# Problem sets for MACE-11010 Engineering Mechanics

Please note that according to the New Grading System, no marks can be gained from answering the following questions. (1 October 2007)

Q1: vector operations
Q2: kinematics
Q3: What will happen if the Newton’s law is not applied to a particle, but to a body? In the formula F=ma, what is F, what is a?
Q4: Show a case that needs the concept of dot production of two vectors.
Q5: Show a case that needs the concept of cross production of two vectors.

Problems set
R.C. Hibbeler (Engineering Mechanics, 10th Edition).

AttachmentSize Q1.pdf13.66 KB Q2.pdf12.81 KB

### HI

I'm Karl Conlon on Aerospace Engineering

give me the points haha

### Great! Thanks for the support

Great! Thanks for supporting my teaching.

why you choose Aerospace Engineering?

1 point

### Q1

A + B = (1-1,1+1,0+0) = (0,2,0)

A - B = (1+1,1-1,0-0) = (2,0,0)

2A + 3B = (2-3,2+3,0+0) = (-1,5,0)

A.B = (1x-1,1x1,0x0) = (-1,1,0)

i think lol

### 1st

Thanks, but not the first. One more question need to be answered. Only 3 point for you treated as a regular comment.

### wheres the last question

they're all right except you missed the last question off. Which is annoying because its the only one i cant remember how to do. just going to check on the net now.

### dot product

the dot product one is wrong.

A.B = (1x-1)+(1x1)+(0x0)=0

### good job!

good job! 3 points.

### yup

im doin aerospace engineering. all of them are right except the last one which is meant to be 0 i think. its funny as i was showing my brother how to do these the other day

A + B = (1-1,1+1,0+0) = (0,2,0)

A - B = (1+1,1-1,0-0) = (2,0,0)

2A + 3B = (2-3,2+3,0+0) = (-1,5,0)

A.B = (1x-1,1x1,0x0) = 0

A*B = (0,0.2)

I GOT SANE ANSWER FOR NO 2. i couldnt do the 3rd question could u pls show me n do i get ny points?

### hi

hi i am Rameez.M...doin mech eng

### Q2

v^2=u^2+2as a=9.81/6=1.635  v^2=2^2+(2x1.635x5)=18.35   v=4.51m/s

are you sure?

### Hi

Hi im Alex Proctor n im also doin aerospace engineering

### points

how do you keep track of your points? are they automatically given to you or do you have to keep note yourself? if you track them yourself how do you make sure people are bein honest

### You keep the record of yourself

You keep the record of your self. Each week you report the total points collected from the last Thursday 1pm to this Thursday 1pm.

### Introduction

Hi I'm Sam Ratcliffe and I'm also doing Aerospace

### This should better be put in

This should better be put in http://imechanica.org/node/1972

### The cross product

The cross product one:

AxB = (0,0,2)

true or false?

### hey

hey i'm Henry Stewart. i'm doing mech eng . heres my answers for the questions

1) A+B = 0,2,0

2) A-B = 2,0,0

3) 2A + 3B = -1,5,0

4) A.B = 0,0,0

5) AxB = -1,1,0

Q2) i got the same as kean for this one and can't see where he went wrong

V^2 = U^2 + 2AS

V = SQR(2^2 + 2x(9.81/6)x5)

V = SQR(20.35)

V = 4.51m/s

Does that make me the first to post all the answers (even if they are wrong)

### there is at least one error

there is at least one error

Question 1

2-  0.2.0

3-  2.0.0

4-  -1.5.0

5-  0

6-  0-0.0-0.1-(-1) = 0.0.2

Question 2:

5=1/2*1/6*9.81t^2+2t

t=1.54  -->  V=9.81/6*1.54+2   ---> V=4.52

Question 1 - Vectors

A+B=(0,2,0)

A-B=(2,0,0)

2A+3B=(-1,5,0)

