Mr. Study's 3 and 4 Question.

[Mr. Study]

Hey,

Thought I would start my own little thread about my dilemmas in Physics.

First question from Jacaranda Physics: A car travelling north at a speed of 40 km h -1 turns right to head due east at a speed of 30 km h -1. This change in direction and speed takes 2.0 s . Calculate
the average acceleration of the car in:

(a) km h-1 s-1 (b) m s-2.

I can get the answers but I am having trouble visualising the process. I also don't know how the hell the answer has South East.

Could someone draw me a vector diagram of this question and label initial and final velocity?

Thanks.

[HERculina]

It's cause it's a vector subtraction: to calculate acceleration = change in velocity/time . And change in velocity = v - u. Initial velocity here is north but has to be changed to south so itd be Change in velocity = v + (-u) , -u in this case beig the opposite of north which is south. Hope i made sense!

[Mr. Study]

Heres another question:

<--------- 100N                       <---------- 400N
______________                    _________                 
|                          |--------____/                  \____
|_____________|          |_________________|
      600 KG                                   1400 KG

It is supposed to be a car pulling a trailer.

a) Find the forward thrust.
b) Magnitude of tension, in the bar between the car and trailer.

Could someone show me how to do it and also explain what forward thrust is.

Thanks!

[b^3]

Firstly note that the 100 N force is the friction force between the wheels of the trailer and the road, the 400 N force is the friction force between the wheels of the car and the road.

In this case the forward thrust force will be the force that the car's motor has to apply so as turn the wheels and propel the car forward.

Now I'm assuming there is more information given?

If it is travelling at a constant speed then the net force will be 0 N (as the acceleration will be 0 m/s^2 and the car will be moving with a constant velocity)

a) Forward Thrust - frictional forces=F_net
Let F be forward thrust.
F-100-400=0
F=500 N
So the frictional force is 500 N.

b) Now the tension will be what is pulling back on the car at the front. In other words in this case the tension will be the frictional forc e of the boat only and the force requried to accelerate the boat.

Since the acceleration is 0 m/s^2, then the force required to accelerate the boat is 0 N.
So here
T-F_r=F_net
T=F_R+F_net
F_net=0 N as a=0 m/s^2
T=F_R
T=100 N

This will all be different if it is not travelling at a constant speed, as then you would have to factor in the force required to accelerate the masses aswell

Hope that helps and hoefully I haven't made any mistakes.

EDIT: Best way to look at it is just to look at the vectors you have and add/minus them accordingly, equalling F_net, then knowing your acceleration and masses you can find F_net (F_net=ma) and put them in and solve for what you are looking for.

[Mr. Study]

Ah! I get it! Thanks so much!

[Mr. Study]

Here are some questions I am stuck on.

This is from Jacaranda Physics (Yuck)



Forgot to mention, speed is constant.

This is from Heinemann. ()

Question 1: Ball, mass 10g, is falling at a speed of 8.2 m/s. Find the magnitude and direction of air resistance.

I am not too sure what formula to use but I only know that a=9.8 m s^-2.

Question 2: A golder, with an elevated tee, 4.9 m above the fairway, strikes a ball and sends it at 20 m s^-1. Find the speed of the ball, 0.8 s after being hit.

This is part D of the question. I could work out the horizontal distance (20m) and time until the ball lands (1.0s). I just made a few assumptions but I worked out part E and NOT part D.

Could someone give me a hint for this one?

Thanks is all my questions for today.

[Phy124]

Here are some questions I am stuck on.

This is from Jacaranda Physics (Yuck)

(Image removed from quote.)

Forgot to mention, speed is constant.

This is from Heinemann. ()

Question 1: Ball, mass 10g, is falling at a speed of 8.2 m/s. Find the magnitude and direction of air resistance.

I am not too sure what formula to use but I only know that a=9.8 m s^-2.

Question 2: A golder, with an elevated tee, 4.9 m above the fairway, strikes a ball and sends it at 20 m s^-1. Find the speed of the ball, 0.8 s after being hit.

This is part D of the question. I could work out the horizontal distance (20m) and time until the ball lands (1.0s). I just made a few assumptions but I worked out part E and NOT part D.

Could someone give me a hint for this one?

Thanks is all my questions for today.

Diagram:
Is it driving upwards? Does it give a mass?

Question 1 - I'm not to sure about this, but here goes:
F_net = F_g - F_a
It has a constant speed so F_net = 0
0 = mg - F_a
0 = (0.01)(9.8 ) - F_a
F_a = 9.8 x 10^-2

Question 2:
Horizontal velocity doesn't change, therefore 20 m/s
The vertical velocity can be calculated by v = u + at (u=0, a=gravity[9.8 or 10], t=0.8s)
Add vectors and you will get the velocity/speed - since it is speed you don't need to state the direction

[Mr. Study]

Diagram:
Is it driving upwards? Does it give a mass?

