. U = 13.9ms^-1???......

On the left, S = 25.95......so to get how far away the other stone is, i did 50-that and so, from there i get 13.9....is that the right method?
I haven't done the calculations but the idea is correct, and if you think about it logically, if the particle on the right was projected with a speed less the particle on the left then the particle on the left would have travelled a greater distance to the right than the particle on the right to the left. That was a bit convoluted but correct.
0
#16
(Original post by Clarity Incognito)
I haven't done the calculations but the idea is correct, and if you think about it logically, if the particle on the right was projected with a speed less the particle on the left then the particle on the left would have travelled a greater distance to the right than the particle on the right to the left. That was a bit convoluted but correct.
Very good point mate.....just checked now and the answer is correct, so thanks mate!.....questions done for now...if i have any more i shall post them on here
0
8 years ago
#17
(Original post by someperson)
Damn you, I have been revising M1 for the last 80minutes or so, decided to open this thread as I thought it might help with the revision but then I disocovered... dum dum dum Ripple Tank Simulation and now.. well... bye bye revision ;D
0
8 years ago
#18
(Original post by Clarity Incognito)
and yours is a mind****!

(Original post by gummers)
Damn you, I have been revising M1 for the last 80minutes or so, decided to open this thread as I thought it might help with the revision but then I disocovered... dum dum dum Ripple Tank Simulation and now.. well... bye bye revision ;D
sorry...but that makes me feel less guilty about not revising lol
0
#19
Problem 4

Spoiler:
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http://www.mei.org.uk/files/papers/ouikyuj45y.pdf
question 5 (iii)......i solved the above by using triangle forces but not sure how to approach this???
0
#20
anyone?
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