A.B=0

AxB=(0,0,2) (Assuming Cross Product)

Question 2 - Kinematics

Initial Velocity = 2 m/s, Acceleration = 9.81/6 m/s^2, Distance = 5m, Final Velocity = V

V^2=(Initial Velocity)^2 + 2(Acceleration x Distance)

V^2= 4 + 2((9.81/6) x 5)

V^2= 20.35

V = 4.51 m/s

10 Points Henry??

### Question 1 2-

Question 1

2-  0.2.0

3-  2.0.0

4-  -1.5.0

5-  0

6-  0-0.0-0.1-(-1) = 0.0.2

Question 2:

5=1/2*1/6*9.81t^2+2t

t=1.54  -->  V=9.81/6*1.54+2   ---> V=4.52

A+B=(1,1,0)+(-1,1,0)=(0,2,0)

A-B=(1,1,0)-(-1,1,0)=(2,0,0)

2A+3B=2(1,1,0)+3(-1,1,0)=(-1,5,0)

A.B= (1,1,o).(-1,1,0)=0

Cross product of AB=(0,0,2)

And i don't support your system of evaluating students as; this is not the proper way. I hope University does not allow you to play havoc with its students.

### Just like everyone else...

Q1:

A+B = (1 - 1, 1 + 1, 0 + 0) = (0,2,0)

A-B = (1 + 1, 1 - 1, 0 - 0) = (2,0,0)

2A + 3B = (2 - 3, 2 + 3, 0 + 0) = (-1,5,0)

A.B = 1.-1 + 1.1 + 0.0 = 0

AxB = (1.0 - 0.1) - (1.0 - 0.-1) + (1.1 - 1.-1) = (0,0,2)

Q2:

v^2 = u^2 + 2as

u = 2m/s

a = 9.81/6m/s/s

s = 5m

Therefore:

v^2 = 4 +  16.35 = 20.35

v = 4.51m/s

But everyone has already posted this so...?

### Hi

I'm doing aersopace eng

### I know lots of people have

I know lots of people have already posted their answers but I may aswell also post them.

Q1

A + B

(1) + (-1) = 0
(1) + (1) = 2
(0) + (0) = 0

(0,2,0)

A-B

(1) - (-1) = 2
(1) - (1) = 0
(0) - (0) = 0

(2,0,0)

2A+3B

2(1) + 3(-1) = -1
2(1) + 3(1) = 5
2(0) + 3(0) = 0

(-1,5,0)

A.B

(1)*(-1) = -1
(1)*(1) = 1
(0)*(0) = 0

1-1=0

A*B

(1)*(-1) = -1
(1)*(1) = 1
(0)*(0) = 0

(-1,1,0)

Q2

U= 2 ms^-1
S= 5 m
a= 9.81/6 ms^-2

Finding V.

Use

v^2=u^2+2as
v^2= 4+2*1.635*5
v= 4.51 ms^-1

hi there... I know other ppl have posted their answers, but that shouldn't be a reason for me not to do it, rite? so, below are my working and answers.

Q1:

A = (1,1,0)
B = (-1,1,0)

A + B = (0,2,0)

A - B = (2,0,0)

2A + 3B = (2,2,0) + (-3,3,0) = (-1,5,0)

A . B = (1 x -1) + (1 x 1) + (0 x 0) = -1 + 1 + 0 = 0

A x B = [(1x0)-(1x0)]i - [(1x0)-(-1x0)]j + [(1x1)-(-1x1)]k
= (0,0,2)

Q2:

u = 2 m/s ; s = 5m ; a = (9.81/6) m/s^2 ; v = ?

v^2 = u^2 + 2as
v^2 = (2)^2 + 2(9.81/6)(5)
v^2 = 20.35

v = 4.511 m/s

Hi my name is Farhan bin Zainol, currently in Mechanical Engineering (BEng) course.

Here are my answers to the questions.

