TOPIC B: MOMENTUM ANSWERS SPRING 2018personalpages.manchester.ac.uk/staff/david.d... · Mechanics...

Mechanics Answers to Examples B (Momentum) - 1 David Apsley

TOPIC B: MOMENTUM – ANSWERS SPRING 2018

(Full worked answers follow on later pages)

Q1. (a) 2.26 m s–2

(b) 5.89 m s–2

Q2. 8.41 m s–2

and 4.20 m s–2

; 841 N

Q3. (a) 1.70 m s–1

(b) 1.86 s

Q4. (a) 1 s

(b) 1.5 s; 7.36 m s–1

; 19.6 m s–2

Q5. (a) 456 N

(b) 227 N

Q6. 1.41 m s–2

Q7. (a) 0.24

Q8. 10.1 s

Q9. (a) 1

1 sm0

5.12

v ; 1

2 sm83.10

25.6

v ;

1

1 smkg0

12500

vm ; 1

2 smkg8120

4690

vm

(b) 1sm64.4

82.9

(c) 17.2 m

Q10. 3.54 m from the wall; speed 15.0 m s–1

Q11. (a) 24.7 m s–1

(b) 20.0 m

Q12. (a) 0.764 m s–1

;

(b) sN0.15

3.18

(c) 0.779

Q13. (a) 10.2 m s–1

(b) 1sm26.7

49.3

(c) 0.693


Q14. (a) 2.55 m s–1

(b) 0.631

(c) 45.1%

Q15. 1

1

2

1v

ev

r

r

Q16. 3.17 m s–1

at 40.9° to right of original direction; 4.16 m s–1

at 30° to left of original

direction; percentage loss of energy = 24.0%

Q17. (a) 35.6 m s–1

(b) 34.3º; 55.7º

(c) 2.59 m

Q18. (a) 7.00 m s–1

(b) (i) 9.81 m s–2

downward; (ii) 19.62 m s–2

upward

(c) 4.20 m s–1

; 2 kg

(d) 2.40 m

Q19. (across, up) = (0.275, 0.331) m, relative to the bottom left corner

Q20. (across, up) = (7.31, 1.88) cm, relative to the bottom-left corner

Q21. (a) TPA = 10 N; TQB = 20 N

(b) 14.0°

Q22. 1.62 m

Q23. (a) topples

(b) no movement

Q24. (a) 1.25105 Pa

(b) 2.38 m


Question 1.

(a) Let a be the acceleration of the 50 kg mass upward or the 80 kg mass downward. Let T be

the tension in the rope.

Resolving upward for the 50 kg mass:

agT 5050

Resolving downward for the 80 kg mass:

aTg 8080

Adding, to eliminate T:

ag 13030

2sm264.281.9130

30

130

30 ga

Answer: 2.26 m s–2

.

(b) Resolving upward for the 50 kg mass:

agT 5050

However, this time, T = 80g. Hence:

agg 505080

2sm886.581.950

30

50

30 ga


.


Question 2.

Let x1 be the distance moved to the right by the 100 kg mass and x2 be the distance moved

down by the 300 kg mass. Let T be the tension in the rope.

Considering the amount of rope passing over the pulley from each side,

21 2xx

Hence,

121

2 xx

There must be the same proportionality between accelerations:

121

2 aa

Resolving to the right for the 100 kg mass:

1100aT (A)

Resolving downward for the 300 kg mass:

23002300 aTg

or

11502300 aTg (B)

Eliminating T by 2(A) + (B):

1350300 ag

Hence,

2

1 sm409.881.9350

300

350

300 ga

Then,

2

121

2 sm204.4 aa

N9.840100 1 aT

Answer: accelerations of 100 kg and 300 kg masses are 8.41 m s–2

, 4.20 m s–2

respectively;

tension = 841 N.


Question 3.

(a) Let x and v be the upward displacement and velocity of the weight. The net pull from the

two sides of the cable on the weight is 2F. Hence,

mgFt

vm 2

d

d

tt

t

vm e630)e315(2

d

d

Integrating from v = 0 when t = 0,

ttvm 0e6)0(

)e1(6 t

mv

Putting t = 2, and noting that weight mg = 30N, so that kg058.381.9/30 m ,

12 sm696.1)e1(962.1 v


.

