Question 1 (Projectile motion)

You throw a ball such that, at the top of its path, it

just clears the 3 m high net at B. The angle at which

the ball is thrown θA = 50º.

a. Determine the initial speed of the ball vA.

b. Determine the distance from the net d that

you need to stand.

c. Determine the speed at which the ball hits

the ground at C.

d. State all assumptions you make.

h m

θA

vA

1 m C

A

d m

B

Question 2 (Rectilinear motion)

Stone A is thrown vertically with velocity 20 m/s, and stone B – with velocity 25 m/s 1

s later.

a. Determine distance between stones in 2 s after second stone is thrown.

b. Determine maximum height and time traveled before hitting the ground for each

stone.

c. State all assumptions you make

Question 3 (Relative motion)

If the end of the cable at A is pulled down with

acceleration a = 2t m/s2

1. Determine the acceleration at which block B

rises (as a function of time)

2. Find velocity of B in 2 seconds

Question 4 (Conservation of energy, force-acceleration)

A man has a mass of 100[kg] and from rest swings

from the cliff by rigidly holding on the tree vine,

which is 10[m] measured from supporting limb A

to his centre of mass. Determine his velocity just

after the vine strikes to the lower limb at B. Also,

with what force must he hold on the vine just

before and just after the vine contacts the limb at

B?

Question 5 (work-energy, impulse-momentum or force acceleration)

The masses A 40 kg and B 60 kg. The collar A

slides on smooth horizontal bar. The system is

released from rest.

1. Determine the velocity of collar A when it

has moved 0.5 m to the right.

2. Determine velocity of A when it has moved

for 1 s.

3. State all assumptions you made.

Question 6 (Cons of energy, impact, cons of energy)

The 3 kg block A is released from rest at 60

degrees position shown and subsequently strikes

the 1 kg cart B. If the coefficient of restitution for

collision is e=0.7, determine the maximum

displacement s of cart B beyond point C. Neglect

friction.

Question 1 (Projectile motion)

You throw a ball such that, at the top of its path, it

just clears the 3 m high net at B. The angle at which

the ball is thrown θA = 50º.

• Determine the initial speed of the ball vA.

• Determine the distance from the net d that

you need to stand.

• Determine the speed at which the ball hits

the ground at C.

• State all assumptions you make.

h m

θA

vA

1 m C

A

d m

B

Solution:

Initial speed of the ball

Analyse in y direction from A to B

2

002

1tatvss c++=

At max height there is zero component of velocity in y direction

( ) ( ) 281.92

150sin13 ttvA −++= (1)

tavv c+= 0

( )tvA 81.950sin0 −+= (2)

Solving simultaneously

smvA /18.8= and st 639.0=

Distance to the net

Time to net st 639.0=

Analyse in x direction from A to B

tvss 00 +=

( )( )639.050cos18.80 +=d md 36.3=

Velocity at point C

Analyse from A to C

( )0

2

0

2 2 yyavv c −+=

( ) ( )( )1081.9250sin18.822

−−+=v ( ) smvyC /67.7=

Total velocity

( ) ( ) ( ) ( ) smvvvyCxCc /29.967.750cos18.8

2222=+=+=

Assumptions:

Motion of particle

No air-resistance

Motion with constant acceleration

Question 2 (Rectilinear motion)

Stone A is thrown vertically with velocity 20 m/s, and stone B – with velocity 25 m/s 1

s later.

• Determine distance between stones in 2 s after second stone is thrown.

• Determine maximum height and time traveled before hitting the ground for

each stone.

• State all assumptions you make

Solution:

Distance

Stone A

?

/81.9

/20

3

2

0

=

−=

=

=

s

sma

smv

st A

( )( ) ( )( )

ms

s

tatvss

A

A

c

81.15

381.92

13200

2

1

2

2

00

=

−++=

++=

Stone B

?

/81.9

/25

2

2

0

=

−=

=

=

s

sma

smv

stB

( )( ) ( )( )

ms

s

tatvss

B

B

c

4.30

281.92

12250

2

1

2

2

00

=

−++=

++=

Difference m6.1481.154.30 =−

Max Height:

Stone A

?

/81.9

0

/20

2

0

=

−=

=

=

s

sma

v

smv

( )

( ) ( )( )

ms

s

yyavv

A

A

c

4.20

081.92200

2

2

0

2

0

2

=

−−+=

−+=

Time:

( )

st

t

atvv

08.4

81.92020

0

=

−+=−

+=

Stone B

?