Q1:

Since A = (1, 1, 0) and B = (-1, 1, 0)

--> A + B = (0, 2, 0)

--> A - B = (2, 0, 0)

--> 2A + 3B = (-1, 5, 0)

--> A.B = (1*-1) + (1*1) + (0*0)

= 0

--> A x B is a cross-product. Hence,

A x B = [(1*0) - (1*0)] i - [(1*0) - (1*0)] j + [(1*1) - (-1*1)] k

= 2 k.

Therefore, A x B = (0, 0, 2)

Q2:

Initial velocity = 2 m/s

Height = 5 m

Acceleration = (9.81/6) m/s²

By using v² = u² + 2as,

v² = (2)² + 2 (9.81/6) (5)

= 20.35

v   = 4.51 m/s  (3 significant figures)

at least 1 error

### Q2

I get the same as everyone else for question 2 and dont understand how it could b wrong

v² = u² + 2as,

v² = 2² + 2 (9.81/6x5)

= 20.35

v   = 4.51 m/s

### Questions from Engineering Mechanics, by R.C Hibbeler

Questions from Engineering Mechanics, tenth edition - by R.C Hibbeler.

1. Solve the following equation for x, y, and z:

xy + z = –1   –x + y + z = –1   x + 2y – 2z = 5

x - y + z = -1 -- ( 1 )
-x + y + z = -1 -- ( 2 )
x + 2y - 2z = 5 -- ( 3 )

Eq. (1) + Eq. (2):
2z = -2
hence, z = -1

Eq. (2) + Eq. (3):
3y - z = 4
3y - -1 = 4
3y = 3
y =1

From Eq. (1):
x - y + z = -1
x -1 + -1 = -1
x = 1

2. Solve the following equation for the two roots of x:   x2 — 16 = 0

x² = 16
x = ±√16
x = +4, -4

3.
Solve the following equation for the two roots of x:   — x2 + 5x = — 6

x² - 5x -6 =0
(x-6)(x+1)=0
x = 6, x = -1

4.   Determine the angle :

θ + 40° = 90°
θ = 50°

5.   Determine the angle :

θ + 40° = 90°
θ = 50°

6.  Determine the length of side AB if right angle ABC is similar to right angle A'B'C'

B'C' =  √(15² + 36²)
B'C' = 39

AB/A'B' = BC/B'C'

AB = A'B' x BC / B'C'
=15 x 13/39
= 5

7. Using the basic trigonomic functions, determine the length of side AB of the right triangle.

AB/AC = sin 30°
AB = 10x0.5
= 5

8. Using the basic trigonomic functions, determine the length of side AB of the right triangle.

AB/AC = sin 45°
AB = 10 x (√2) / 2
= 5√2
= 7.071

9. Using the basic trigonomic functions, determine the length of side AB of the right triangle.

AB/BC = tan 45°
AB =  10 x 1
= 10

10. Determine the angles and and the length of side AB of the triangle. Note that there are two
possible answers to this question and we have provided only one of them

6 / sin 40° = 7 / sin Φ
sin Φ = (7 / 6) sin 40°
= 0.7499
Φ = 48.6°

40° + Φ +  θ = 180°
θ = 180° - 40° - 48.6°
= 91.4°

AB / sin 91.4° = 6 / sin 40°
AB = 6 sin 91.4° / sin 40°
= 9.33

### true / false questions - engineering mechanics chapter 1

1. Static's deals with bodies that have constant velocity.
- False, statics means not moving

2. Equilibrium means at rest or moving with constant velocity.
- True

3. A particle has a mass but its size can be neglected.
- True

4. The particles that compose a rigid body remain at a fixed distance from one another both before and after a load is applied.
- True

5. A concentrated force is assumed to act at a point on a body
- True

6. A particle subjected to a system of forces will remain at rest provided the force system is balanced.
- True

7. The weight of a body remains constant regardless of its location.
- False, it depends on the gravity force which depends on location.