(b) From part (a),

)e1(6

d

d t

mt

x

Integrating from x = 0 when t = 0:

tttx 0e962.10

)1e(962.1 ttx (*)

This is not invertible analytically. Instead, solve numerically to find t when x = 2 m.

There are many possible methods. Two such are given below.

Method 1: (repeated trial)

Denoting the RHS of (*) by f(t), we find

f(1) = 0.7218, f(2) = 2.228

x increases monotonically with t (as v is always positive) and so a solution to f(t) = 2 must lie

between t = 1 and t = 2. Homing in on a solution by repeated trial gives

t = 1.864 s

Method 2: (iteration)

When x = 2, one way of rearranging (*) is

tt e962.1

21

This can be iterated (use the [ANS] button on your calculator) from, e.g., t = 1, to give

t = 1.000, 1.651, 1.828, 1.859, 1.863, 1.864, 1.864, ...

Answer: 1.86 s.


Question 4.

(a) The tension throughout the cable is F. The mass is subject to force 2F upward (F from

each side of the loop of cable) and weight mg downward.

Taking x, v and a positive upward (though you are quite at liberty to adopt positive

downward if you prefer), “force = mass acceleration” gives (in kg-m-s units):

t

vmmgF

d

d2

)12(81.981.92

62.1922

d

d

t

tg

m

F

t

v

When t is 0, v = 0 and the acceleration is negative, which is actually enough to show that the

initial motion will be downward. However, to find for how long we need to integrate and find

v explicitly.

)12(81.9d

d t

t

v

Integrating from the initial conditions (v = 0 when t = 0) to general (v, t):

0)(81.90 2 ttv

)1(81.9 ttv

From its factors, v goes negative (i.e. downward motion) immediately after t = 0 and stays

negative until t = 1.

Note that downward motion continues until velocity v = 0, not the equilibrium point where

a = 0 or forces are in balance. At the latter time v is actually maximal!

Answer: downward motion until t = 1 s.

(b) In terms of upward displacement x:

)(81.9d

d 2 ttvt

x

Integrating:

)6

32(81.90)

23(81.90

2323 ttttx

)32(635.1 2 ttx

Hence, x = 0 when t = 0 (start time) and t = 3/2 s. At the latter time:

1

21

23 sm358.781.9)1(81.9 ttv

2sm62.19281.9)12(81.9 ta

Answer: returns after 1.5 s; speed = 7.36 m s–1

; acceleration = 19.6 m s–2

.


Question 5.

m = 50 kg

μ = 0.25

a = 2 m s–2

(a)

Resolving perpendicular to the plane:

30cos30sin mgTR

30sin30cos TmgR

Since the block is moving, friction is given by

)30sin30cos(μμ TmgRF

Resolving along the plane: maFmgT 30sin30cos

maTmgmgT )30sin30cos(μ30sin30cos

)30cosμ30sin()30sinμ30(cos ggamT

Hence,

N5.455

30sin25.030cos

30cos81.925.030sin81.9250

30sinμ30cos

30cosμ30sin

ggamT

Answer: 456 N.

(b) As above for R and F. This time, resolving along the plane and noting that there are pulls

of magnitude T (but different directions) from either side of the pulley:

maFmgTT 30sin30cos

maTmgmgTT )30sin30cos(μ30sin30cos

)30cosμ30sin()30sinμ30cos1( ggamT

Hence,

N7.226

30sin25.030cos1

30cos81.925.030sin81.9250

30sinμ30cos1

30cosμ30sin

ggamT

Answer: 227 N.


Question 6.

m = 40 kg

μ = 0.4

T = 100 N

(a)

Resolving perpendicular to the plane:

mgTR 30sin2

N4.29230sin100281.94030sin2 TmgR

The maximum friction is then:

N0.1174.2924.0μmax RF

The motive force along the plane is

N2.17330cos100230cos2 T

This exceeds the maximum friction force. Hence, the block moves, friction is maximal

(F = Fmax) and the acceleration a is given by

maFT 30cos2

2sm405.140

0.1172.17330cos2

m

FTa


.


Question 7.

(a) At the limiting state (minimum coefficient of friction) the friction force F is maximal and

given by F = μR and exactly balances the horizontal hydrostatic pressure force on the front

wall of the dam.

