A cart collides with a bumper -- Find the final displacements, etc.

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In summary, the conservation of energy can be used to solve for both the velocity and compressed spring distance of a cart released from rest on an inclined plane and colliding with a bumper. The velocity can be found by equating the initial gravitational potential energy to the final kinetic energy, and the compressed spring distance can be found by equating the initial kinetic energy to the final spring potential energy. The mass of the cart does not affect the time it takes for it to hit the ground or its final velocity, as long as air resistance is neglected.
  • #1
paulimerci
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Homework Statement
The cart is released from rest and slides from the top of an inclined frictionless plane of height h. Express all algebraic answers in terms of the given quantities and fundamental constants.
a) Determine the speed of the cart when it reaches the bottom of the incline.
b) After sliding down the incline and across the frictionless horizontal surface, the cart collides with a bumper of negligible mass attached to an ideal spring k. Determine the distance x_m the spring is compressed before the cart and bumper come to rest.
Relevant Equations
Conservation of energy
E_i = E_f
I'm using conservation of energy to solve for both a) and b).
a) Initially, the cart is released from rest, which has maximum GPE at the top of the inclined plane. U_g is converted to K.E as it reaches the bottom of the incline.
$$ E_i = E_f $$
$$ U_g = K.E_f$$
$$ 2mgh = \frac {1}{2}2mv^2$$
$$ v = \sqrt {2gh}$$
b) After sliding down the incline, the cart collides with the bumper, which transforms K.E in to spring P.E.
$$ E_i = E_f$$
$$\frac{1}{2}2mv^2 = \frac {1}{2}kx_m^2$$
$$ x_m = \sqrt \frac {2m}{k} v$$
where ##x_m## is the compressed spring distance.
In part a), I don't know how to find velocity in terms of mass because mass gets canceled out in the equation.
 

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  • #2
paulimerci said:
In part a), I don't know how to find velocity in terms of mass because mass gets canceled out in the equation.
Neglecting air resistance, if you release two different masses from the same height above ground, which mass will hit the ground first? Which mass will have the greater velocity when it reaches the ground? Does your answer depend on whether the masses fall straight down or on a frictionless incline?
 
  • #3
paulimerci said:
In part a), I don't know how to find velocity in terms of mass because mass gets canceled out in the equation.
If you drop a 1kg stone and a 2kg stone the same height (and ignoring air resistance), how does the mass affect the final velocity of each stone?

By the way, the diagram in the attachment shows the mass of the cart as m and there’s a curved arrow labelled m/4 to the left of the cart. The text of the question states the cart’s mass is 2m. So that’s very confusing!.

For your answer in part a) you have taken the mass as 2m. But in part b) you have taken the mass as m. So that’s inconsistent.
 
  • #4
Steve4Physics said:
By the way, the diagram in the attachment shows the mass of the cart as m and there’s a curved arrow labelled m/4 to the left of the cart. The text of the question states the cart’s mass is 2m. So that’s very confusing!.
Each wheel is ##\frac{m}{4}##, that's how they imply the mass is ##2m##.
 
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  • #5
kuruman said:
Neglecting air resistance, if you release two different masses from the same height above ground, which mass will hit the ground first? Which mass will have the greater velocity when it reaches the ground? Does your answer depend on whether the masses fall straight down or on a frictionless incline?
Both masses will hit the ground at the same time because falling objects accelerate at the same rate. It doesn't depend on the masses. It depends only on g.
 
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  • #6
Since the masses of the wheels are given, I guess you are expected to take into account rotational KE of the wheels. More guessing: the wheels can be modeled as uniform solid disks and they roll without slipping.
 
  • #7
TSny said:
Since the masses of the wheels are given, I guess you are expected to take into account rotational KE of the wheels. More guessing: the wheels can be modeled as uniform solid disks and they roll without slipping.
I think it says it is a frictionless track though. I think they are just trying to be sneaky by providing the mass of the wheel.
 
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  • #8
erobz said:
I think it says it is a frictionless track though. I think they are just trying to be sneaky by providing the mass of the wheel.
Yes. You’re right. I missed “ frictionless”. I also missed “sliding” in part (b). :blushing:
 
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  • #9
paulimerci said:
Both masses will hit the ground at the same time because falling objects accelerate at the same rate. It doesn't depend on the masses. It depends only on g.
So, did you figure out what that implies for the velocity of the cart w.r.t. its mass?
 
  • #10
kuruman said:
Neglecting air resistance, if you release two different masses from the same height above ground, which mass will hit the ground first? Which mass will have the greater velocity when it reaches the ground? Does your answer depend on whether the masses fall straight down or on a frictionless incline?
Are the expressions correct? I understood that my answer doesn’t depend on masses falling down or on a frictionless incline.
 
  • #11
paulimerci said:
Are the expressions correct? I understood that my answer doesn’t depend on masses falling down or on a frictionless incline.
Your answers and work in post #1 look correct.
 
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  • #12
paulimerci said:
I don't know how to find velocity in terms of mass because mass gets canceled out
When a problem says to write the answer in terms of certain given variables, it only means you can’t use any others. There is no requirement to use all of them if some are redundant.
But btw, your answer to part b violates the instruction by using v in the answer. And, technically, I don't think g counts as a "fundamental constant", but there's nothing you can do about that.
paulimerci said:
I understood that my answer doesn’t depend on masses falling down or on a frictionless incline.
No, it does depend on its being frictionless, or you would have to account for the rotational KE of the wheels. That would reduce v but not affect the spring compression.
 
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  • #13
TSny said:
Your answers and work in post #1 look correct.
Thank you!
 
  • #14
TSny said:
More guessing: the wheels can be modeled as uniform solid disks and they roll without slipping.
Then you need to know their radii
 
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  • #15
malawi_glenn said:
Then you need to know their radii
You don't, it turns out.
 
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  • #16
haruspex said:
You don't, it turns out.
Yeah that is true, I had a brain-hickup :oldbiggrin: you only need the mass-distrubution "form factor", for a solid cylinder it is 1/2
 

1. What is the definition of displacement in physics?

Displacement is the distance and direction of an object's change in position from its starting point. It is a vector quantity, meaning it has both magnitude and direction.

2. How is displacement different from distance?

Distance is the total amount of ground an object has covered, while displacement is the straight-line distance between the starting and ending points. Distance is a scalar quantity, meaning it only has magnitude.

3. How do you calculate displacement?

Displacement can be calculated by subtracting the initial position from the final position. This can be represented by the equation: Δx = xf - xi, where Δx is displacement, xf is final position, and xi is initial position.

4. Can displacement be negative?

Yes, displacement can be negative if the final position is behind the initial position. This indicates that the object has moved in the opposite direction from its starting point.

5. How does a cart colliding with a bumper affect its displacement?

When a cart collides with a bumper, its displacement will change depending on the direction and magnitude of the force applied by the bumper. If the force is in the opposite direction of the cart's initial motion, the displacement will decrease. If the force is in the same direction, the displacement will increase.

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