8. The SI system considers force to be a derived unit.
- True

9. A newton is equal to the force required to accelerate 1 kg at 9.81 m/s2.
- False, it is the force to accelerate a mass of 1 kg at 1 m/s²

10. The prefix giga ( G ) indicates 109.
- True

11. The prefix mega ( M ) indicates 10-6.
- False, should be 10^6

12. The unit ms is a milli-second.
- True

13. A milli-meter squared is written as mm2.
- True

14. It is satisfactory to write k µ s as an appropriate composite unit.
- True

15. The set of composite units s/kg are allowed.
- True

16. The accuracy of a number is indicated by the number of significant figures it contains.
- True

17. 2.16 rounded off to two significant figures is 2.2.
- true

18. 2.25 rounded off to two significant figures is 2.3.
- True

19. 2.05 rounded off to two significant figures is 2.0.
- False, should be 2.1

20. 2.35 rounded off to two significant figures is 2.4.
- True

### at least 1 error

I found that there is at least 1 error.

### invitation for short lectures

Since so many of you have already known Vectors and some Basic kinematics, is there anyone volunteer to give short lectures tomorrow (10-15 minutes about vector operations and 10-15 minutes about kinematics). No points for that in order to be fair for those who haven't learnt about this, but it is an honor.

Hi! I'm Tee Soon Teck from Mechanical Engineering class (BEng).

Below are my solutions and answers:

Vector

A + B = (i,j) + (-i,j) = 2j or (0,2,0)

A - B = (i,j) - (-i,j) = 2i or (2,0,0)

2A + 3B = 2(i,j) + 3(-i,j) = -i,5j or (-1,5,0)

A . B = (i.j) . (-i,j) = -1+1 = 0

A x B = [(1 x 0) - (1 x 0)]i - [(1 x 0) - (1 x 0)]j + [(1 x 1) - (1 x 1)]k = 2k or (0,0,2)

Kinematics

u = 2 m/s

v = ?

a = (9.81/6) m/s^2

s = 5 m

v^2 = u^2 + 2as

v^2 = 2^2 + 2 (9.81/6) (5)

v^2 = 20.35

v = 4.511 m/s

### Let's break the problem down

Okay, here's the notoriously difficult-to-solve but apparently easy Question number two. The strange thing is, I think all the above posts have the correct answer.

The descent engine stops at h = 5m, which means that it doesn't have any force acting on it, except for the lunar gravity. At that point of time, t, its velocity is u = 2m/s. Just before the landing gear impacts, (presuming that the landing gear is attached to the space vehicle) the final velocity will obviously be given by the formula v^2 = u^2 + 2ah.

And hence, v^2 = 20.35 m^2/s^2 Hence, v = 4.51 m/s

But after it impacts the groud, its velocity will obviously be zero. Isn't that correct?

### what will happen to its velocity after it impacts the groud?

Discussions invited about what will happen to its velocity after it impacts the groud, please.

### decceleration and rebound

Obviously there is no such thing as a instantaneous stop. The craft would instead deccelerate at a very fast rate when it came into contact with the contact force from the moon. At the very first point of contact, when the contact force from the moon has not yet begun to affect the decceleration of the craft, the craft would still have the velocity of 4.51 m/s.
Shortly after it would reach 0 m/s and then depending on whether it bounces up again, it will either stay at 0 m/s or become subject to a (non-zero) velocity in the other direction, back off of the moon...

I'm unsure of the answer you are looking for... what have we missed?

### Irrelevant

The question states: "compute the impact velocity", therefore it is irrelevant to the answer what happens to the ship. The answer only ask for the speed of the rocket at the time it hits the moon. But in real life, some of the kinetic energy of the landing will be lost throught vibrations on the surface of the moon and the rocket's landing gear. Some energy will be transfered to heat, some to other particles on the moon. However, the rocket is most likely going to bounce slightly, as not all the energy can be transfered, as the rocket nor the moon are completely elastic.

