Question

A student of mass 50 kg swings on a playground swing, which is very light compared to the student. A friend releases the seat of the swing from rest at a hieght of 1 m above the lowest point of of the motion. The student swings down and, at the lowest point, grabs a jug of water of mass 4 kg. the jug is initially at rest on a small table to the right of the swing, so it does not move vertically as the student grabs it. the student keeps swinging forward while holding the jug, and the seat reaches a maximum height of H above the lowest point. air resisitance and friction are negligible.

Answer 1

A) Indicate whether H is greater than, less than, or equal to 1 m.
Justify your answer qualitatively, with no equations or calculations.

Answer 2

The student starts with potential energy (energy due to position) since he is raised up by a friend.

Answer 3

But this energy is for only him (50 kg).

Answer 4

Would you think it would take more, less, or the same amount of energy to lift 54kg to the 1 meter mark as the 50 kg?

Answer 5

more energy

Answer 6

Right! :)

Answer 7

Right but im going into AP Physics.... dont i need a little more than that?

Answer 8

Sometimes simple is best. :)

Answer 9

You could mention also that the potential energy is converted into kinetic energy at the bottom of the trajectory of the swing.

Answer 10

And that when you add the 4 kg that it has no potential energy or kinetic energy associated with it.

Answer 11

Thank you. i have a couple other things for my teacher like the collision causes the final velocity of the jug and student to be less. therefore, the swing would not reach 1 m in the same amount of time.

Answer 12

could you maybe help me out with part b and c? i am seriously struggling with this problem.

Answer 13

I can try. :)

Answer 14

B) Explain how H can be calculated. You need not actually do the calculations, but provide complete instructions so another student could use them to calculate H

Answer 15

This would be an inelastic collision, so kinetic energy is not conserved. However, momentum is conserved.

Answer 16

ok.....? how does that help find h?

Answer 17

Still working on that...

Answer 18

okay....

Answer 19

Using conservation of energy, you can find the velocity of the student at the lowest point.

Answer 20

Then using conservation of momentum, you can find the new velocity when the student picks up the jug.

Answer 21

but i thought you said that KE is not conserved here

Answer 22

And how can i use the momentum equation if i dont have velocities of accerlations for anything?

Answer 23

All the potential energy is converted into kinetic energy before the student picks up the jug.

Answer 24

gotcha. so how would that look mathematically?

Answer 25

mgh = 1/2 mv^2?

Answer 26

Correct.

Answer 27

m's cancel....
g = 9.8
h is 1?

Answer 28

Right.

Answer 29

then v = 4.4271887

Answer 30

Right. What velocity is that though?

Answer 31

the fastest? just before he grabs the jug?

Answer 32

Perfect! :) Just before he grabs the jug.

Answer 33

Can you now calculate his velocity the moment he grabs the jog?

Answer 34

*jug

Answer 35

Using conservation of momentum.

Answer 36

4.099?

Answer 37

You got it!

Answer 38

now to find h, do i do another mgh=1/2mv^2?

Answer 39

Correct!

Answer 40

hold on..... i think i did something wrong in my calcuations.....

Answer 41

No worries

Answer 42

but i have the right idea?

Answer 43

Yup.

Answer 44

Make sure you square your v before doing anything else.

Answer 45

do you wanna take a look at c while i fix this?

Answer 46

Sure.

Answer 47

C) the student now swings back towards the starting point. at the lowest point, the student drops the jug. Indicate whether the new maximum height will be greater than, less than, or equal to H. Justify your answer.

Answer 48

What do you think for c?

Answer 49

not less than......

Answer 50

my best guess would be greater than....

Answer 51

What would the velocity at the bottom be?

Answer 52

wouldn't it be the same is when the student had the jug?

Answer 53

or no because the mass decreases.....?

Answer 54

Right.

Answer 55

It would be back to 4.099 since it would all come from the potential energy.

Answer 56

but didnt the jug contribute to the potential when it because part of the motion?

Answer 57

Didn't the masses cancel out?

Answer 58

in the energy equation... not the momentum one

Answer 59

Right, that is how we got the new velocity.

Answer 60

but why doesnt the velocity change now? if the jug was used to find the new velocity before, why not again?

Answer 61

Because when the student drops the jug it doesn't collide again.

Answer 62

that kinda makes sense......

Answer 63

so would the answer be equal to because there isnt a change in the magnitude of the velocity?

Answer 64

Yup :)

Answer 65

im still a little confused. if v doesnt change and momentum = mv... and m decreased, should p decrease?

Answer 66

No because the jug has a velocity this time.

Answer 67

Some of the momentum goes to the jug.

Answer 68

ooooooooooh. that makes sense now!

Answer 69

:)

Answer 70

do you ahve time to help me through my next homework problem? i ahve some stuff written on scratch bu i think i need help connecting the dots. like with this problem, i had all the inform and equations written on my scratch paper but couldnt quite figure out how they fit together.

Answer 71

I can try, but can you open a new question, please? This one is getting pretty long.

Answer 72

yup :)

Answer 73

Thanks.