/81.9

0

/25

2

0

=

−=

=

=

s

sma

v

smv

( )

( ) ( )( )

ms

s

yyavv

A

B

c

9.31

081.92250

2

2

0

2

0

2

=

−−+=

−+=

Time:

( )

st

t

atvv

1.5

81.92525

0

=

−+=−

+=

Assumptions:

Motion of particle

No air-resistance

Motion with constant acceleration

Question 3 (Relative motion)

If the end of the cable at A is pulled down with

acceleration a = 2t m/s2

• Determine the acceleration at which block B

rises (as a function of time)

• Find velocity of B in 2 seconds

Solution:

Position equation

( )CBB

CA

sssL

ssL

−+=

+=

2

1 2

Differentiate once to get velocities:

( )CBB

CA

vvv

vv

−+=

+=

0

20

Differentiate twice to get accelerations:

( )CBB

CA

aaa

aa

−+=

+=

0

20

BA aa 4−=

taA 2= m/s2 taB

2

1−= m/s

2

Now have to integrate to get velocity of B in 2 seconds:

taB2

1−= m/s

2

tdt

dvaB

2

1−==

dtt

tdv ∫∫

−=

−=

4

2tvB smvB /1−=

Question 4 (Conservation of energy, force-acceleration)

A man has a mass of 100[kg] and from rest swings

from the cliff by rigidly holding on the tree vine,

which is 10[m] measured from supporting limb A

to his centre of mass. Determine his velocity just

after the vine strikes to the lower limb at B. Also,

with what force must he hold on the vine just

before and just after the vine contacts the limb at

B?

Solution:

Find velocity in vertical position – just before hitting B

Conservation of energy-

( )( )( ) ( )

smv

v

mghmvmghmv

VTVT

/67.7

01002

145cos101081.91000

2

1

2

1

2

2

2

2

2

21

2

1

2211

=

+=−+

+=+

+=+

Force just before hitting limb B

smv /67.72 =

( )( ) ( )( )

NT

T

r

vmmgT

maF nn

1569

10

67.710081.9100

2

2

=

=−

=−

=∑

Force just after hitting limb B

smv /67.72 =

( )( ) ( )( )

NT

T

r

vmmgT

maF nn

2942

3

67.710081.9100

2

2

=

=−

=−

=∑

Question 5 (work-energy, impulse-momentum or force acceleration)

The masses A 40 kg and B 60 kg. The collar A

slides on smooth horizontal bar. The system is

released from rest.

• Determine the velocity of collar A when

it has moved 0.5 m to the right.

• Determine velocity of A when it has

moved for 1 s.

• State all assumptions you made.

Solution:

Determine the velocity of collar A when it has moved 0.5 m to the right

Work-energy

( ) ( )

( )( )( ) ( )

smv

v

ghmvmmghmvmm

VTVT

BBABBA

/43.2

060402

15.081.9600

2

1

2

1

2

2

2

2

2

21

2

1

2211

=

++=+

++=++

+=+

Determine velocity of A when it has moved for 1 s.

Impulse-momentum

( ) ( ) ( )

( )( ) ( )

smv

vdt

vmmdtgmvmm BABBA

/89.5

604081.9600

2

2

1

0

2

1

0

1

=

+=+

+=++

∫

∫

Assumptions

No friction

Motion with constant acceleration

Constant tension

Question 6 (Cons of energy, impact, cons of energy)

The 3 kg block A is released from rest at 60

degrees position shown and subsequently strikes

the 1 kg cart B. If the coefficient of restitution for

collision is e=0.7, determine the maximum

displacement s of cart B beyond point C. Neglect

friction.

Solution:

Velocity of A just before impact

Conservation of energy

( ) ( )

( )( )( ) ( )

smv

v

ghmvmghmvm

VTVT

AAAA

/2.4

032

160cos8.18.181.930

2

1

2

1

2

2

2

2

2

21

2

1

2211

=

+=−+

+=+

+=+

Impact

Conservation of linear momentum

2211 BBAABBAA vmvmvmvm +=+

( )( ) ( ) ( ) 22 1302.43 BA vv +=+ (1)

Coefficient of restitution

21

22

BA

AB

vv

vve

−

−=

02.47.0 22

−

−= AB vv

(2)

Solving (1) and (2) simultaneously smvA /42.22 = smvB /36.52 =

Velocity of block B at point C

Conservation of energy

( ) ( )

( )( ) ( )( )( )

smv

ghmvmghmvm

VTVT

BBBB

/74.4

30cos4.24.281.910036.512

1

2

1

2

1

2

2

2

2

21

2

1

2211

=

−+=+

+=+

+=+

Max displacement of block B from point C

Conservation of energy

( ) ( )

( )( ) ( )( )( )

mh

s

h

h

ghmvmghmvm

VTVT

BBBB

3.230sin

15.1

30sin

15.1

81.910074.412

1

2

1

2

1

2

2

2

21

2

1

2211

===

=

+=+

+=+

+=+