Conservation of Momentum Lab PDF

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Lab report, received letter grade A...

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STONY BROOK UNIVERSITY DEPARTMENT OF PHYSICS AND ASTRONOMY PHYSICS 133 SECTION 12

Conservation of Momentum

Experiment Date: 4/3/2020 Report Date: 4/12/2020

Introduction: In this lab, we were tasked with studying the principles of conservation of momentum and energy in three different collisions. There is a large cart and a small cart placed on an air track which collided with one another. We measured the mass of each cart and the length of the flags on top of each cart. In each trial, the carts went under the photogates to measure the time it took to go through it to determine the velocity of each cart before and after the collisions. After completing each trial, we calculated the initial and final velocities of each cart. After calculating the velocities, we also calculated the initial and final momentum of each cart and we found the initial and final kinetic energy of each cart as well. Each trials’ values help us see if the momentum and kinetic energy was conserved during the trials.

Theory: ‘Dimensionless Formulas’ In this lab we are faced with two different types of collisions, both of which are elastic and inelastic. The elastic collisions have two equations that both show the velocities of each object after a collision has occurred. The first equation is v 1f = equation is v 2f =

2m 1 m 1 +m 2 v 1i

−

m 1 −m 2 m 1 +m 2v 2i .

m 1 −m 2 v m 1+m 2 1i

+m

2m 2 +m 2 v 2i 1

and the second

Using the dimensionless parameter λ =

m2 m1

, it can be

used to remove the masses and create a dimensionless equation in both equations mentioned prior. In the first equation, if we divide the whole equation by m

equation as v 1f = v 1f =

1−λv 1+λ 1i

+

1− m 2

1 m 1+ m 2 1

m1 m1

we end up with our first

m

v

1i

+

2λ v which 1+λ 2i

2m2

1 m

1+ m 2

v

2i

and then we can simply substitute λ to get

1

is what we are striving to get. As for the second equation, we can

m

simply start by dividing by

m1 m1

again and we received the equation v 2f =

2 m 1+ m 2

v 1i −

1

2 v 1+λ 1i

and when we substitute λ , we get the equation v 2f =

−

1−λ v 1+λ 2i

1− m 2

1

m

1+ m 2

v

2i

1

which is a success.

As for the inelastic equation, there is only one equation mainly because both objects after they collide stick together. The inelastic equation is v f =

m1 v m 1 +m 2 1i

+

m2 v m 1 +m 2 2i

and if we also

m2

divide this by

m1 m1

we end up with the equation, v f =

1 m 1+ m 2

v 1i +

1

equation, we can plug in λ and we end up with the equation v f =

m1 m

1+ m 2

v 2i . Once we have this

1

1 1+λ v 1i

+

λ 1+λv 2i

which is a

success. After we replace all three equations for both inelastic and elastic, we end up with dimensionless equations that do not require masses.

‘Initially Stationary Cart ’

Type of Collision

m1=m2

m1>>m2

m1...