### Is there a point in me posting this?

Hey, I'm Dave and I'm studying Aerospace Engineering. I am HUGELY sceptical of this assement system and honestly think it will never work! I've posted on another page (http://www.imechanica.org/node/1931#new) explaining all this - please go on and tell me if you agree (that's to all students as well as Mr Tan). I'm really not looking forward to a year of me being marked and assesed in this way!
Here are my answers, I did do them myself but I could have just as well copied from someone else - this system makes NO SENSE!

Q1

A + B = (1 + -1), (1 + 1), (0 + 0) = (0,2,0)

A - B = (1 - -1), (1 - 1), (0 - 0) = (2,0,0)

2A + 3B = 2(1,1,0) + 3(-1,1,0) = (2,2,0) + (-3,3,0) = (-1,5,0)

A.B = (1 * -1) + (1 * 1) + (0 * 0) = -1 + 1 + 0 = 0

AxB = i[(1 * 0) - (0 * 1)] + j[(1 * 0) - (-1 * 0)] + k[(1 * 1) - (1 * -1)] = 0i + 0j + 2k = (0,0,2)

Q2

v^2 = u^2 + 2as
s = 5, a = 9.81/6 = 1.635, u = 2

v^2 = (2^2) + (2 * 1.635 * 5)
v^2 = 4 + 16.35
v^2 = 20.35
v = 4.51 m/s (3 s.f.)

Does anyone else find it take ages to type this stuff out, or do I just need more practice?

### teaching experiment

I am allowed by the School two weeks to test this method of teaching. Please let me know if you support this way of teaching, or prefer the traditional way, or have some other suggestions. Thanks a lot!

### Q3 has been posted

Newton's law is not that simple, although most of you said that you have learnt about it.

My name is Yu Wu, a aerospace engineering student

A+B=(1,1,0)+(-1,1,0)=(0,2,0)
A-B=(1,1,0)-(-1,1,0)=(2,0,0)
2A+3B=2(1,1,0)+3(-1,1,0)=(-1,5,0)
A.B=(1,1,0).(-1,1,0)=1*(-1)+1*1+0=0
AxB=| i j k |
| 1 1 0 |
|-1 1 0 |

=0-0+(1*1+1*1)k
Written as (0,0,2)

With the formula: 2as=v^2-u^2
so v=(2as+u^2)^(1/2)
=(2*9.81*5/6+2^2)^(1/2)
Approximiately=4.51m/s

Hi i'm chirag, i'm definitly late with these answers but oh well.

Q1:

A + B = (0,2,0)

A - B = (2,0,0)

2A + 3B = (2,2,0) + (-3,3,0) = (-1,5,0)

A . B = (1 x -1) + (1 x 1) + (0 x 0) = -1 + 1 + 0 = 0

A x B = [(1x0)-(1x0)]i - [(1x0)-(-1x0)]j + [(1x1)-(-1x1)]k
= (0,0,2)

Q2:

u = 2 m/s

s = 5m

a = (9.81/6) m/s^2

v = ?

v^2 = u^2 + 2as
v^2 = (2)^2 + 2(9.81/6)(5)
v^2 = 20.35

v = 4.511 m/s

Q1

A+B : (0,2,0)

A-B : (2,0,0)

2A+3B : (-1,5,0)

A.B : 0

AxB : (0,0,2)

Q2

v=5m/s (1sf)

### at least 1 error

at least 1 error

A+B= (0,2,0)

A-B= (2,0,0)

2A= (2,2,0)  3B= (-3,3,0)       2A+3B= (-1,5,0)

A.B= (1x-1)+(1x1)+(0x0)=0

AxB= ((1x0)-(0x1)), ((0x-1)-(1x0)), ((1x1)-(1x-1))

= 0,0,2

Q2

s=5, a=9.81/6, u=2, v=?

v^2=u^2+(2as)

v=4.51 m/s at the exact moment of impact after which the velocity will decrease to ZERO as the energy is disappated.

