How far apart were the centres of the two tunnels?

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Homework Help Overview

The problem involves a competitive mini-golf scenario where a player strikes an opponent's ball, leading to two balls moving towards different tunnels. The context includes concepts from momentum conservation and vector analysis in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum as a key principle to analyze the collision and subsequent motion of the balls. There are questions about the application of gravitational force and the correct interpretation of momentum in vector form.

Discussion Status

Some participants have provided guidance on using conservation of momentum and breaking it into components, while others express confusion about the relationship between the given velocities and the calculations being attempted. There is ongoing exploration of how to apply these concepts to find the distance between the tunnel centers.

Contextual Notes

Participants note discrepancies in the problem statement regarding the velocities of the balls, which raises questions about the assumptions being made. There is also mention of a diagram that may not accurately represent the situation, contributing to the confusion.

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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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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?
 
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?
 
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!)
 
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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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.
 
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. :(
 

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