Need Help With Some Basic Physics Questions (1 Viewer)

[ConHeo]

Hello, my teacher has been MIA for the past few days or so and I really need help with some of this work. I don't quite understand it, and I'm not even sure if I'm getting some of these right. I would really appreciate the help, so thank you in advance.


1. A Physics student drops a coin into a wishing well and takes just 3.0s to make a wish. The coin splashes just as she finishes making her wish. The coin accelerates towards the water at a constant 9.8ms^-2.
a) What is the coin's velocity as it strikes the water?
b) How far does the coin fall before hitting the water?

2. A car of mass 1600 kg left parked on a steep but rough road begins to roll down the hill. After as short while it reaches a constant speed. The road is inclined at 15 degrees to the horizontal. Its speed is sufficiently slow that the air resistance is insignificant and can be ignored. Determine the magnitude of the road friction on the car while it is rolling at constant speed.

Showing your working out so I may understand would also be very much appreciated. Thank you to anyone who helps me out in this dire situation!

 

[someth1ng]

1a. v=u+at
v=0(3)+9.8(3)
v=29.4ms-1

1b. s=ut+.5at^2
s=.5(9.8)(3^2)
s=44.1m

 

[ConHeo]

Cheers mate, you're a legend! I got 1a. right, but I did it a bit differently. Can you please tell me if this is err, scientifically correct?
a=s/t
9.8=s/3
29.4=s
and at this point in time s=v. Is there something wrong with using that?
And for b, what I did was sort of the same thing and it's double what you had... I did this:
s=d/t
29.4=d/3
88.2=d
...yeah I don't know.

I appreciate the help though, cheers mate!

 

[D94]

Cheers mate, you're a legend! I got 1a. right, but I did it a bit differently. Can you please tell me if this is err, scientifically correct?
a=s/t
9.8=s/3
29.4=s
and at this point in time s=v. Is there something wrong with using that?
And for b, what I did was sort of the same thing and it's double what you had... I did this:
s=d/t
29.4=d/3
88.2=d
...yeah I don't know.

I appreciate the help though, cheers mate!

No, don't do that.

How can 's' which is distance equal 'v' which is velocity? The units are different. It's like saying your weight is 60 metres; it doesn't make sense. You have to use the relevant kinematic equations of motion. You're not given distance, and since there is a known formula without the use of distance, you should immediately use that formula (i.e. the one someth1ng used)

In your part b, what is 's' and what is 'd'? Regardless, you have to use the kinematic equations of motion. It's a syllabus dot point, so it means you need to understand them.

 

[ConHeo]

Ohh okay, thanks a lot, I really need the help. Not particularly fond of my teacher's way of doing things so far. Could you please help me with question 2?

Edit: Woops, I meant s as in speed, which is ridiculous as displacement=s. My apologies. And in part b, my apologies, another derp. I rushed into things, s is supposed to be speed again, and d is displacement. Sorry if it's confusing, my errors are very silly already.

 

[D94]

For (2):

You probably should draw a diagram, but it's essentially an inclined plane at 15°, with a car of mass m = 1600 kg. There is a weight force, mg, in the downwards direction, a normal force, N, perpendicular to the plane at where the car is in contact with the plane/hill, and a frictional force, F, in the direction opposing the direction of motion.

Now, if we take the reference axis to be parallel to the plane and perpendicular to the plane, we only need to resolve the mg force into its x and y components. So, you get mgcos(15) in the y direction (perpendicular to the plane downwards) and mgsin(15) in the x direction (parallel to the plane pointing downwards).

The direction of motion is obviously down the plane in this case, so there is a force, ma, in that direction. However, the question states constant velocity, so if the velocity is constant, acceleration is zero (0).

So the sum of forces in the x direction is N - mgcos(15) = 0, and in the y direction, F = mgsin(15). I assume the question just asks for the Friction, and not the coefficient of friction, so it's just F = mgsin(15) = 1600 * 9.81 * sin(15) = 4062 N.

 

[cutemouse]

I don't think (2) is in the syllabus for the preliminary or HSC course... but the school might ask it in their assessment tasks... So it's worth checking with your teacher.

 

[someth1ng]

Yeah, part 2 is NOT in the syllabus for physics (only in MX2) and so, you don't need to know it.