My GMA Question Thread

[alchemy]

1. What is the unit vector perpendicular to 5i + j -2k?
It's a multiple choice question and I can get the answer by checking each option. However, I'm not sure how to complete the problem from scratch.

[Nato]

Have you tried 1/|5i+j-2k| multiplied by the original vector? that gives you the unit vector.

|5i+j-2k|, is the magnitude so just sqroot(5^2 + 1^2 + (-2)^2)=root(30)

so the unit vector would be 1/root(30) multiplied by 5i+j-2k

[b^3]

Normally you'd make another vector, lets say , then for the two vectors to be perpendicular, the dot product of the two vectors should be zero. As we want a unit vector, the magnitude of this new vector should be , i.e. , i.e. .

The problem is we have three unknowns and only two equations/pieces of information. This is because we only have a single vector that we're given, so there is more than one unit vector that is perpendicular to it, i.e. there is a set of vectors in a plane that is perpendicular to the vector. You could solve the two equations in terms of a parameter but it looks like more effort than it's worth for this question, (if it were not VCE maths I'd say something different here though ).
i.e. When we have a single vector, all the vectors in that plane will be perpendicular to the vector we're given.

Normally with these problems we'd have two vectors, and be asked to find a unit vector perpendicular to both, which would lock it down into one or two vectors (the second including a unit vector in the opposite direction to the first).

In this case it's probably easier to just take the dot product of the MC options with the vector you have, and choose the one that gives zero.
Anyways, if you are asked for the vector perpendicular to two vectors that you're given, then you'd do this.
1. .
2. Take the dot product of and the first vector, set it to zero (as they are prependicular) and form equation [1].
3. Take the dot product of and the second vector, set it to zero (as they are perpendicular) and form equation [2]/=.
4. As we have a unit vector, the magnitude is one, so


5. Solve the three equations simultaneously.

[alchemy]

The question information is attached.
Question: What is T₁ and T₂. Give answer in terms of g.
   

[M_BONG]

The question information is attached.
Question: What is T₁ and T₂. Give answer in terms of g.
   

Do you have the solutions?

I got some weird decimal answer; where T2=6.71082g and T1=3.3541g :s

[alchemy]

Do you have the solutions?

I got some weird decimal answer; where T2=6.71082g and T1=3.3541g :s

The solutions are:



Feel free to have another go.

[RKTR]

pythagoras theorem,  (T1)^2 + (T2)^2 = (6g)^2  (1)
                                   
from question, T2 = 2T1
sub it into (1)
you will get (T1)^2 + (2T1)^2 =(6g)^2
                   5(T1)^2=(6g)^2
                     (T1)^2 = (6g)^2 /5
                     T1 = 6g / root 5
                      T2=2T1 =12g / root 5

EDIT : added an image

[M_BONG]

The solutions are:



Feel free to have another go.

This is the weird thing. My answers work as well, if you sub it in. Here's how I did it.
Btw, I did it Tech active (ie. via CAS solve function).
1. Draw diagram.

Resolve in i direction
T2 sin(x) + T1 sin (y) = 6g (Equation 1)

As T2=2T1

x= tan^-1 (T1/2T1)
  =tan^-1 (1/2)

CAS gives the angle (X)  to be 26.57 degrees. Keep it on CAS unrounded.

Thus, Y= 90-26.57 degrees = 63.43 (unrounded on CAS)

Therefore, use CAS to solve:

Solve [2x sin (26.57) + x sin (63.43) =6, x)
x= 3. 3541
Thus, 2x = 7.01

BUT if you sub:  T1= 6g/sqroot(5) etc into Equation (1) , as above, with angles I obtained, the answer is also 6g.

Can someone explain why this is possible; is my answer still correct?

[b^3]

Solve [2x sin (26.57) + x sin (63.43) =6, x)

Should be equal to 6g, not 6, which changes your tension values.

Also when you do this, rather than actually finding the angle, keep it as , then let , and , which means , which then lets you get exact values for and for that angle, that is in this case and .

BUT if you sub:  T1= 6g/sqroot(5) etc into Equation (1) , as above, with angles I obtained, the answer is also 6g.

Also I'm not sure how you're getting the same values?
http://www.wolframalpha.com/input/?i=7.01sin(26.57)%2B3.3541sin(63.43)%3D6*9.8&t=crmtb01

[alchemy]

An object of mass 3kg on a smooth plane inclined at 30 degrees to the horizontal is attached by a light inelastic string passing over a smooth pulley at the top of the plane to a 7kg object hanging freely.
After the 7kg object has fallen 5 meters, it strikes the ground without rebounding.