If L is the length of the dam in metres then the mass of the dam is

Lm 5103

Balancing forces vertically:

mgR

Balancing forces horizontally:

average pressure area = friction force

RhLp μ

If h is the depth of water then ghp ρ2

1 (the pressure at the centroid). Hence,

mgLgh μρ 2

21

24.0103

121000ρμ

5

2

212

21

gL

gL

mg

Lgh

Answer: minimum coefficient of friction = 0.24.

(b) In reality, water will seep underneath the dam, producing upthrust and reducing R. The

coefficient of friction would then have to be higher to provide the same friction force. It is

useful to key the dam into the underlying bedrock, both to provide resistance to horizontal

motion and reduce the seepage underneath it.


Question 8.

Initial velocity:

1sm25s3600

m100090

hour

km90

Using:

impulse = change of momentum

mutmg 0)60005sin(

s09.105sin81.918006000

251800

5sin6000

mg

mut

Answer: 10.1 s.


Question 9.

(a) 1000 kg vehicle:

1

1 sm0

5.12

v

1

1 smkg0

12500

vm

750 kg vehicle:

1

2 sm83.10

25.6

60sin5.12

60cos5.12

v

1

2 smkg8123

4688

vm

(These can be rounded for a final answer).

(b) Total momentum before = total momentum after

Hence,

v17508123

4688

0

12500

which gives:

1sm642.4

822.9

v

(This can be rounded for a final answer).

(c) Magnitude of velocity in part (b) is

122 sm86.10642.4822.9 u

The acceleration is

2sm429.31750

6000

m

Fa

Using

asuv 222

m18.17)429.3(2

86.100

2

222

a

uvs

Answer: 17.2 m.


Question 10.

Horizontally, the ball travels toward the wall at 20 m s–1

and travels back from it at

1sm14207.0

Vertically, the ball’s motion is not changed by the wall (it is the component parallel to the

surface), so, since uy = 0 and ay = –g, its vertical displacement from initial position is given

for all t (i.e. before and after hitting the wall) by

2

2

10 gty

Setting y = –1.5 m gives time to first bounce on the ground: t = 0.5530 s.

Time taken to reach wall at 20 m s–1

horizontally is

s3.020/6

The remaining time to the first bounce on the ground is 0.2530 s. The horizontal distance

travelled back at 14 m s–1

is then

m542.32530.014

Velocity and speed when it first hits the ground:

1sm14 xv

1sm425.55530.081.90 gtvy

122 sm01.15 yx vvv

Answer: 3.54 m from the wall; speed 15.0 m s–1

.


Question 11.

(a) Initial velocity:

θcos0vux

θsin0vu y

Displacement (as a function of t) by constant-acceleration formulae:

tvx )θcos( 0

2

21

0 )θsin( gttvy

Substitute θcos0v

xt in the latter displacement formula to get the trajectory:

2

0

21

θcosθtan

v

xgxy

Substituting x = 25 m, y = 10 m, θ = 35°,

2

0

456951.1710

v

v0 = 24.67 m s–1


.

(b) Time to hit the wall,

s237.135cos67.24

25

θcos0

v

xt

The velocity components at this time are:

1

0 sm21.2035cos67.24θcos vuv xx

1

0 sm015.2237.181.935sin67.24θsin gtvgtuv yy

Restitution affects only the velocity component perpendicular to the wall (i.e. the horizontal,

x, component). Hence, the constant-acceleration motion in the second part of the motion has:

1sm13.1221.206.0 xu

1sm015.2 yu

The time for a vertical displacement of y = –10 m (from the height at which it hit the wall to

when it hits the ground) is given by

2

21 gttuy y

2

21 81.9015.210 tt

010015.2905.4 2 tt

s648.1905.42

10905.44015.2015.2 2

t (or a meaningless negative time)

The horizontal position (relative to the wall) is then

m99.19648.113.12 tux x

Answer: 20.0 m back from the wall.


Question 12.

(a) Conservation of x-momentum:

v3430cos310

1sm7641.0 v


.

(b) Use:

Impulse = change of momentum

sN15

34.18

30sin3

30cos3

0

7641.010)(Impulse

uvm

Answer: sN0.15

3.18

.

(c)

J45310KE 2

21 before

J925.97641.034KE 2

21 after

Fraction of energy lost:

7794.045

925.945

KE original

KE of loss

Answer: 0.779.


Question 13.

Take the origin of coordinates at the initial point. Then x = 6 m and y = 2 m at C.