### Q1 & Q2

I know I'm certainly not first to answer but it's been my first chance to get to a computer all day

Q1

A+B = (0,2,0)

A-B = (2,0,0)

2A+2B = (2,2,0) + (-3,3,0) = (-1,5,0)

A.B = (1x-1) + (1x1) + (0x0)  = 0

AxB = ((-1x0)-(0x1)) , ((0x-1)-(1x0)) , ((1x1) – (1x-1)) = (0,0,2)

Q2

V^2 = u^2+2as            s=5 , a=9.81/6 = 1.635 ,  u=2

v^2 = (2^2)+(2x1.635x5) = 20.35

Therefore: v = 4.51 ms

### Am an aerospace student. hope i get marks 4 this

Q1

A + B

(1) + (-1) = 0
(1) + (1) = 2
(0) + (0) = 0

A-B

(1) - (-1) = 2
(1) - (1) = 0
(0) - (0) = 0

2A+3B

2(1) + 3(-1) = -1
2(1) + 3(1) = 5
2(0) + 3(0) = 0

A.B

(1)x(-1) = -1
(1)x(1) = 1
(0)x(0) = 0

so -1+1+0=0

AxB

(1)x(-1) = -1
(1)x(1) = 1
(0)x(0) = 0

Q2

U= 2 m/s
S= 5 m
a= 9.81/6 m/s2

v2=u2+2as
v2= 4+2x1.64x5
v= 4.51 m/s

here you go!

Q1
A + B = (0,2,0)
A - B = (2,0,0)
2A + 3B = (2,2,0) + (-3,3,0) = (-1,5,0)
A . B = (1 x -1) + (1 x 1) + (0 x 0) = -1 + 1 + 0 = 0
A x B = (1x0)-(1x0) - (1x0)-(-1x0) + (1x1)-(-1x1) = (0,0,2)

Q2:
u = 2 m/s    s = 5m     a = (9.81/6) m/s^2    v = ?
v^2 = u^2 + 2as
v^2 = 2^2 + 2(9.81/6)5
v = 4.511 m/s

### Q1. (A+B)=

Q1.

(A+B)= (1-1),(1+1),(0+0)= (0,2,0)

(A-B)= (1+1),(1-1),(0-0)= (2,0,0)

2A+3B= 2(1,1,0) + 3(-1,1,0)=(2,2,0)+(-3,3,0)= (-1,5,0)

A.B = (1x-1)+(1x1)+(0x0)=(-1+1+0)= 0

AxB = [(1x0)-(0x1)],[(1x0)-(-1x0)],[(1x1)-(1x-1)] = (0,0,2)
^This is the cross product. Some people seem to be confusing it with the scalar/dot product.

Q2.

v2 = u2 + 2as
u=2, a=9.81/6=1.635, s=5

v2 = 2x2 + 2x1.635x5
v2 = 20.35
v2 = 4.51 ms-1 (to 3 s.f.)

 In response to Q3:

Newtons 2nd law states that, observed from an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum. Apply this to a body and the simple equation is F=ma; F is the force needed to be applied to a body of mass m in order to produce an accceleration a.

Hi, I am Aman Mehra, Mechanical Engineering.

Q1:

A+B =  (1-1,1+1,0+0) = (0,2,0)

A-B =    {1-(-1),1-1,0-0} = (2,0,0)

2A + 3B = (2-3,2+3,0+0) = (-1,5,0)

A.B = (1*-1) + (1*1) + (0+0) = 0

A x B = {0-0}, {0-0}, {1-(-1)} = (0,0,2)

Q2:

s = 5m   u = 2m/s    a = 9.8/6    v = ?

Formula used: v² = u² + 2as

v² = 4 + 5(9.8/3)

v² = 61/3

v = 4.51 (3s.f.)

