How far apart were the centres of the two tunnels?

  • Thread starter Voltman123
  • Start date
In summary, the player's ball was knocked 20 degrees to the South of East with a velocity of 1.7m/s into the tunnel leading to the bunker. Assuming the ball weighs 40g, the distance between the centres of the two tunnels is 20.03 cm.
  • #1
Voltman123
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0

Homework Statement


A competitive mini-golf player notices an opponent's ball situated 50cm in front of a wall containing two tunnels. One tunnel leads to the hole, the other to a bunker. The player skillfully putts their ball due East at the opponent's ball with a velocity of 4m/s. The opponent's ball is knocked 20 degrees to the South of East with a velocity of 1.7m/s into the tunnel leading to the bunker. Meaning the player's ball travels into the second tunnel located North of East (to great applause from spectators). If the player's ball weighed 40g, and the opponent's ball weighed 45g, how far apart were the centres of the two tunnels?


Homework Equations



Velocity = displacement/time

Force = G*(mass1*mass2)/(distance^2)

The Attempt at a Solution



Using the above formula, transpose so distance can be found.
(distance^2)=G*(mass1*mass2)/force

I'm not sure where to get the value for force though if this is the correct method.

I have tried drawing a diagram myself but am stuck and don't know what to do.

I hope someone can help me with this problem, work through it with me or point me in the right direction.

Any help would be appreciated!
 
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  • #2
Voltman123 said:
Using the above formula, transpose so distance can be found.
(distance^2)=G*(mass1*mass2)/force
I seriously doubt that you need to worry about the gravitational force between the two balls. :smile:

Instead, consider the collision. What's conserved?
 
  • #3
Doc Al said:
I seriously doubt that you need to worry about the gravitational force between the two balls. :smile:

Instead, consider the collision. What's conserved?

Momentum?
 
  • #4
Voltman123 said:
Momentum?
Good. Use conservation of momentum to figure out the angle that the player's ball traveled after the collision. The rest is a bit of trig.

I assume that the wall is oriented along a North-South line and is due east of the balls. (You should have a diagram!)
 
  • #5
Does the attached diagram represent all the above information correctly?

If so, what do I need to do next?

The question mentions that the ball hits the opponent's ball while traveling at 4m/s.
Using momentum formula:
p=mv
=40g*4m/s
=160g m/s

Conservation of momentum:

mv (opponent's ball)=160g m/s
v=(160 g m/s)/(45 g)
v(opponent's ball) = 3.6 m/s

So, the opponent's ball should be traveling 3.6 m/s? Why does it say 1.7 m/s in the question?

Sorry, I'm still lost on how to calculate the distance between the centres of the two tunnels.

How does knowing the conservation of momentum help in answering the final question.

Thanks
 

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  • #6
Voltman123 said:
Does the attached diagram represent all the above information correctly?
Almost. The 20° angle is in the wrong place.

The question mentions that the ball hits the opponent's ball while traveling at 4m/s.
Using momentum formula:
p=mv
=40g*4m/s
=160g m/s
That's the momentum of the player's ball and is thus the total momentum of the system. Note that momentum is a vector. The direction is east.

Conservation of momentum:

mv (opponent's ball)=160g m/s
v=(160 g m/s)/(45 g)
v(opponent's ball) = 3.6 m/s
You are told the speed and thus momentum of the opponents ball.

So, the opponent's ball should be traveling 3.6 m/s? Why does it say 1.7 m/s in the question?
You are misapplying conservation of momentum.

Since momentum is a vector, break it into components. x (east) and y (north) components of total momentum must be separately conserved.

Sorry, I'm still lost on how to calculate the distance between the centres of the two tunnels.

How does knowing the conservation of momentum help in answering the final question.
Once you have the angle that the player's ball bounces off at, you can use trig to find where the straight line trajectories hit the tunnel.
 
  • #7
Doc Al said:
Since momentum is a vector, break it into components. x (east) and y (north) components of total momentum must be separately conserved.


Once you have the angle that the player's ball bounces off at, you can use trig to find where the straight line trajectories hit the tunnel.

Sorry,
could you give me more clues/hints on what to do?

...please.

I need more help. :(
 

1. How were the centres of the two tunnels measured?

The centres of the two tunnels were measured using precise surveying techniques, such as laser scanning and GPS technology. These methods allow for accurate measurement of distances and coordinates between two points.

2. What was the purpose of measuring the distance between the two tunnel centres?

Measuring the distance between the two tunnel centres is important for ensuring that the tunnels are built at the correct distance apart to avoid any collisions or structural issues.

3. How far apart were the two tunnel centres?

The distance between the two tunnel centres can vary depending on the specific project, but typically they are built at a distance of 5-8 meters apart.

4. Can the distance between the two tunnel centres change over time?

Yes, the distance between the two tunnel centres can change over time due to factors such as ground movement or erosion. Regular monitoring and maintenance can help maintain the desired distance between the two tunnels.

5. What happens if the distance between the two tunnel centres is incorrect?

If the distance between the two tunnel centres is incorrect, it can lead to collisions or structural issues which can be dangerous and costly to fix. That's why precise measurement and monitoring is crucial in tunnel construction.

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