Initial velocity:

θcos0vux

θsin0vu y

Displacement (as function of t) by constant-acceleration formulae:

tvx )θcos( 0

2

21

0 )θsin( gttvy

Substitute θcos0v

xt in the latter displacement formula to get the trajectory:

2

0

21

θcosθtan

v

xgxy

Rearranging:

)θtan(2θcos

0yx

gxv

Set x = 6 m and y = 2 m to get

1

0 sm21.10)270tan6(2

81.9

70cos

6

v

Answer:

10.2 m s–1

.

(b) The time taken is given by

s718.170cos21.10

6

θcos0

v

xt

Then

1

0 sm492.370cos21.10θcos vvx

1

0 sm259.7718.181.970sin21.10θsin gtvvy

Answer: 1sm26.7

49.3

(c) After oblique collision, only the component of velocity perpendicular to the wall is

changed. After collision:

1sm492.3 evx

1sm259.7 yv

Since


y

x

v

vθtan

we have

259.7

492.3

3

1 e

6929.0e

Answer: 0.693.


Question 14.

(a) The speeds of the driver before and after may be determined from energy (Topic C) or

from the constant-acceleration formula asuv 222 .

Taking positive direction downward:

Before (u = 0, a = 9.81 m s–2

, s = 2 m): velocity hitting pile = 6.264 m s–1

.

After (v = 0, a = 9.81 m s–2

, s = –0.1 m): velocity leaving pile = –1.401 m s–1

.

By momentum conservation:

pilev2400401.18000264.6800

1sm555.2 pilev


.

(b) Coefficient of restitution:

6315.0264.6

401.1555.2

speedapproach

speedseparation

e

Answer: 0.631.

(c)

J15700264.6800KE 2

21 before

J8619555.22400401.1800KE 2

212

21 after

Percentage of energy lost

%10.4510015700

861915700100

energyinitial

lostenergy

Answer: 45.1%.


Question 15.

Consider the collision between the (r –1)th

and the rth

sphere. The first of these has velocity

vr–1 before collision and wr–1 after.

Momentum:

rrr mvmwmv 11 0

11 rrr vvw (A)

Restitution:

11 rrr evwv (B)

Adding (A) + (B) to eliminate wr–1:

1)1(2 rr vev

Hence,

12

1

rr v

ev

and so, inductively,

1

1

2

2

12

1

2

1

2

1v

ev

ev

ev

r

rrr

Answer: 1

1

2

1v

ev

r

r


Question 16.

By trigonometry (see the diagram) the line of centres when the balls

collide obliquely is 30° from the original direction of motion. Adopt

new x and y coordinates along the tangent to the surfaces at collision

and along the lines of centres respectively. The second particle must

move off in the direction along the line of centres (v2x = 0).

Note that u1 = 6 m s–1

.

Momentum along tangent surface:

xmvmu 11 30sin

31 xv (A)

Momentum along line of centres:

yy mvmvmu 211 30cos

3321 yy vv (B)

Restitution:

30cos112 uevv yy

38.112 yy vv (C)

Adding (B) + (C) gives:

38.42 2 yv

34.22 yv

(B) then gives

36.033 21 yy vv

In the new coordinate system, sphere 1 has velocity

36.0

31v . The magnitude and

direction are, respectively:

122

1 sm175.3)36.0(3 v

89.7036.0

3tanθ 1 to right of line of centres (or 40.89° to original direction)

In the new coordinate system, sphere 2 has velocity

34.2

02v , with magnitude

1

2 sm157.434.2 v

and is directed along the line of centres.

Percentage of energy lost:

y

x

r

r

r

u1

30o


%00.24

1006

157.4175.31

1001

100)(

100energyinitial

lostenergy

2

22

2

1

2

2

2

1

2

121

2

2212

1212

121

u

vv

mu

mvmvmu

Answer: Sphere 1: 3.17 m s–1

at 40.9° to the right of the original direction; sphere 2:

4.16 m s–1

at 30° to left of original direction; percentage loss of energy = 24.0%


Question 17.

(a) Impulse = change in momentum

)0(045.06.1 V

V = 35.56 m s–1


.

