M2 Exam Questions From Board Works

8/7/2019 M2 Exam Questions From Board Works

1/25

A ball is thrown from a point O with a velocity of 10 ms1 at

an angle ofU to the horizontal.

At time tseconds after projection the ball is at the point (x, y).

a) i) Find expressions in terms ofU and tforx andy.

ii) By eliminating t, find the equation of the trajectory of theball.

In the case where U = 45, find:

b) i) the maximum height of the ball above the level ofprojection.

ii) the time of the flight.


2/25

a) i) Using s = ut+ at2 o with s = y, u = 10 sinand a = ggives,

1

y = 10tsin 4.9t2

Horizontal distance travelled

= horizontal component of velocity time

= 10 cosU t

@ x= 10tcosU t=x

10 cosU2

a) ii) Substituting into : 1

2

1

2

2=10 sin 4.9

10cos 100cos

x xy U

U U

2

2

0.049t

c s

xy x U

U

!


3/25

b) i) Using v2 = u2 + 2as o with u = 10 sin45, v = 0 and a = ggives,

2gs = 50

02 = (10 sin45)2 2gs

s = 2.55 (to 3 s.f.)

Therefore the maximum height above the level of projection reached by the ball is 2.55m.

b) ii) Using s = ut+ at2 o with s = 0, u = 10 sin45, and a = ggives,

12

0 = 10tcos45 4.9t2

t(4.9t 10cos45) = 0

@ t= 0 or t= 1.44 (to 3 s.f.)

Therefore the time of flight is 1.44 seconds.


4/25

a) Find the velocity of the particle at time t.

b) Find the acceleration of the particle at time t= 3.

Question 1: A particle has position vector at time tseconds given by

r= (2t2

3t)i + (t3

t2

+ 4)j


5/25

a) Find the velocity of the particle at time t.

b) Find the acceleration of the particle at time t= 3.

Question 1: A particle has position vector at time tseconds given by

r= (2t2

3t)i + (t3

t2

+ 4)j

a) Differentiate to find v:

v = (4t 3)i + (3t2 2t)j

d

dt

rv =


6/25


7/25

Question 2: A particle is moving with acceleration a = 3ti 2j at time tseconds.

Given that at time t= 0 the particle is at 4i +2j and has velocity i 4j, find:

a) the velocity of the particle at time t.

b) the position vector of the particle at time t.

c) the distance of the particle from the origin when t= 3.


8/25


Given that at time t= 0 the particle is at 4i +2j and has velocity i 4j, find:




2 t dt! v i ja) Integrate to find v:

23

22

tt c d! v i j i j


9/25

23

1 2 42

tt

!

v i j

b) Integrate to find r:

3

2 42

tt t t c d

!

r i j i j

When t= 0: r = 4i +2j

3

24 4 22

tt t t

!

r i j

@

@ c = 4 and d= 2

@

When t= 0: v = i 4j

@ c = 1 and d= 40 + 0 + ci + dj = i 4j

231 2 42

tt dt

! r i j

4i +2j = ci +dj


10/25

c) When t= 3:

@ The distance from the origin = (20.52 + 192)

= 28.0 m (3 s.f.)


Given that at time t= 0 the particle is at 4i +2j and has velocity i 4j, find




27 3 4 9 12 22 i j = 20.5i 19jr =


11/25

A uniform plane lamina is in the shape of a square next to a rectangle. The length of a side

of the square isx cm and the rectangle is cm by 2x cm.

Find the distance of the centre of mass of this lamina from A in terms ofx.

A

x

2xx

2x

2x


12/25

The centre of mass of both the square and the rectangle are found by symmetry.

Area of square =x2

2

2 2

22

2

x x x x

x x

v v

!

2 2

2

2 4

2

x xx x

y x

v v

!

2 25 3

Distance4 8

x x @ !

Area of rectangle =x2

5

4

x

!

3

8

x

!

2 225 9

16 64

x x!

109

8x!

2x

y

x

x2x2mass

2x

2

x4x


13/25

A uniform rectangular laminaABCD has a circular hole cut out of it. The rectangle has

length 80 cm and width 40 cm and the radius of the circle is 10 cm.

The centre of the circle is 20 cm from bothAB andAD.

Find the centre of mass of this lamina.

40

80

2

0

A

B C

D


14/25

To find the centre of mass of this lamina consider the circle, the rectangle and the given

lamina separately.

Area of rectangle = 3200 cm2

The centre of mass of the rectangle is 40 cm from AB and 20 cm fromAD.

