Summary Mathematics A Level - an A* student's notebook (statistics&mechanics)

Statistics & Mechanics

, Motion :




SB) 1)
25N a
1




to
plane Y REY 3gcos20 =
R

(parallel
- v
R
resultant
_ =
27 . 655N
col ↑ % 27 7N
& CoS20-30 sins
3a
=
20

R)
=
.




F



RE] bysinto 3
=
a


30V Sa
1) Parallel to plane

2)
:


T 35ms-2
is) 2 Scos20-30 sin IS
a = 3 .




V
30) R =
Sgcos30 = 42 4 N .




30N
2) Perpendicular to the plane
g
:




2)
S0-SgsinSo
25 Sin 20-30 sos IS

-
R 5 = 12 46 .




F
3)
=
Ma
49ms-2



-
2
cosan a

tana)
R = 0 .
x y8 a = .




R



4)
M

> 30N
R 92N b) R 6x9 8xCoS15

o
= 3 .
=
.




b) 0 5 .
x 9 8 .
x Sin (tan") =
F
Fr =
56 .
80N



F =
2 .
94N 15 % Parallel to slope 30-Fr-6x9
: . Sxsin 15
0 .




59 -
a =
5 88 ms
Fr
.



= 14 78 N

N
.




Friction< Reaction Fc- > z8M
-




a) 9 SN

↓
.




Sg
6) 2 .
04 ms -
2




R
Sg 4 = 0 4




↑ find
.




uR
Fr =

2 .
% .


P ↑ 9) F = 10

Fr
> 18
-


when the friction at it's
greatest Fr
↓
is + P =
0 So a =
0
X


↓
FXuR Fr = 0 4R b) limiting equili-
Sg
.




earth muis <1 Brium
on ,




limiting friction 3g
-
limiting equilibrium

92 09 .


= 6




-F >
-
parallel F-lOgSin : 20 = 10a



↑ perpendicular : R-
10g(os 20 = 0
↳
R 33 518
= .




5) SN 20) TOg
F MR =




~
M 5 =
9 8 .
.


m x 60530


30 (3 S tan"
*
m = 0 .



589kg s .

f)
)




2 =


26N
2 =
38
acceleration F 1



di
:
= 0 .

589x9 8 x Sin 36 . a =




F 2 88675N




o
=
.




F
· = m / 26 20545-12-mx9 83in30
= a
12N Parallel to slope : .
= F
m
je =

4 9
3
30 Mx9 SSin30
6)
= .
ms-
6
.




. 385 12 + Br
-


9 850s30
Ssin3a
=
mx m
- -


.




m( + 9 .




15
=

12+
59
-638
-




08kg
m


Parallel to 30cos30- 5g Sin 30 = 1
plain
.

:




F
=


-


N Challenge :




&
2
1 9
-



a ms
=

=
.




Pr
/1 9 ms 2
Gu .
down slope mysinG
=
ma Using gsin(0 60
=
+


R

7) Mg sin(0 60) Sin(0 60)
~
+ =
Uma 4 sinc = +




F
constant velocity ms- of identifies
0
gsinD
L So a * 4
=
= a · . .
. . .




&
20) F =
0 V
gSin(0 60) + = 4a

V
my
3 x 9 8 Sin40- 6Sin40- Fr ON
39
=
.




Fo =
15 04 N
.

, 1) y =
7
i)
52)a)
a
2) So 14


e)
R R14N
Sg

↑ ↑ ↑ Y
> 3N F + 3
> 12N gu > GN
-




-

↓
-
-




↓
↳ t 59
>
5g Sg
F =
ESg = 7N

F = SN
in this case F 3N=
F =
7N R =
14 + Sx9 8 .
R =
35N
,
2
a = 0 2 ms-
.




block is
stationary
a = 1ms - F =

E .



63 =
9N F =
SN ,
this case 3N,



k) limiting equilibrium stationary
i)
. so
R14N

2
1 ms-
T -




Stationary >
-




F

F
S




2)a
↓
R



!
R
L




I
4) ↑
39

↓
F


R R =
Sg-28sin 38 =
35N
Sg- 14sin30 42 N
=


Sg
=



S6N 10g
F =
6N F =
SN

14 cos 38 = 753 2810530. SN =
Resultant R =
Sx9 8 + 56sin4s
.
=
(uS + 28(2)N R =

10g- 20sin30 =
SSN


a
=
2 42ms
.
(3 5 .

8 )
)
a = 3 .
85ms(35 f) .
F =
7 + 4E = 12 .
66N(2d .


p )
.
E =
88 ·

2



. SMS-2
6
> >
-


bosYS-F =
a


b)
-




S30-F
R . Sms-2
8

R
2012 N
2
↑ (35 f)
Me
Y ZON a = 5 .
39ms-2 .




F
Ta?
F
Fu 10 2020330




·
=




↓
-




↳
-




S L

3) Fr =
7 .
32N(35 G ) . .




10g log
7 = 0 .
0832/35 f ) . .




R =
log-20sin60 =
(98-1053) M R =

log + 201 sin 48 = 118N
o& ta
F
P

0 3
al

a
8
Sg
= .



.




Fr =
Ru Fr =
1181
.S
O
- Fr
& R 0
SgcosIS 4 733 N R Zgcos20 18 42N
&
10260 Cosus - Fr
= = = = =
a .
.
.




SysinS-E
= Fr S SZSN
O
= .




Fr SN F ISN 25
=
Egsin20-F- p = Ma
=




u =
0 . 0620(35F .



y
= 0 .
127 (3sf)
E = 1 . 143N P = 0 .
778N(3s f) .




u 02

5) 242/35f)
=



6)
.




R P R =

Sgcos30 = 42 435N u = 0 .




inst
.




!
= 8 47
pR
Fr .




↑ Y 1 =
0 1 .




7) V = u + at
h F a 0 3ms
-2
=z

Resin30-F
2
=
30a

↓
0 20
.




= = +
7 ↑ 50s
a
=
E (decelerating)
p
=
42 .
987 . .
10g 1
R


43 0(35 f)
P
.

.




R Psin4S F < R
10g
=

mg
- =




F =
0 .



1(10g -



Ep) =
(5 - p)N
. 8 V F =


ymg
mg




-Fo
P F - = ma
-umg =
-Emb) unchanged -
it's



P-umg =
-




Em ug =
33 independent of air resistance




59/
p =
0 u = -


depends on wheels & rails
0




# +p
= 3 + 9 8 .




p
Constant acceleration .
16 4563 .
= .




16 5 (35 f )
t(u v)t p
=

at
.
.
.




V= u + s = +




?
S = ut (at
+
v2 u = + 2as