Momentum and Impulse during a collision

In summary, a 500 g cart is released from rest at a height of 1.0 m on a frictionless, 30 degree ramp. It rolls down the ramp and bounces off a rubber block at the bottom. By using the formula for velocity at the bottom of an incline (v^2_f = v^2_i + 2ad), the velocity of the cart at the bottom was calculated to be 2.21 m/s. The figure also showed the force during the collision, which can be used to calculate the impulse (product of average force and time) and the change in momentum of the cart. By applying the principle of conservation of energy (mgh = 1/2 mv^2),
  • #1
kirbykirbykirby
21
0
A 500 g cart is released from rest 1.0 m from the bottom of a frictionless, 30 degree ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure shows the force during the collision.



knight_Figure_09_30.jpg


Here's what I did. Which is probably wrong for the very first part so I'll only show my work for first part:

[tex]\[ v^2_f = v^2_i + 2ad \]
\[ v^2_f = 0+ 2(0.5)(gsin30)(1)\]

so \[v_f = 2.21 m/s \] [/tex]

I took the area of the graph as [tex]\ [J=200\times26.7\times10^-^3\times0.5\][/tex]

So can somebody please help with the first part, I don't understand.
 
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  • #2
Find the velocity of the cart at the bottom of the incline from energy conservation, i.e. mgh = 1/2 m v^2.
 
  • #3
The area of this graph has dimensions of Newtons*seconds, which is the same as the the dimensions of momentum. The area of the graph is called Impulse. How is that related to the momentum of the object that experiences this force? What is the formula for the area of a triangle? You can find the velocity before the bounce using your method, or you can look at energy conservation as radou suggested. The two are equivalent.

I think this problem was done very recently in another thread.
 
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  • #4
I think usually when you see a graph like that it's an impulse question - that is they want you to use Impulse ( Av Force x time) = change in momentum.

So the area under the graph in this case will give you the chenge in momentum of the cart and from that you can calculate its final velocity.
 
  • #5
Okay thanks, I did that

[tex] \[mgh = 0.5mv^2\]
\[g\sin30 = 0.5v^2\]
v= 3.13 m/s

So

\[J= p_f - p_1\]

\[2.67 = 0.5(v_f + 3.13)\]

so \[v_f= 2.21 m/s\]

then
\[mgh = 0.5mv^2\]

\[gh = 0.5(2.12)^2\]

so h= 0.229m??!
[/tex]
Looks like something went horribly wrong somewhere.
Durr... I took out my fractions because I can never do them.

Can you tell me what is my problem?

EDIT: OKAY I did not see the two last posts as it took me forever to write that out. I am bad at it. Also I know this is solved on another thread but it didn't help at all and NO ONE answered when I posted on it yesterday.

Hi, also can someone anser my question why did I get a different answer when I did conservation of energy and kinematics. For that I am like 'wtf?'.
 
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  • #6
What are you trying to find? You've confused me by ending up with h.

What was the question asking you to do?
 
  • #7
Where did the minus sign get lost? impulse = final momentum - initial momentum. Further on, does the length or the height of the incline equal 1? According to your calculation, it would be the length..
 
  • #8
Durr... I am supposed to find how high the cart bounces back. Or maybe it is length.

The 1m is apparently length.

Edit: "After the cart bounces, how far does it roll back up the ramp?"

Oh and I took the minus sign out because the final momentum is opposite the intial so the minus cancels?
 
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  • #9
AH, Ok

I agree with your velocities and the value for Impluse.

I also agree with the 0.229 by conservation of energy.

Why do you think this is wrong? - this is a height remember so you need to think about sines or cosines to get the length of track it rolls back up
 
  • #10
Just read your edit.

Ok, so convert your 0.229 height to length - does that give you the right answer?
 
  • #11
Durr... I believe you two just gave me opposing information.

and I calculate length to be l=0.459m but I lose marks for the wrong answer... durr...

EDIT: Durr... the opposing info is gone and now I look like a crazy person. ;'( tear.
 
  • #12
Now you've really lost me - opposing information?

When I said "why do you think this is wrong?" it was because I think it's right - but you seemed to think in Post 5 that it was horribly wrong.
 
  • #13
radou said something about not taking direction of momentum into account.

I KNEW I LOOKED CRAZY :cry:

Anyways, that's the right answer.

Thanks so much. I am sastified customer.
 
  • #14
Ok, well done.
 

Related to Momentum and Impulse during a collision

What is momentum?

Momentum is a physical quantity that describes an object's motion. It is calculated by multiplying an object's mass by its velocity.

How is momentum conserved during a collision?

Momentum is conserved during a collision, meaning that the total momentum before the collision is equal to the total momentum after the collision. This is because there is no external force acting on the system, and therefore the total momentum remains constant.

What is impulse?

Impulse is a measure of the change in an object's momentum. It is calculated by multiplying the force applied to an object by the time it is applied.

How does impulse affect momentum during a collision?

During a collision, the impulse applied to an object will result in a change in its momentum. The greater the impulse, the greater the change in momentum will be.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, only momentum is conserved while some kinetic energy is lost to other forms of energy, such as heat or sound.

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