EK Physics vertical ramp

[naphthalene]

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A ball is rolled down a 1m ramp placed at an angle of 30degrees to the horizontal. The same ball is rolled down a 1m ramp placed vertically. which of the following statements is true?

a. the ball required the same amount of time for both trips.
b. the ball had the same displacement at the end of both trips.
c. the ball accelerated at the same rate for both trips.
d. the ball reached approximately 1.4 times the speed of the second trip as it did on the first trip.

by 1m ramp, do they mean the hypotenuse or the width of the ramp? What is a vertical ramp? any help will be great.

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A forum. Also, you should list where this question came from.

 

[naphthalene]

This is from EK physics

 

[liveoak]

i think hypotenuse.

 

[DrFit]

A ball is rolled down a 1m ramp placed at an angle of 30degrees to the horizontal. The same ball is rolled down a 1m ramp placed vertically. which of the following statements is true?

a. the ball required the same amount of time for both trips.
b. the ball had the same displacement at the end of both trips.
c. the ball accelerated at the same rate for both trips.
d. the ball reached approximately 1.4 times the speed of the second trip as it did on the first trip.

by 1m ramp, do they mean the hypotenuse or the width of the ramp? What is a vertical ramp? any help will be great.

yes to hypotenuse (it is 1m). The vertical ramp is a "ramp" at 90* -pointing straight upward- with a length of 1m. So essentially the ball would be in free fall. I remember this questions from EK physics.

 

[naphthalene]

yes to hypotenuse (it is 1m). The vertical ramp is a "ramp" at 90* -pointing straight upward- with a length of 1m. So essentially the ball would be in free fall. I remember this questions from EK physics.

I am still unable to picture a vertical ramp having 90 degrees

 

[DrFit]

I am still unable to picture a vertical ramp having 90 degrees

It looks like a wall. Imagine tilting the original ramp (that is 30* from horizontal) upwards until the hypotenuse is straight up and down. This is the ramp at 90 degrees. If you put a ball against your wall and let it drop it would be moving down a ramp at 90*.

 

[naphthalene]

Ok.. how does one arrive at the answer?
The answer is D.

 

[plzNOCarribbean]

Ok.. how does one arrive at the answer?
The answer is D.

process of elimination.

First, C cannot be correct since the ball will NOT accelerate at the same rate down both ramps. Essentially, by vertical ramp, they mean the ball just fell, or was dropped straight down, as in free fall, so you can ignore the ramp. It equivalent to putting a ball up to the wall and letting go. Ignoring friction against the ball, the ball would just free fall down. So, its acceleration would be equal to gravity, at 9.8 m/s^2. However, the ball rolling down the first ramp accelerates at mgsin theta, since it is at an incline. Therefore, it does not accelerate at 9.8 m/s^2, but at a value less than that.

Since the acceleration is not the same, the time taken to reach the bottom cannot be the same for the different ramps, so eliminate A. Now, the hypotenuse of the ramp is greater than the opposite leg (which is essentially the distance the ball falling from the "vertical" ramp traveled) so the distance traveled by the ball is different for the different trips. Eliminate B.

This is the reasoning we use to arrive at answer D

 

[SKation]

process of elimination.

First, C cannot be correct since the ball will NOT accelerate at the same rate down both ramps. Essentially, by vertical ramp, they mean the ball just fell, or was dropped straight down, as in free fall, so you can ignore the ramp. It equivalent to putting a ball up to the wall and letting go. Ignoring friction against the ball, the ball would just free fall down. So, its acceleration would be equal to gravity, at 9.8 m/s^2. However, the ball rolling down the first ramp accelerates at mgsin theta, since it is at an incline. Therefore, it does not accelerate at 9.8 m/s^2, but at a value less than that.

Since the acceleration is not the same, the time taken to reach the bottom cannot be the same for the different ramps, so eliminate A. Now, the hypotenuse of the ramp is greater than the opposite leg (which is essentially the distance the ball falling from the "vertical" ramp traveled) so the distance traveled by the ball is different for the different trips. Eliminate B.

This is the reasoning we use to arrive at answer D

Hey,

How would we be able to prove this with a formula?

Thanks!

 

[30somethingyearold]

I think the formula to get to answer D is:
First step:
Inclined ramp:
ma = mgsin theta
a = (10m/s^2) (sin 30)
= 5m/s^2
Vertical drop:
a = g
a = 10m/s^2

Second step:
Inclined ramp:
(final velocity)^2 = (initial velocity)^2 +2ad
where d=1m, a=5m/s^2, initial velocity = 0m/s
final velocity is about 3m/s

Vertical drop:
(final velocity)^2 = (initial velocity)^2 +2ad
where d=1m, a=10m/s^2, initial velocity = 0m/s
final velocity is about 4.5m/s = which is 1.4x is inclined ramp final velocity

Please correct me if I'm wrong!

 

[NextStepTutor_1]

you can also answer this using energy. So, the height of the first 1m ramp at 30 degrees is:

h = 1m * sin30 = .5

PE = mgh = m*10*.5 = 5m
KE = 1/2 mv^2 = PE
cross of masses to get:
v^2 = 10
v = sqrt(10)

for the second case h = 1, and we repeat the above steps to get:

PE = 10m = 1/2mv^2
v = sqrt(20)

So we can see that the speeds differ by a factor of sqrt(2) = 1.4

 

[desperadoflakes]

The above methods using the formulas are correct, and you should definitely know how to use them. However, if you want to score well you'll need to learn how to do each problem as quickly and efficiently as possible. This problem is fairly straightforward; if you understand the physics, it is fairly obvious that Options A,B, and C are all false, leaving you only answer D. Answer D makes sense intuitively because an object in freefall will have a somewhat faster final velocity than an object on a ramp. The important thing here is being confident in ruling out options A, B, and C.