Calculating Average Acting Force: Impulse and Interaction Time

[noname1]

What would be the average force acting on the glider if the interaction time of the collision with the bumper is 0.1 sec

I am kind of confused how i shall proceed here, I calculated the impulse from a question before which was

I = pf -pi .0999-.168 = -0.0681

but now i am confused on how to solve the next question, any help would be greatly appreciated

i can do it by this formula correct?

f = (I/T)

 

[rock.freak667]

That formula should work.

 

[noname1]

I did several experiments with the 2 types of collisions

Perfectly inelastic
elastic collision where:
m1=m2
m1>m2
m1<m2

to prove that the momentum was conserved pi has to be equal to pf no matter what type of collision correct?

 

[rock.freak667]

I did several experiments with the 2 types of collisions

Perfectly inelastic
elastic collision where:
m1=m2
m1>m2
m1<m2

to prove that the momentum was conserved pi has to be equal to pf no matter what type of collision correct?

Yes. Well close or near to.

 

[noname1]

well on

elastic collision
m1=m2 - i got a difference of 4.86%
m1>m2- i got a difference of 52.4%
m1<m2- i got a difference of 25.24%

I can say that the momentum on these were not conserved right?

 

[rock.freak667]

well on

elastic collision
m1=m2 - i got a difference of 4.86%
m1>m2- i got a difference of 52.4%
m1<m2- i got a difference of 25.24%

I can say that the momentum on these were not conserved right?

Well your results would say that, you should know that momentum is always conserved in collisions. You might just need to account for why you had such large differences.

 

[noname1]

And to verify Kinectic energy conservation it is

inicial ke = 2 x final ke

right?

 

[rock.freak667]

And to verify Kinectic energy conservation it is

inicial ke = 2 x final ke

right?

No, you'd need to have final KE = initial KE for it to be conserved.

 

[noname1]

The weird thing is my professor put this question

He asked us to calculate the kinetic energy before and after the collision for

elastic collision where:
m1=m2
m1>m2
m1<m2and then below is the question, is the kinetic energy conserved for each of these elastic collisions and then has this formula

(.5*m1*v1²i = .5*m1*v1²f + .5*m1*v1²f) ?

the formula is throwing me off, do you know perhaps what it means?

 

[rock.freak667]

and then below is the question, is the kinetic energy conserved for each of these elastic collisions and then has this formula

(.5*m1*v1²i = .5*m1*v1²f + .5*m1*v1²f) ?

the formula is throwing me off, do you know perhaps what it means?

Kinetic energy is 1/2mv². Essentially what the formula is saying is that the initial (terms with i) kinetic energy is equal to the final kinetic energy (terms with th f).

 

[noname1]

but why is it adding the final twice?

does it mean the the initial is just one of the masses and the final is adding both masses since they entering collision?

 

[rock.freak667]

but why is it adding the final twice?

does it mean the the initial is just one of the masses and the final is adding both masses since they entering collision?

Not sure, how was your collision done? The right side should have two different 'm's

 

[noname1]

it was done on a glider where we would push a mass from right to left where it would impact with another mass and we had photogates measuring times initial times and final times

 

[rock.freak667]

it was done on a glider where we would push a mass from right to left where it would impact with another mass and we had photogates measuring times initial times and final times

the second mass is initially at rest, so the formula should be


$$\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_1v_{1f}^2$$


'1f' denotes the final of 1
'1i' is the initial of 1.

 

[noname1]

so for initial kinetic, i can only use mass 1 and not add both masses correct?

 

[rock.freak667]

so for initial kinetic, i can only use mass 1 and not add both masses correct?

Yes, that is how you described the experiment to be.

 

[noname1]

that would make sense, in reality kinetic energy is always conserved in elastic collisions?

and on the calculation of the kinetic final energy

$$\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_1v_{1f}^2$$

the first one i should use

.5*m1*v1²F + .5*m2*v2f² and not how i wrote it initially

 

[rock.freak667]

that would make sense, in reality kinetic energy is always conserved in elastic collisions?

and on the calculation of the kinetic final energy

$$\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_1v_{1f}^2$$

the first one i should use

.5*m1*v1²F + .5*m2*v2f² and not how i wrote it initially

Sorry, I should of rechecked what I typed

$$\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$$


But the definition of an elastic collision is one in which kinetic energy is conserved.

 

[noname1]

i must be doing something wrong because i am off on all of them

 

[rock.freak667]

Off how?

 

[noname1]

These were the values i got from the tests

Like you can see momentum is not conserved on none of them and when calculating the kinetic energy before and after i get these values, i know they are still really close but its showing that its not conserved

D
initial - .5 x .2106 x .719² = .0541
Final - .5 x .2106 x 0² + .5 + .2096 x .685² = .0492

E
initial - .5 x .2106 x .327² = .0113
Final - .5 x .2106 x .0936² + .5 + .4092 x .209² = .00986

F
initial - .5 x .4102 x .943² = .182
Final - .5 x .4102 x .313² + .5 + .2096 x .862² = .0980

 

[rock.freak667]

How much accuracy are you required to have? (decimal places)

 

[noname1]

professor didnt say but i am leaving it to three, trying to do significant figures

 

[rock.freak667]

You should realistically get them to be fairly close. But your results show them to be quite off to 3sf.

 

[noname1]

that why i was wondering if i did something wrong, elastic collisions are supposed to conserve kinetic energy

 

[rock.freak667]

that why i was wondering if i did something wrong, elastic collisions are supposed to conserve kinetic energy

Well they in the same range to decimal place, but in practical situations, you will have losses due to friction. But since it is a lab, you have to put what you get.

 

[noname1]

yea i guess, i appreciate all of your help