- #1

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I tried using m1v1=m2v2

where 1*18=x*1 which is not right.

I do not know what to do with the height.

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- Thread starter Sneakatone
- Start date

- #1

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I tried using m1v1=m2v2

where 1*18=x*1 which is not right.

I do not know what to do with the height.

- #2

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Try thinking about it in terms of energy

- #3

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1/2m1v1^2=1/2m1v'1^2+1/2m2v2'^2

- #4

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1) out of the cannon

2)grabbing the other performer

3) at the height of the launch

Also, what are the two energy equations you should be using?

- #5

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should the second equation be m1v1+m2v2=m1v'1+m2v'1+m2v'2?

- #6

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What is the equation for the kinetic energy of an object?

What about potential energy?

- #7

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KE=1/2mv^2

PE=mv

PE=mv

- #8

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almost. PE = mgh

try now

try now

- #9

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I would assume mass is 1 since they have the same mass so

1*9.81*4.5=44.145J

1*9.81*4.5=44.145J

- #10

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That's a terrible idea!

use m for mass and see where you get

use m for mass and see where you get

- #11

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PE=44.14m

- #12

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OK. Try first finding the height at which the pair of the start to fall.

To start off:

The kinetic energy as the clown leaves the cannon: .5*m*v^2

The potential energy of the pair before they drop: 2m*g*h

note it is 2m, not just m, because he grabbed the performer on the way up.

Also recall that the energy is conserved. How would you get h, the height before they drop?

- #13

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so can I set PE=KE to find mass(m)?

- #14

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You can set PE = KE, but look at what happens to the masses

- #15

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the masses cancel out,

- #16

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- #17

Mentor

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- #18

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I did not understand what you said in the last part .@dinospamoni

- #19

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Out of the cannon:

KE=.5*m*(v_1)^2

PE=0

He grabs the person:

PE = mgh_1

KE = .5 (2m)*(v_2)^2

Solve for v_2

This is the speed of just the clown before he grabs the person

Now you can use m1v1=m2v2, Where the left side is the clown and the right side is the clown+person

They now have KE .5 (2m)*(v_3)^2

The new KE gives a PE (2m)g*h_2

Now you can solve for height h_2

- #20

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V_2=sqrt[mgh/.5(2m]

H_2=(0.5*2m*v_3^3)/2mg

H_2=(0.5*2m*v_3^3)/2mg

- #21

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Almost. When he picks up the other person he has both KE and PE

I have this:

.5*m*(v_1)^2 = m*g*h_1+.5*m*(v_2)^2

You know everything except for v_2

I have this:

.5*m*(v_1)^2 = m*g*h_1+.5*m*(v_2)^2

You know everything except for v_2

Last edited:

- #22

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So we know what the mass is?

- #23

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Whoops forgot to include mass in a term on that post. Fixed now.

They still cancel though

They still cancel though

- #24

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If the masses cancel then v2= 0.46

- #25

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That's not what I'm getting. Don't forget to subtract the g*h_1 from both sides

- #26

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I accidentally divided instead,

my new answer is 15.35 m/s=v2

my new answer is 15.35 m/s=v2

- #27

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Nailed it. Now use that velocity in .5*2m*(v_2)^2 to find the maximum height they reach using 2mgh

- #28

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He grabs the person:

PE = mgh_1

KE = .5 (2m)*(v_2)^2

Solve for v_2

This is the speed of just the clown before he grabs the person

Why is the mass of the clown before he grabs the person 2m?

- #29

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It isn't. That was an error.Why is the mass of the clown before he grabs the person 2m?

- #30

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Whoops sorry about that

- #31

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my max height will be 6m

- #32

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- #33

Mentor

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That's the correct speed of the clown just as she's about to grab the performer. (When she first reaches y = 4.5 m.) What's their speed after she grabs him?I accidentally divided instead,

my new answer is 15.35 m/s=v2

There's no need to calculate the highest point reached (but you can if you like), since what you want is their speed when they fall back to the starting point (y = 0).

- #34

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1/2mv^2=mgh

v=sqrt(gh2)

v=15.34

is this the right method?

v=sqrt(gh2)

v=15.34

is this the right method?

- #35

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No.1/2mv^2=mgh

v=sqrt(gh2)

v=15.34

is this the right method?

What is the speed of the pair right after the clown grabs the performer ?

Then what is the kinetic energy of the pair at that moment ?

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