How far will the ball travel horizontally?

[tman]

Could you please help, I don’t understand this problem. I thought it would rise 20cm because it raises the same height according to Newton’s first law, but that is wrong.

could not import diagram.
but it is like a half pipe and a ball is on the left side and it is 20 cm from the ground to the top of the half pipe.


Referring to the above diagram, if the incline on the right has a slope of 5 cm of rise for each 2 cm of run, how far will the ball travel horizontally?
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[CeeAnne]

Tman, I don't believe it has to do with Newton. I believe the question is asking for the distance from the point on the ground beneath where the ramp or trough is at 20cm to where it meets the ground at 0cm. On the xy grid, where x=0, y=20, that's where the ball starts ... it rolls down the ramp and where x=2, it has come down to where y=15 ... when it rolls to where x=4, it's down to where y=10 and so on. When it rolls to where y=0, then however far x is at that point is how far the ball has rolled horizontally. Can you figure it now?

 

[tman]

Newtons

the section is about Newtons law.

 

[tman]

is it 16 then

 

[Stargate]

Nope! it is 18 cm squared!

 

[tman]

No It's Not, it is 8 but I don't know why

 

[VietDao29]

Hi,
Your question is certainly not clear,... and lots of people have different answers.
I suggest you posting your question in a clearer way.
You can use:
a = \frac{F}{m}
F = m \times g \times \sin{\alpha}
\Leftrightarrow a = g \times \sin{\alpha}
Viet Dao,

 

[Severian596]

Hi,
Your question is certainly not clear,... and lots of people have different answers.
I suggest you posting your question in a clearer way.
You can use:
a = \frac{F}{m}
F = m \times g \times \sin{\alpha}
\Leftrightarrow a = g \times \sin{\alpha}
Viet Dao,

Those are some nice equations for the original poster, VietDao29.

I believe this problem is an issue of figuring out the force with which the ball begins to roll up the incline, then figuring out the distiance it will roll before sufficient force has been applied against its direction of motion to bring the net velocity to zero. Here are a couple questions to ask yourself:

- How fast is the ball traveling when it starts up the incline?
- What's the net acceleration being applied to the ball in the opposite direction of its motion? (note: if the incline were straight up this acceleration would be equal to gravity's acceleration, but it's dependent on the angle of ascent and this incline doesn't go straight up)
- How much time will pass before the ball's velocity is overcome by the net force found in the previous question? Another way to ask this is how far will the ball travel under the "negative" acceleration before its velocity becomes zero?