(b) Horizontal and vertical displacements from constant-acceleration formulae:

tVx )θcos(

)θsin()θsin(212

21 gtVtgttVy

Range (y = 0) corresponds to time

g

Vt

θsin2

and distance

g

V

g

V

g

VVx

θsinθcosθsin2θsin2)θcos(

22

θsin81.9

56.35120

2

9310.0θsin

2θ = 68.59º or 111.41º

θ = 34.30º or 55.71º

Answer: 34.3º or 55.7º.

(c) Time to first bounce:

s085.430.34cos56.35

120

θcos

V

xt

Velocity components just before first bounce:

1sm38.29θcos Vvx

1sm03.20θsin gtVvy

Actually, the latter is obvious. By energy or symmetry the downward velocity on return is

equal to initial upward velocity: 1sm04.20θsin V (last digit different due to rounding).

Velocity components after first bounce (oblique collision with e = 0.6):

1sm38.29 xu (parallel component unchanged)

1sm02.1203.206.0 yu (perpendicular component subject to restitution)

Further time to hit the wall:

s212.265

xu

t

m588.22

21 gttuy y

Answer: 2.59 m.


(d)

Horizontal component: constant (no acceleration,

and unaffected by the bounce)

Vertical component: constant downward slope

(acceleration dvy/dt = –9.81 m s–2

); reverses sign

and magnitude changes by factor e = 0.6 at the

bounce on the ground.

v (m/s)

t (s)

x

v (m/s)

t (s)

y

bounce

29.38

20.03

-20.03

12.02

0

04.085 6.297

hits wall


Question 18.

(a)

Gain in KE = loss in PE

mgLmv 2

21

gLv 22

1s m004.75.281.922 gLv


.

(b) Particle A is moving in a circle, so:

radial acceleration is v2/r and can be found directly from the speed v;

tangential acceleration is dv/dt and can be found as tangential force/mass.

(i) Initially, the speed v is zero, so acceleration is purely tangential:

2sm81.9 gm

mga (tangential; i.e. downward here)

(ii) At the lowest point there is no tangential force, so no tangential acceleration. The

acceleration is purely centripetal (radially inward) of magnitude:

22

sm62.19)(a)partfrom(2 gL

va (centripetal; i.e. upward here)

Answer: (i) 9.81 m s–2

downward; (ii) 19.62 m s–2

upward.

(c)

Just before collision, particle A has speed uA = 7.004 m s–1

and particle B is at rest.

Restitution:

speed of separation = e speed of approach

0(0 AB uev )

1sm202.4004.76.0 AB euv

Conservation of momentum:

BBAA vmum 00

kg26.0

2.112.1

ev

umm

B

AAB

Answer: speed = 4.20 m s–1

; mass = 2 kg

(d) Particle B now undergoes two-component motion under gravity. Since all the subsequent

motion is downward take x horizontal forwards and y positive downward.

1sm202.4 xu ; 0xa

0yu ; 2sm81.9 gay


Find the time of impact:

2

212

21 0 gttatuy yy

s5711.081.9

6.122

g

yt

Horizontal distance travelled:

m400.205711.0202.42

21 tatux xx

Answer: horizontal distance travelled = 2.40 m.


Question 19.

For a uniform plane lamina the centre of mass coincides with the centre of area.

Measure coordinates x, y horizontally and vertically from (e.g.) the bottom-left corner. The

frame can be obtained by subtracting rectangle 2 from rectangle 1 where the areas and

centroids of the original rectangles are as follows. So that they can be used in a summation

formula, subtracted areas are marked negative.

Rectangle 1 (outer rectangle):

a1 = 0.5 × 0.7 = 0.35 m2

x1 = 0.25 m

y1 = 0.35 m

Rectangle 2 (inner rectangle):

a2 = –0.2 × 0.35 = –0.07 m2

x2 = 0.15 m

y2 = 0.425 m

Then,

2m28.007.035.0 iaA

m275.028.0

15.007.025.035.0

A

xax

ii

m33125.028.0

425.007.035.035.0

A

yay

ii

Answer: (across, up) = (0.275, 0.331) m, relative to the bottom left corner.


Question 20.


