Hard work
Badges: 8
Rep:
?
#1
Report Thread starter 4 years ago
#1
Anyone here plzzz need help with mechnics....
0
reply
eden3
Badges: 19
Rep:
?
#2
Report 4 years ago
#2
....Okay? What exactly do you need help with? You haven't provided us with a question.
1
reply
Hard work
Badges: 8
Rep:
?
#3
Report Thread starter 4 years ago
#3
I am confused about this question......Name:  m3.png
Views: 410
Size:  148.9 KBAttachment 516003516005
Attached files
0
reply
Hard work
Badges: 8
Rep:
?
#4
Report Thread starter 4 years ago
#4
so the bit I am confused about is why did they times it by 2? and also ehy h=0 and they divided by the acceleration why???
0
reply
Hard work
Badges: 8
Rep:
?
#5
Report Thread starter 4 years ago
#5
HELOOOOOOOOOOOOOOOOOOOO any one???????????????????
0
reply
Hard work
Badges: 8
Rep:
?
#6
Report Thread starter 4 years ago
#6
I NEED HELP!!! plsssss someone help me with mechanics I am failing
0
reply
Hard work
Badges: 8
Rep:
?
#7
Report Thread starter 4 years ago
#7
(Original post by eden3)
....Okay? What exactly do you need help with? You haven't provided us with a question.
I posted the question so can u help now?
plssssssss
0
reply
eden3
Badges: 19
Rep:
?
#8
Report 4 years ago
#8
(Original post by Hard work)
I posted the question so can u help now?
plssssssss
Calm down.. I haven't done A-levels since last June so it's taking me a while to figure out.
1
reply
eden3
Badges: 19
Rep:
?
#9
Report 4 years ago
#9
(Original post by Hard work)
I NEED HELP!!! plsssss someone help me with mechanics I am failing
I'm sorry I only understand the alternative answer.
0
reply
Hōseki
Badges: 14
Rep:
?
#10
Report 4 years ago
#10
(Original post by Hard work)
I am confused about this question......Name:  m3.png
Views: 410
Size:  148.9 KBAttachment 516003516005

Vertical Motion (initial speed) = 40sina

Recall the equation s=ut+1/2at^2 but as it is decelerating it becomes s=ut-1/2at^2.

Note that in this case a is interchangable with g as it is acceleration due to free fall.

So next you rearrange for t:

s=0 as there is no displacement so you can rule that out.

Add 1/2gt^2 on both sides leaves 1/2gt^2=ut
The t's on both sides cancel out leaving 1/2gt=u
Times both sides by 2 to give you gt=2u
Then divide both sides by g so t=2u/g
Then just substitute T= 2x40xsina/g


T=80sina/g
1
reply
TheTechnoGuy
Badges: 7
Rep:
?
#11
Report 4 years ago
#11
Name:  M1-Projectiles.png
Views: 417
Size:  312.0 KBTry this, I hopefully simplified enough. Ignore my terrible handwriting.
2
reply
Hōseki
Badges: 14
Rep:
?
#12
Report 4 years ago
#12
As for the second half of the question it is just Speed=Distance/Time
So we rearrange for D giving D=ST where D is Rm.

Use cos to calculate the horizontal motion (initial speed) which is 40cosa.

R=ST so you now have R=40cosaxT
You know that T=80sina/g from the first part of the question so you just substitute giving...
40xcosax80xsina/g
40x80=3200
Final Answer: 3200sinacosa/g
1
reply
TheTechnoGuy
Badges: 7
Rep:
?
#13
Report 4 years ago
#13
(Original post by Hōseki)
Vertical Motion (initial speed) = 40sina

Recall the equation s=ut+1/2at^2 but as it is decelerating it becomes s=ut-1/2at^2.

Note that in this case a is interchangable with g as it is acceleration due to free fall.

So next you rearrange for t:

s=0 as there is no displacement so you can rule that out.

Add 1/2gt^2 on both sides leaves 1/2gt^2=ut
The t's on both sides cancel out leaving 1/2gt=u
Times both sides by 2 to give you gt=2u
Then divide both sides by g so t=2u/g
Then just substitute T= 2x40xsina/g


T=80sina/g
Beat me to it, aha
0
reply
Hōseki
Badges: 14
Rep:
?
#14
Report 4 years ago
#14
(Original post by TheTechnoGuy)
Beat me to it, aha
:happy2:
1
reply
Hōseki
Badges: 14
Rep:
?
#15
Report 4 years ago
#15
(ii) Calculate the acceleration of the car and trailer. Calculate also the force in
the coupling, stating whether this is a tension or a thrust. [6]

Newton's second Law F=ma

Add the masses: 1500+900=2400kg
Driving force = 600N
F=ma
600N=2400kgxa
A=0.25m/s^2

For calculating the Tension you also use F=ma
Where F is the tension.
T=900kgx0.25m/s^2
T=225N
0
reply
Hard work
Badges: 8
Rep:
?
#16
Report Thread starter 4 years ago
#16
Thanks so much guys really appreciate it.
I have another question I am struggling with,
1. Ann and Beryl are both pushing a piano. Ann pushes in the direction 4i + 3j andBeryl in the direction 12i + 5j, where i and j are the standard unit vectors.Together they produce a force of (68i + 33j) N so that p(4i + 3j) + q(12i + 5j) = 68i + 33j, where p and q are constants.Calculate the force with which Beryl pushes.
0
reply
Hōseki
Badges: 14
Rep:
?
#17
Report 4 years ago
#17
(Original post by Hard work)
Thanks so much guys really appreciate it.
I have another question I am struggling with,
1. Ann and Beryl are both pushing a piano. Ann pushes in the direction 4i + 3j andBeryl in the direction 12i + 5j, where i and j are the standard unit vectors.Together they produce a force of (68i + 33j) N so that p(4i + 3j) + q(12i + 5j) = 68i + 33j, where p and q are constants.Calculate the force with which Beryl pushes.
I believe it's a simultaneous -
4p + 12q = 68
3p + 5q = 33

Take it from here and find the values for p and q.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (41)
16.08%
I'm not sure (8)
3.14%
No, I'm going to stick it out for now (92)
36.08%
I have already dropped out (4)
1.57%
I'm not a current university student (110)
43.14%

Watched Threads

View All