Marble Rolling/Projctle Motion

In summary, the problem involves finding the location where a coin should be placed so that a ball, traveling at an average velocity of 1.5m/s, will strike it directly on impact with the ground. This can be solved using the equations for projectile motion, with the given time of 0.2 seconds between photogates and the height of the ramp at 92cm. The time for the ball to travel through the photogates is not needed, but can be used to calculate the initial velocity of the ball. The horizontal component of gravity is also needed to solve the problem.
  • #1
Let It Be
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1. A student finds that it takes .20s for a ball to pass through photogates places 30cm apart on a level ramp. The end of the ramp is 92cm above the floor. Where should a coin be placed so that the ball strikes it directly on impact with the ground?


2. ΔX=ViTf+1/2 ATf^2
v= ΔX/ΔT



3. Tf=.2s
h=92cm (.92m)
I don't understand why the 30cm is there or how to complete the problem.
 
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  • #2
Let It Be said:
1. A student finds that it takes .20s for a ball to pass through photogates places 30cm apart on a level ramp. The end of the ramp is 92cm above the floor. Where should a coin be placed so that the ball strikes it directly on impact with the ground?


2. ΔX=ViTf+1/2 ATf^2
v= ΔX/ΔT



3. Tf=.2s
h=92cm (.92m)
I don't understand why the 30cm is there or how to complete the problem.


I can't understand the question but you might use 30sm to find radius ...
 
  • #3
cupid.callin said:
I can't understand the question but you might use 30sm to find radius ...

It's a projectile motion problem...you don't use radius ever.
 
  • #4
attachment.php?attachmentid=42328&stc=1&d=1325300657.gif
 

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  • #5
gneill said:
attachment.php?attachmentid=42328&stc=1&d=1325300657.gif

Yep, that's the picture I have! However, I have NO CLUE where to start it. Where does the 30cm come into play? I'm thinking find the time, then average velocity, to then both put into the ΔX=blah blah blah equation
 
  • #6
Let It Be said:
Yep, that's the picture I have! However, I have NO CLUE where to start it. Where does the 30cm come into play? I'm thinking find the time, then average velocity, to then both put into the ΔX=blah blah blah equation

You're GIVEN the time for it to pass between the photogates. No need to find it. If you know the time for the projectile to travel the 30 cm between the gates, what's its speed?

If you then have its speed, what are the formulas that describe its trajectory when it leaves the edge of the level surface?
 
  • #7
gneill said:
You're GIVEN the time for it to pass between the photogates. No need to find it. If you know the time for the projectile to travel the 30 cm between the gates, what's its speed?

If you then have its speed, what are the formulas that describe its trajectory when it leaves the edge of the level surface?

This is how I did the problem-
H=1/2gt^2
.92m=1/2(9.8)t^2
T=.43s

V=ΔX/ΔT
30cm/.2s
V=1.5m

Then I plugged in both answers into:
ΔX=ViTf+1/2ATf^2
ΔX=1.5(.43)+1/2(0)tf^2
ΔX=.65m

Does this look right?
 
  • #8
Let It Be said:
It's a projectile motion problem...you don't use radius ever.

woops ... I didnt read whole topic, I read rolling and thought and ...
 
  • #9
Let It Be said:
This is how I did the problem-
H=1/2gt^2
.92m=1/2(9.8)t^2
T=.43s

V=ΔX/ΔT
30cm/.2s
V=1.5m

Then I plugged in both answers into:
ΔX=ViTf+1/2ATf^2
ΔX=1.5(.43)+1/2(0)tf^2
ΔX=.65m

Does this look right?

Yep its right ... sorry for my above confusing post ... this problem was not in the language i am used to ...
 
  • #10
No worries!
 
  • #11
Let It Be said:
This is how I did the problem-
H=1/2gt^2
.92m=1/2(9.8)t^2
T=.43s

V=ΔX/ΔT
30cm/.2s
V=1.5m /sec

Then I plugged in both answers into:
ΔX=ViTf+1/2ATf^2
ΔX=1.5(.43)+1/2(0)tf^2
ΔX=.65m

Does this look right?
Your method looks right.
 
  • #12
Let It Be said:
This is how I did the problem-
H=1/2gt^2
.92m=1/2(9.8)t^2
T=.43s

V=ΔX/ΔT
30cm/.2s
V=1.5m

Then I plugged in both answers into:
ΔX=ViTf+1/2ATf^2
ΔX=1.5(.43)+1/2(0)tf^2
ΔX=.65m

Does this look right?

Couple questions on the problem still-
1)First, why did I have the use the H=1/2gt^2 equation to get T=.43s? I think it has something to do with the horizontal and vertical relationship to each other? But if the problem gives me .2s already-what is that for?

2) Second, I used the average velocity equation, but how can the average velocity switch to being the Vi in the ΔX=ViTf+1/2ATf^2 equation? How does that work?
 
  • #13
Let It Be said:
Couple questions on the problem still-
1)First, why did I have the use the H=1/2gt^2 equation to get T=.43s? I think it has something to do with the horizontal and vertical relationship to each other? But if the problem gives me .2s already-what is that for?

2) Second, I used the average velocity equation, but how can the average velocity switch to being the Vi in the ΔX=ViTf+1/2ATf^2 equation? How does that work?

1)

when marble falls from table, in Y direction there is no initial velocity (right?) ... so for eqn:
y = uyt + 0.5 g t2 .. uy turns out to be 0
 
  • #14
Let It Be said:
Couple questions on the problem still-
1)First, why did I have the use the H=1/2gt^2 equation to get T=.43s? I think it has something to do with the horizontal and vertical relationship to each other? But if the problem gives me .2s already-what is that for?
The 0.2 sec is used to calculate the speed of the ball before it starts falling. The 0.43 sec is the time it spends in midair. While it's free-falling, it also continues to move horizontally away from the table.

2) Second, I used the average velocity equation, but how can the average velocity switch to being the Vi in the ΔX=ViTf+1/2ATf^2 equation? How does that work?
The initial speed of the ball was calculated on a level surface. What is the value of the horizontal component of gravity acting to speed up the ball over this path?
 

1. What is marble rolling/projectile motion?

Marble rolling/projectile motion is a type of motion in which an object, in this case a marble, moves along a curved path due to the influence of gravity and any initial force given to it. This type of motion can be seen in everyday activities such as playing with marbles or throwing a ball.

2. What factors affect the motion of a rolling marble?

The motion of a rolling marble is affected by various factors such as the force applied to it, the angle at which it is released, the surface it is rolling on, and the mass and shape of the marble itself. These factors can determine the speed, direction, and distance the marble will travel.

3. How is the velocity of a rolling marble calculated?

The velocity of a rolling marble can be calculated by dividing the distance traveled by the time taken. This can be represented by the equation v=d/t, where v is velocity, d is distance, and t is time. The velocity of a marble can also be affected by the force applied to it and its mass.

4. What is the difference between linear and curved projectile motion?

Linear projectile motion refers to the motion of an object in a straight line, while curved projectile motion refers to the motion of an object along a curved path. In the case of a rolling marble, it can exhibit both types of motion, as it initially moves in a curved path due to the force applied to it, but then continues in a linear motion due to the influence of gravity.

5. How is the trajectory of a rolling marble affected by different surfaces?

The trajectory of a rolling marble can be affected by different surfaces in various ways. For example, a smooth surface will allow the marble to roll faster and with less friction, while a rough surface will cause the marble to slow down and encounter more resistance. The angle of the surface can also affect the trajectory, as a steeper incline will result in a faster and longer roll for the marble.

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