Find the velocity when the 7kg object strikes the ground.

Solution: 7.34 m/s

[M_BONG]

Should be equal to 6g, not 6, which changes your tension values.

Also when you do this, rather than actually finding the angle, keep it as , then let , and , which means , which then lets you get exact values for and for that angle, that is in this case and .
Also I'm not sure how you're getting the same values?
http://www.wolframalpha.com/input/?i=7.01sin(26.57)%2B3.3541sin(63.43)%3D6*9.8&t=crmtb01

Thanks for the advice! I agree - should have kept everything in exact values. Plus my method of using the solve function.  was dodgy.

Btw Wolfram Alpha said your answer was false because it assumed you meant the degrees were in radians.
"Assuming trigonometric arguments in radians | Use degrees instead"

[b^3]

Since the kg block is heavier and at a greater angle we assume that it will be the mass that moves downwards, so we will take down as positive for this block and up the plane as positive for the other block.
Looking at the motion for the kg block, the blocks weight is acting downwards, that is it is positive and tension acting in the other direction is negative. The sum of the forces on the object gives the net force, .

Now lets look at the other block. Since there is no friction, as we have a smooth plane, we don't need to resolve the forces perpendicular to the surface. But acting on the block we have the blocks weight, acting straight down and the tension in the rope acting up the plane. We take the coordinate system so that we have the i direction up the plane and the j direction perpendicular to the plane. This means that the object's weight is acting at an angle to the plane, that is at an angle , so we resolve this force in the direction of the plane. Since the tension force is larger and we know the block is moving up the plane, we have (we use sine since we want the force that is acting opposite to the angle in the triangle formed),

Npw if we rearrange for and substitute into we get, (we rearrange for rather than as we need to find , we don't need so we can eliminate it).

Now that we have our acceleration we can use our equations of motion for linear motion. We know that the object falls 5 m, and we want to know the final velocity, assuming that the initial velocity was zero.

EDIT: Merged Double Post

Thanks for the advice! I agree - should have kept everything in exact values. Plus my method of using the solve function.  was dodgy.

Btw Wolfram Alpha said your answer was false because it assumed you meant the degrees were in radians.
"Assuming trigonometric arguments in radians | Use degrees instead"

I swear I changed that... It resets to radians every time you do a different calculation -.-
It's still not equal when it's in degrees anyways.
http://www.wolframalpha.com/input/?i=7.01sin%2826.57%29%2B3.3541sin%2863.43%29%3D6*9.8&a=TrigRD_D

-> http://www.wolframalpha.com/input/?i=7.01sin%2826.57%29%2B3.3541sin%2863.43%29&a=TrigRD_D
-> http://www.wolframalpha.com/input/?i=7.01sin%2826.57%29%2B3.3541sin%2863.43%29&a=TrigRD_D

[RKTR]

x= 3. 3541
Thus, 2x = 7.01

erm.. 2 x 3.35 equals 6.7.....but anyway this isnt the main problem

the main problem is that using your method, the 26.57 will be the angle between 6g and T2 for the left triangle in my image.

to find the angle between T2 and 6g for the right triangle is 180-26.57 which is 153.43. then you minus 90 to get 63.43

your equation should be T2 sin 63.43 + T1 sin 26.57 = 6g  instead of the equation you used. took me a long time to figure out lol.. at 1st i thought your answers worked as well

EDIT : your way works when the T1 and T2 in the right triangle are lengths instead of tension.

[RKTR]

does it say g= what??

i tried and it seems that they want g=9.8

9a.   Fnet = ma

          5 x 9.8 - 3 x 9.8 x sin(25) = 8a

           a=4.57

10.  Fnet = ma
       
       8x9.8xsin(20) - 4x9.8xsin(25)=12a
           a=0.85   
   

[alchemy]

1) A mass of 500 grams hangs from a point M by a rope 8cm long.
It is pulled aside by a horizontal force of P kgwt until it is 4cm from the vertical line through M. The value of P(correct to one significant figure is)?

Answer=0.3

2) A hot air balloon is rising vertically with a speed of 9.8m/s. An object is dropped from the balloon at a height of 150m.
Find the maximum height reached by the object.

Answer=154.9m