IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: [Spy] on May 15, 2010, 12:19:09 pm

Title: M.e.c.h.a.n.i.c.s
Post by: [Spy] on May 15, 2010, 12:19:09 pm
Doubts/Predictions/Past Paper Questions


hello people, im amazing at mechanics, i solved alll papers and i got a 48/50 in my mock in mehcanics

so any help im here :)
Title: Re: M.e.c.h.a.n.i.c.s
Post by: mz on May 15, 2010, 01:17:43 pm
awesome! please post any predictions! or pm me!!! i was good at m1 but right now i cannot comprehend any question...can you please give an overview of what we have to keep in mind when solving enery/work related questions???
Title: Re: M.e.c.h.a.n.i.c.s
Post by: Ghost Of Highbury on May 15, 2010, 03:19:22 pm
One horse pulls with a force of X N, a cart of mass 800 kg along a horizontal road at constant speed. Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8m/s^2. Find the time it wd take for 2 horses to increase the speed of the cart from 2m/s to 5m/s, given that each horse pulls with a force of X N and that the resistance to motion has the same constant values.
Title: Re: M.e.c.h.a.n.i.c.s
Post by: [Spy] on May 15, 2010, 03:36:20 pm
One horse pulls with a force of X N, a cart of mass 800 kg along a horizontal road at constant speed. Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8m/s^2. Find the time it wd take for 2 horses to increase the speed of the cart from 2m/s to 5m/s, given that each horse pulls with a force of X N and that the resistance to motion has the same constant values.

is the answer 7.01 seconds?
Title: Re: M.e.c.h.a.n.i.c.s
Post by: cooldude on May 15, 2010, 03:45:58 pm
One horse pulls with a force of X N, a cart of mass 800 kg along a horizontal road at constant speed. Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8m/s^2. Find the time it wd take for 2 horses to increase the speed of the cart from 2m/s to 5m/s, given that each horse pulls with a force of X N and that the resistance to motion has the same constant values.

let the resistance to the cart=R
therefore X-R=0, X=R
3X-R=ma, 3X-R=800*0.8
2R=640
R=X=320
2X-R=800a
320=800a
a=0.4
v=u+at
5=2+0.4t
t=7.5 sec
Title: Re: M.e.c.h.a.n.i.c.s
Post by: Ghost Of Highbury on May 15, 2010, 03:55:27 pm
Thanks a lot.
Title: Re: M.e.c.h.a.n.i.c.s
Post by: DaYmN on May 15, 2010, 07:59:11 pm
Quote
let the resistance to the cart=R
therefore X-R=0, X=R
3X-R=ma, 3X-R=800*0.8
2R=640
R=X=320
2X-R=800a
320=800a
a=0.4
v=u+at
5=2+0.4t
t=7.5 sec

That's genius! thank yu very much!  ;)
Title: Re: M.e.c.h.a.n.i.c.s
Post by: preity on May 16, 2010, 09:18:15 am
i need help for Q6 O/N 03 paper?plssss
Title: Re: M.e.c.h.a.n.i.c.s
Post by: cooldude on May 16, 2010, 10:11:11 am
i need help for Q6 O/N 03 paper?plssss

a) resolve horizontally and vertically about M-->
horizontally - let tension in BM=T
                  AM=K
Ksin30+Tsin30=10
K+T=10
vertically - Tsin30-Ksin30=0
T=K=5

b) let coefficient of friction=c
resolving vertically about B
Tcos30-F=0.2*10
F=cR
R=T*sin30=2.5
F=2.5c
5cos30-2.5c=2
therefore c=0.932
 
c) T=5
F+5cos30=2+10m
F=Tsin30=2.5
5cos30+(2.5*0.932)=2+10m
m=0.466 kg

Title: Re: M.e.c.h.a.n.i.c.s
Post by: chocolatesss on May 16, 2010, 12:10:54 pm
One horse pulls with a force of X N, a cart of mass 800 kg along a horizontal road at constant speed. Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8m/s^2. Find the time it wd take for 2 horses to increase the speed of the cart from 2m/s to 5m/s, given that each horse pulls with a force of X N and that the resistance to motion has the same constant values.
did u make up this question or is it a prediction.???
Title: Re: M.e.c.h.a.n.i.c.s
Post by: Ghost Of Highbury on May 16, 2010, 12:25:02 pm
did u make up this question or is it a prediction.???

M1 book, Chapter 2, Ex 2B, Question 15 i suppose.
Title: Re: M.e.c.h.a.n.i.c.s
Post by: halosh92 on May 16, 2010, 09:16:27 pm
let the resistance to the cart=R
therefore X-R=0, X=R
3X-R=ma, 3X-R=800*0.8
2R=640
R=X=320
2X-R=800a
320=800a
a=0.4
v=u+at
5=2+0.4t
t=7.5 sec

for the first step u did.........shouslnt we be using F=ma
??(X-R=0).....or is it because its moving at constant speed so we dont use it?
Title: Re: M.e.c.h.a.n.i.c.s
Post by: cooldude on May 17, 2010, 03:16:24 am
for the first step u did.........shouslnt we be using F=ma
??(X-R=0).....or is it because its moving at constant speed so we dont use it?

yeah we're using F=ma only but becuase it is moving at a constant speed so a=0 thus ma=0
Title: Re: M.e.c.h.a.n.i.c.s
Post by: halosh92 on May 17, 2010, 06:33:39 am
yeah we're using F=ma only but becuase it is moving at a constant speed so a=0 thus ma=0

okek thx  :)
Title: Re: M.e.c.h.a.n.i.c.s
Post by: halosh92 on May 17, 2010, 09:04:46 am
jan 2008 edexcel
Q1a)
for the impulse isnt it
I=mv-mu
I= 4*1-4*5
= -16

how come its negative..in the ms....its +16
Title: Re: M.e.c.h.a.n.i.c.s
Post by: chocolatesss on May 17, 2010, 11:30:58 am
M1 book, Chapter 2, Ex 2B, Question 15 i suppose.
ohh...okaaayyyyy