Area of circle = 100T cm2

Area of lamina = 3200 100T cm2

The centre of mass of the circle is 20 cm from bothAB andAD.

y

x

32003200100T100Tmassrectanglelaminacircle

2020

4020 x

y


15/25

Forming an equation for :

x

100 20 3200 100 3200 40xT Tv ! v

3200 40 100 20=

3200 100x

T

T

v v@

100 20 3200 100 3200 20yT Tv ! v

3200 20 100 20

3200 100

yT

T

v v@ !

y

Therefore the centre of mass of the lamina is 42.2 cm fromAB and 20 cm fromAD.

Forming an equation for :

= 42.177 (5 s.f.)

= 20


16/25

Examination-style question 1: A car of mass 1000 kg moves along a straight,

horizontal road. The car engine is working at a constant rate of 45 kW and thetotal resistance to the motion of the car is 500 N.

a) Find the acceleration of the car when its speed is 15 ms1.

The car comes to a hill which is inclined at an angle ofE to the horizontal, where

sinE = 0.1. The resistance to the motion of the car is unchanged.

b) Find the maximum speed of the car up the hill when it is working at a rate

of 45 kW.


17/25

Power = Force Velocity

Applying Newtons Second Law, F= ma

3000 500 = 1000a

1000a = 2500

@ a = 2.5 Therefore the acceleration of the carwhen it is travelling at 15 ms1 is 2.5 ms2.

500 N D1000

aa) To calculate the acceleration it is first

necessary to calculate the drivingforce.

Po er orce =

Velocity

So,

45 000= = 3000

15D


18/25

R

D

1000g500 N

E

Applying Newtons Second Law up the slope,

D 500 1000gsinE = 0D = 500 + 1000g 0.1 = 1480

The driving force of the car is 1480 N.

Power = Force Velocity @

b) Before calculating the maximum

speed of the car it is first necessaryto calculate the driving force.

The maximum speed of the car is 30.4 ms1.

Powerelocity = Force

. ( )45 000

30 4 3 s.f.1480

v ! !


19/25

Examination-style question 2: A stone of mass 0.5 kg is sliding down a rough

plane inclined at an angle of 15

to the horizontal.The stone passes a point A with a speed of 10 ms1 and a point B with a speed of

6 ms1.

a) Find the loss of mechanical energy of the stone as it moves from point A to

point B.

b) Calculate the coefficient of friction between the stone and the plane.


20/25

a) Loss of energy of the stone = loss of K.E. + loss of G.P.E.

Initial K.E. = 0.5 102

Final K.E. = 0.5 62

Loss of K.E. = 25 9 = 16 J

15

10h

Loss of G.P.E. = 0.5 9.8 10sin15 = 12.7 (3 s.f.)

16 + 12.7 = 28.7

@ h = 10sin15

Therefore the loss of energy of the stone is 28.7 J.

sin15 10

h!


21/25

R = 0.5gcos15 = 4.73 (3 s.f.)

b) To calculate the coefficient of friction, it is

necessary to calculate both the normalcontact force and the frictional force.

@ the coefficient of friction is 0.61.

Loss of energy of the stone found in part a) is equal to the work doneagainst friction.

The work done against friction over a distance of 10 m is 28.7 J.

Work done = force distance

@force = 28.7 10 = 2.87

R F

0.5g15

.. ( )

.

2 87 0 61 2 s.f.

4 73

F ! ! !


22/25


23/25

a) Work done = force distance

Since the force acts at an angle of 30 to the

slope, the component of force acting along the

slope must be calculated.

300 N

30c

@c = 300cos30o = 260 (3 s.f.)

Work done = 260 10 = 2600

So, work done by the force is 2600 J.

c s30=300

c


24/25

10

20

h

b) so, h = 10sin20

To find gain in K.E. it is necessary to

find the speed of the parcel after it

has travelled 10 m.

Gain in K.E. = 30 2 = 30

sin20 =

10

h

Gain in G.P.E. = 30 9.8 10sin20o

= 1006 (4 s.f.)

v2 = u2 +2as

= 0 +2 0.1 10 = 2

v = 1.41 ms1 (3 s.f.)


25/25

The difference between the work done by the force acting up the slope and the gain in

energy of the system is2600 1036 = 1564 J.

This is the work done by the resistance force.

Work done = force distance

@ R = 1564 10 = 156.4 N

Therefore the increase in energy of the system is 1036 J (4 s.f.).

1006 + 30 = 1036

M2 Exam Questions From Board Works

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