Question 1 - I'm not to sure about this, but here goes:
F_net = F_g - F_a
It has a constant speed so F_net = 0
0 = mg - F_a
0 = (0.01)(9.8 ) - F_a
F_a = 9.8 x 10^-2

Question 2:
Horizontal velocity doesn't change, therefore 20 m/s
The vertical velocity can be calculated by v = u + at (u=0, a=gravity[9.8 or 10], t=0.8s)
Add vectors and you will get the velocity/speed - since it is speed you don't need to state the direction

Thanks for that! I forgot to mention it gave a mass of 90KG and that it was driving upwards.

[Phy124]

Thanks for that! I forgot to mention it gave a mass of 90KG and that it was driving upwards.

Ok well, I haven't done one of these in ages, but I believe that you need to work out the force that gravity is providing down the slope (make a triangle with angle 15* underneath the block):

I believe this force would be mg sin(15*) = 882 sin (15*) = 228.3 N (It can also be 882 cos(75*) if you prefer to do it that way)

So the forces down the slope would add up to 248.3 N

Because velocity is constant F_net = 0, meaning forces up the slope = forces down, so the driving force would be 248.3 N

Given this is correct, I can draw a diagram if you need more assistance.

[Mr. Study]

Hey, Thanks for that! I get it, wasn't too sure what formula to use.

UPDATE: Hey, heres a question that is tearing my hair out.

A dodgem car, mass 200kg, strikes a barrier head on, at a speed of 8.0 ms^-1, due west, and rebounds in the opposite direction, with a speed of 2.0 ms^-1. What is the impulse?

I know impulse is mass X changein velocity but I'm confused about assigning positives and negatives. Could someone reword the question in a way that says, the car went this way but after bumping into something, it went the other way. I still don't get the 'due west'.

[Lasercookie]

Hey, Thanks for that! I get it, wasn't too sure what formula to use.

UPDATE: Hey, heres a question that is tearing my hair out.

A dodgem car, mass 200kg, strikes a barrier head on, at a speed of 8.0 ms^-1, due west, and rebounds in the opposite direction, with a speed of 2.0 ms^-1. What is the impulse?

I know impulse is mass X changein velocity but I'm confused about assigning positives and negatives. Could someone reword the question in a way that says, the car went this way but after bumping into something, it went the other way. I still don't get the 'due west'.

The car is moving at 8 m/s west . It hits something. It rebounds. It is now going the other way - east. It is now moving at 2 m/s east (.

Due west means that it's going directly west, not on an angle a tiny bit to the left of west or anything. On a compass it would be 270°T.

Looking at the question +8m/s is west, so the opposite direction is east, so I'll assign east to be negative. You could do it the other way around (west = negative, east = positive), you'll get the same answer out, so it doesn't really matter.

Impulse =


Since the mass stays constant here:


So the Impulse is 2000 N s east.

[Mr. Study]

Ahh! Thanks for that! It actually got me excited for physics.

[Mr. Study]

Two new questions with two new concepts. I wasn't taught these type of concepts at school and am having trouble with it.

1. A person pushes a lawn mover, of mass 20kg, at a constant speed, with a force of 100N along the handle. The handle is at 35 degrees, to the horizontal.
.
(I draw diagrams to help me visualise the question but with this one, I am not so sure my diagram is correct. :S, If it is wrong, could someone point out why and then explain).

    a) What is the horizontal retarding force?
    b) What is the normal force by the ground on the mower?



2. 4.5 kg and 2.5kg are connected by a light rope that passes over a smooth pulley.


    a) What is the magnitude of the acceleration, of the two masses?
    b) What is the tension between the rope?

Thank you.

[trinh]

For 2.

On the left (2.5kg mass), there is a tension force (let's call it ) directly upwards, while there is a force of 2.5g N downwards.

Likewise, on the right (4.5 kg mass), there is also a tension force of N directly upwards, while there is a force of 4.5g N downwards.

Since the right block is heavier, it is obvious that that the 4.5 kg mass will move downwards, while the 2.5 kg mass will move upwards (the pulley will turn clockwise). Thus I (personally) choose clockwise movement about the pulley to be positive. Also just for reference, , where F is the net force.

So for the 2.5 kg mass (which moves up), [equation 1]

For the 4.5 kg mass (which moves down), [equation 2]

Equating equations 1 and 2,


You can then find t by substituting a into either equation 1 or 2

EDIT: did a bit of editing lol

[Phy124]

For question 1; in your question you say it is 35 degrees but in the diagram you say 45, I used 45 since it was easier but you should be able to work out what to do if it is 35 degrees.

Working out here: http://i39.tinypic.com/8yx0cg.jpg

Sorry, too lazy to do latex at the moment

edit: never mind with question 2, good work trinh