You are on page 1of 6

# AQA Mechanics 1 Newtons Laws of motion

## Section 2: Connected particles

Solutions to Exercise
1.

T
T

3 kg

9 kg
3g
9g

T 3g 3a
9g T 9a
6g 12a

Considering 3 kg mass:
Considering 9 kg mass:

(1)
(2)

a 21 9.8 4.9
The acceleration of the system is 4.9 ms-2.
(1) gives T 3g 3a

3( 9.8 4.9)
44.1
The tension in the string is 44.1 N.
2. Resistance experienced by engine 60000 80 4800000
Resistance experienced by truck 12000 50 600000
Train is travelling at constant speed, so acceleration is zero.

R1

R2
12000 kg

60000 kg

4800000

600000

60000g

12000g
Considering whole system:

X 4800000 600000 0
X 5400000

## The driving force of the engine = 5400 kN.

Considering truck:

T 600000 0
T 600000

1 of 6

10/01/13 MEI

## AQA M1 Newtons Laws 2 Exercise solutions

3.

R
T

4 kg

4g

1.5 kg
1.5g
For the 4 kg mass:
For the 1.5 kg mass:

T 4a
1.5 g T 1.5 a
1.5 g 5.5 a

(1)
(2)

1.5 9.8
2.67
5.5
The acceleration of the system is 2.67 ms-2 (3 s.f.)

## Substituting into (1):

T 4a

1.5 9.8
10.7
5.5
The tension in the string is 10.7 N (3 s.f.)

T 4

4.

R2
200000 kg
7000

R1

a
T

50000 kg

60000

3000
50000g

200000g

## Considering the whole system: 60000 3000 7000 250000a

50000 250000a
a 0.2
The acceleration of the system is 0.2 ms-2.
Considering the truck:

T 7000 40000
T 47000

2 of 6

10/01/13 MEI

5. (i)

T2

T1

T2
a

T1

10 kg

10g

5 kg

8 kg

5g

8g

## (ii) Considering 8 kg mass:

8 g T 1 8a

T 1 8 g 8a
(1)
Considering 5 kg mass:
T2 5 g 5a
T2 5 g 5a
(2)
Considering the 10 kg mass:
(3)
T 1 T 2 10a
Substituting (1) and (2) into (3): 8 g 8a (5 g 5 a ) 10a
3g 23a
3 9.8
1.28
23
The acceleration of the system is 1.28 ms-2 (3 s.f.)

## Substituting into (1):

68.2
Substituting into (2):

3 9.8

23

T 1 8 9.8 8

3 9.8

23

T 2 5 9.8 5
55.4

6.

R
m

mg

T
3 of 6

10/01/13 MEI

## Considering particle on table: T ma

Considering hanging particle: mg T ma
mg 2ma

a 4.9, u 0, t 0.5

a 21 g 4.9

s ut 21 at 2
0 21 4.9 0.5 2
0.6125
Distance travelled = 0.6125 m

v u at
0 4.9 0.5
2.45
Speed = 2.45 ms-1.

7.

R 6g 0
R 6 9.8 58.8
Friction is limiting so F R
F 0.4 58.8 23.52
Horizontally for the box: T F 6a
T 23.52 6a
(1)
3g T 3a
For 3 kg mass:
(2)
Vertically for the box:

4 of 6

10/01/13 MEI

## AQA M1 Newtons Laws 2 Exercise solutions

u 0, t 3, a
s ut 21 at 2

1.96
3

3 9.8 23.52 9a
5.88 9a
1.96
a
3

1 1.96

32
2
3
2.94
It travels 2.94 m in the first three seconds of motion.
0

8.

M kg
Mg

T
m kg
mg

## (i) If the system is at rest:

For mass m: T mg 0

T mg

T F 0
F T mg
R Mg 0
R Mg

## For mass M horizontally:

For mass M vertically:

F R
mg 31 Mg
M 3m

(ii) If the mass M is not to slide down the plane, friction acts upwards.

R
T
F
T
30

Mg
5 of 6

10/01/13 MEI

mg

## AQA M1 Newtons Laws 2 Exercise solutions

T mg 0
T mg
Resolving for mass M parallel to plane:
For mass m:

T F Mg sin 30 0
F 21 Mg T

## Resolving for mass M perpendicular to plane:

R 21 Mg 3

F R
1
2

F 21 Mg mg
R Mg cos 30 0

Mg mg 31 21 Mg 3

3M 6m M 3

M(3 3) 6m
M 4.73m
(iii)

As in (ii), R 21 Mg 3 and T mg .
Resolving for mass M parallel to plane:

T F Mg sin 30 0
F T 21 Mg
F mg 21 Mg

F R
mg 21 Mg 31 21 Mg 3
6m 3M M 3

M(3 3) 6m
M 1.27 m

6 of 6

10/01/13 MEI
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}