Regarding the system debate, I highly doubt this would do the students any good. I would prefer to go back to the traditional system.

### Putting it on the web:

Hey all, heres a little email i sent mr. Tan about what i think of the system:

Dear Prof. Tan,

Firstly, I would like to introduce myself, my name is Rafael, shortly
known as Raf. Secondly, I would like to comment on this method of
learning. At the same time of which I found it interestingly innovative, I
realized its great unfairness. Now that so many people have answered, one
could argue that there is no reason for me to answer, as it could seem
that I am simply copying the previous answer. Similarly, others could
always wait for the first answer to be posted, and gain points for copying
the right answer, and unfortunatelly, not study/learn. I believe that if
each one of us were to send u a personal message with our answer, it would
be fairer. Now, as everyone can see eachothers working, answering to the
blog is like doing a test with the answers; you can work them out, but you
dont have to, you can see them. I dont want to be negative, but I believe
that this system is fairly naive. Whereas, if we were all to mail you the
answer, and you reply with how many points we gain...it would be fair for
everyone, specially people who have activities, and would take longer to
get to this website than others. Today for example, I took a few hours to
get to my internet, as I had other activities to cater for.

I just wanted to give you my opinion on your interesting system.

Kind Regards,

Rafael Moraes

What do u think about it?

### comment on each other (rather then instructor comment on you)

Good opinions and suggestions.

My motivation is trying to let students comment on each other (rather than let me comment on you), through this way you actually learn even more. You do not have to be the first one to solve the problem, you can always write comments which can bring you credits also. The best comments worth even more.

### The after-impact discussion

I guess the bounce depends on whether the collission is elastic, doesn't it? And I never said it would be instantaneously zero. Or I might have.... oh well, sorry if you took it that way.

The space vehicle may/may not reach the velocity after impacting the lunar surface. It depends on several other factors.

### Dot-product and cross-product.

Uh... what kind of cases do we have to give?

Do we have to state any random formula which requires a dot product?

1. Dot-product: Work = F.s, where F - force applied, and s - dsplacement of body.

2. Vector-product: Torque about an axis = Force x radius of the circle

Again, I might have deeply misunderstood the question. And the answer might be dead wrong.

### you understood correctly

good, you understood correctly. anyone want to comment?

### i cant understand this

i cant understand this method at all....how r we suppose to know how many points have we earned n how do we get to knw how many marks are we being awarded....

if this method is being taught then please make us understand how will it work and what exactly you want from us...

Sheet 1

1) A+B:  (1,1,0) + (-1,1,0) = (0,2,0)

2) A-B: (1,1,0) - (-1,1,0) = (2,0,0)

3) 2A+3B: (2,2,0) + (-3,3,0) = (-1,5,0)

4) A.B: (1*-1) + (1*1) + (0*0) = 0

5) AxB:    i      j      k

1   1   0

-1   1   0

gives  2k

Sheet 2

S= 5m  U= 2ms-1  V=?  a=9.81/6 => 1.635 T= ?

V2 = u2+2as

v2 = 22+(2*1.635*5)

v2=407/20 => v = 4.511ms-1

### Answers - but every little point counts!

Q1.

A+B = 0, 1, 0

A-B = 2, 0 , 0

2A + 3B = -1, 5, 0

A.B = 0

AxB = -1, 1, 0

### Q2. using v^2 = u^2 +

Q2. using v^2 = u^2 + 2as

v being 4.511 m/s

### Questions 3

3) If F=ma is applied to a body, will still work, as long as the body is assumed to be a collection of particles, you just have to take all the vectors to the centre of the objects mass.....F is force and a is acceleration

just realised that when i posted my answers before they didnt appear

A+B : (0,2,0)

A-B : (2,0,0)

2A+3B : (-1,5,0)

A.B : 0

AxB : (0,0,2)

