Complex Kinematics and Dynamics

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SUMMARY

This discussion centers on a physics problem involving two Teflon pucks, each weighing 5 kg, on a Teflon table. Puck A is initially at rest and is pushed with a force of 20 N at a 30° angle after a 1.5-second delay, while Puck B moves towards it at 0.5 m/s. The calculations reveal that Puck A experiences a net force of 7.35 N, resulting in an acceleration of 1.47 m/s². The analysis suggests that the distance Puck A travels before colliding with Puck B must account for the time taken for both pucks to meet, emphasizing the need for a clear understanding of their relative motions.

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Vraj Patel
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Homework Statement


Two pucks (5 kg each) made of Teflon are on a long table, also made of Teflon. Puck A is sitting at
rest on the left end of the table. Puck B is 15 m away at the right hand end of the table, and is
travelling toward Puck A with an initial speed of 0.5 m/s. A person on the left waits 1.5 seconds and
then pushes Puck A forward from rest with a force of 20 N applied at an angle of 30° to the horizontal.
If that force is maintained at a steady rate for the entire question, how far will puck A travel before it
strikes puck B? (The coefficient of friction between the Teflon and Teflon is 0.04.

2. The attempt at a solution:

Puck A
Fax = 20NSin30
Fax = 10N

Fay = 20NCos30
Fay = 17.32N

Fg = mg
Fg = (5kg)(9.8m/s^2)
Fg = 49N


Fn = Fay + Fg
Fn =17.32N + 49N
Fn = 66.32

Ff = uFn
Ff = (0.04)(66.32N)
Ff = 2.65

Fnet = Fax - Ff
Fnet = 10N - 2.65
Fnet = 7.35N

a = Fnet/m
a = 7.35N/5kg
a = 1.47m/s^2

d = v1t + 1/2at^2
15-x = 1/2(1.47)(y-1.5)^24
x = -0.74y^2 + 2.22y +13.33

Puck B
Fg = mg
Fg = (5kg)(9.8m/s^2)
Fg = 49N

Ff = uFn
Ff = (0.04)(49N)
Ff = 1.96N


Since there is no applied force Fnet = -Ff

a = Fnet/m
a = -1.96/5kg
a = -0.39m/s^2

d = v1t + 1/2at^2
x = (0.5)(y) + 1/2(-0.39)(y)^2
-0.74y^2 + 2.22y +13.33 = 0.5 - 0.195y^2
0 = 0.545y^2 - 1.72y-13.33


I then put it in the quadratic formula and got:
y = 6.77s and y = -3.61s(omited)


Puck A Distance
d = v1t + 1/2at^2
d = (0m/s)(6.77s-1.5s) + 1/2(1.47m/s^2)(6.77-1.5s)^2
d = 20.41m(This answer doesn't make sense)
 
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Hi,

First thing that comes to mind is your decomposition of the 20 N in ##F_x## and ##F_y## ... :rolleyes:

Second thing I notice is that you deleted part 2 of the template -- not good !
Third is I miss a list of variable names. What is ##d## ? (and: are you sure that is what you think it is ? -- a drawing might help an awful lot here ! )
 
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Vraj Patel said:
an angle of 30° to the horizontal.
Pushing down or pushing up?
 
Let's assume pushing down :wink:
 
haruspex said:
Pushing down or pushing up?
pushing down
 
BvU said:
Hi,

First thing that comes to mind is your decomposition of the 20 N in ##F_x## and ##F_y## ... :rolleyes:

Second thing I notice is that you deleted part 2 of the template -- not good !
Third is I miss a list of variable names. What is ##d## ? (and: are you sure that is what you think it is ? -- a drawing might help an awful lot here ! )
d represents distance.
 
BvU said:
Hi,

First thing that comes to mind is your decomposition of the 20 N in ##F_x## and ##F_y## ... :rolleyes:

Second thing I notice is that you deleted part 2 of the template -- not good !
Third is I miss a list of variable names. What is ##d## ? (and: are you sure that is what you think it is ? -- a drawing might help an awful lot here ! )
I tried to upload a photo of my diagram but it wouldn't let me, but the pucks are heading toward each other.
 
Vraj Patel said:
d represents distance.
I know. How does your expression for ##d## deal with the fact that the pucks are moving towards each other (as opposed to: moving away from each other, or moving in the same direction) ?
 
Hint: Puck B has no external force affecting it other than friction. So why not find out how far it can coast before coming to rest, and how long it takes to come to rest? Is it less than or greater than the 1.5 s pause before puck A starts to move?
 

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