Divide the shape into three parts (triangle / rectangle / triangle) whose areas and individual

centres of mass / centres of area are:

2

1 cm5.12552

1a

cm333.353

21 x

cm667.153

11 y

2

2 cm70.11)30tan

5516(5 a

cm170.6)30tan

5516(

2

152 x

cm5.252

12 y

2

3 cm65.2130tan

55

2

1a

cm23.1030tan

5

3

2163 x

cm667.153

13 y

Then:

2cm85.4565.2170.115.12 iaA

cm314.785.45

23.1065.21170.670.11333.35.12

A

xax

ii

cm880.185.45

667.165.215.270.11667.15.12

A

yay

ii

Answer: (across, up) = (7.31, 1.88) cm, relative to the bottom-left corner.


Question 21.

(a) The centroid of a triangle lies at 1/3rd

of the altitude from any side; here it is 50 mm from

AB and 100 mm from BC. The weight of 30 N acts through this point.

Taking moments about B:

1.0303.0 PAT

N10PAT

Taking moments about A:

2.0303.0 QBT

N20QBT

Answer: TPA = 10 N; TQB = 20 N.

(b)

25.0200

50θtan

04.14θ

Answer: 14.0°.


Question 22.


Take coordinates x and y from the corner at which the lamina is suspended, with y along the

long side.

By symmetry, m35.0x .

The shape can be formed by subtraction of a circle from a rectangle. Working in metres

throughout we have the following. So that they can be used in a summation formula,

subtracted areas are marked negative.

Part 1 (rectangle):

La 7.01

Ly21

1

Part 2 (circle):

2827.03.0π 2

2 a

35.02 y

Then

2827.07.0

09895.035.0

2827.07.0

35.02827.07.0 2

21

L

L

L

LL

a

yay

i

ii

But, from the angle of suspension,

3640.020tan y

x

whence

9615.03640.0

35.0

3640.0

xy

Hence,

9615.02827.07.0

09895.035.0 2

L

L

2718.06731.009895.035.0 2 LL

01729.06731.035.0 2 LL

Solving a quadratic equation:

305.0or618.135.02

1729.035.046731.06731.0 2

L

Since L is the longer side, here it takes the value 1.618 m.

Answer: 1.62 m.


Question 23.

One must consider, in this order:

toppling, if the vertical through the centre of mass lies outside the base;

sliding, if the downslope component of weight is greater than maximum friction.

(a) The centre of mass is at height m2.0m4.021 . The distance from the cylinder axis at

which this acts on the slope is

m0536.015tan2.0 x

This is greater than the base radius 0.05 m, so that the line of action of the weight lies outside

the base of the cylinder. The weight then produces an unbalanced moment about the bottom

of the cylinder, so the cylinder will topple over.

Answer: topples.

(b) The centre of mass is at height m175.0m35.021 . The distance from the cylinder axis

at which this acts on the slope is

m0469.015tan175.0 x

This lies within the confines of the base of the cylinder and hence the cylinder does not

topple.

The maximum friction force is

15cosμμmax mgRF

The component of weight downslope is

15sinmgFll

The ratio of these two along-slope forces is

12.115tan

3.0

15sin

15cosμmax

mg

mg

F

F

ll

so that friction is sufficient to prevent motion.

Answer: no movement.


Question 24.

(a)

)(2 columnoneofareabase

gvolumedensity

areabase

weightpressureAverage

W

Ag

WD

gADp

2

ρ

2

)(ρ

(*)

where ρ is density, A is frontal area, g is gravity, W is width of one column. Note that the

unknown depth D cancels, because both the weight and the base area are proportional to it.

Divide the front up into two equal-area rectangles (1 and 2), outer semicircle (3) and inner

semicircle (4), where the last is to be subtracted.

Working in metres throughout we have the following. Subtracted areas are marked negative.

4.238.021 aa 5.121 yy

021.46.1π 2

21

3 a 679.36.1π3

433 y

005.18.0π 2

21

4 a 340.38.0π3

434 y

Total area (in m2):

816.7005.1021.44.22 iaA

From (*), the average pressure exerted on the foundations is

Pa10246.18.02

81.9816.72600

2

ρ 5

W

Agp

Answer: average pressure = 1.25105 Pa.

(b) Height of the centre of mass:

384.2816.7

340.3005.1679.3021.45.14.22

A

yay

ii

Answer: height of centre of mass = 2.38 m.

TOPIC B: MOMENTUM ANSWERS SPRING 2018personalpages.manchester.ac.uk/staff/david.d... · Mechanics...

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Transcript of TOPIC B: MOMENTUM ANSWERS SPRING 2018personalpages.manchester.ac.uk/staff/david.d... · Mechanics...