2. using SuVat you get 4.511m/s

V^2=u^2 +2aS

Sheet 1

1. (1,1,0) + (-1,1,0) = (0,2,0)
2. (1,1,0) - (-1,1,0) = (2,0,0)
3. 2(1,1,0) + 3(-1,1,0) = (2,2,0) + (-3,3,0) = (-1,5,0)
4. A.B = 1 x -1 + 1 x 1 + 0 x 0 = 0
5. AxB = (1x0 - 0x1, 0x-1 - 1x0 , 1x1 - 1x-1) = (0,0,2)

Sheet 2

v2 = u2 + 2as

u = 2

a = 9.81/6 = 1.635

s = 5

v2 = (2)2 + 2x1.635x5 = 20.35

v = 4.511 m/s = 5 m/s (1 s.f.)

With regards to motion after this, we would need to model the shock absorption as a lunar module will have some sort of system in place to reduce the damage from impact to the rigid structure as much as possible. This could make an interesting extended question if you could give us a model to work with.

q1)
1 - (0,2,0)
2 - (2,0,0)
3 - (-1,5,0)
4 - 0
5 - (0,0,2)

ok, let me do question 2 now... that's all for now!

question 2!

v2=u2+2as

v2=22+2(10/6)(5)

v=4.55m/s

yea?

### Question 5

Cross products are needed for things such as motor effect using vectors.

10 points?

### Question 1/2

Thought I might aswell join the club and put my answers up.

Part 1

Q1) = (0,2,0)

Q2) = (2,0,0)

Q3) = (-1,5,0)

Q4) = 0

Q5) = (0,0,2)

Part 2

u = 2m/s

s = 5m

a = 1/6*9.81m/s2

Equation  v2=u2+2as

v2=22+(2*5*(1/6*9.81))

v2=20.35m/s

v= 4.51m/s

Job done

### if dot product is zero, the

if dot product is zero, the vectors are perpendicular. There is also a formula involving the dot product which allows u to find the angle between the two vectors.

cross product is used to find the normal of the two vectors, this result can help u find the area of a triangle, the angle between vectors, the resulting vector, and many other things...it all depends on what you are meant to find out.

Q1. (1,1,0)

Q2. (-1,1,0)

Q3. (0,2,0)

Q4. (-1,5,0)

Q5. (0,0,2)

Part 2

v^2=u^2+2as

=2^2+2x(9.8/6)x5

v=4.5ms^-1

Im guessing that if Newtons laws are applied to a body, the answer will be much more complex becuase unfortunately unlike particles the mass is not concentrated on one point so the force on different areas of the body will be different so moments will start to become clear.

If dot product fo vectors=o then they are mutually perpendicular, to expand on Raf's comment a.b=|a||b|cos@  (thats supposed to be a theta btw)

### Answer to Q1 & Q2

(Q1) A+B = (0,2,0)

A-B = (2,0,0)

2A+3B = (-1,5,0)

A.B = 0 AxB = (0,0,2) or 2K

(Q2) h=5m

Initial Velocity= 2m/s

acceleration= 9.81/6 m/s^2

We have to find Impact velocity or Final velocity Using the formula v^2-u^2=2as

where v=final velocity u=Initial velocity a=Acceleration s=Distance

so. v^2-2^2=2*(9.81/6)*5

v^2=20.35

v= 4.51m/s

I am sorry for replying late, as I didnt knew how to use Imechanica.org I had a view about this way of teaching. I know it is good, but it is not a fair way of evaluating the capabilities of a student. We never know that anyone who has posted the answers are fairly done by him, as anyone can copy my work make a few changes in it and post it. What so every you are my teacher, if this is the way you like to teach us then um more then happy atleast i would not have to give my exams :-D

I have noticed the flaw for the initially proposed assessment method. The New Grading System has been posted, and I will talk about that in Thursday's (4/10) lecture. Sorry for the confusing.

Henry. 