What Is the Speed of the Clown and Performer When They Reach the Launch Height?

[Sneakatone]

A circus clown in a cannon is shot vertically upward with an initial speed of 18.0 m/s. after ascending 4.5 m ,she collides with and grabs a performer sitting still on a trapeze. They acend together and then fall. what is their speed when they reached the original launch height. The clown and trapee have the same mass.

I tried using m1v1=m2v2
where 1*18=x*1 which is not right.
I do not know what to do with the height.

 

[dinospamoni]

The equation you used is for momentum, which shouldn't have to come into play here.

Try thinking about it in terms of energy

 

[Sneakatone]

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

 

[dinospamoni]

Sort of on the right track. Break it down into steps:

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?

 

[Sneakatone]

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

 

[dinospamoni]

That is still momentum.

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

 

[Sneakatone]

KE=1/2mv^2
PE=mv

 

[dinospamoni]

almost. PE = mgh

try now

 

[Sneakatone]

I would assume mass is 1 since they have the same mass so
1*9.81*4.5=44.145J

 

[dinospamoni]

That's a terrible idea!

use m for mass and see where you get

 

[Sneakatone]

PE=44.14m

 

[dinospamoni]

Side note: The units of energy are Joules, the units of distance m. Don't just guess!

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?

 

[Sneakatone]

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

 

[dinospamoni]

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

 

[Sneakatone]

the masses cancel out,

 

[dinospamoni]

Exactly. Now you can find the height they are at when they start to fall down, and in turn the velocity

 

[Doc Al]

Careful. When the clown grabs the other performer, energy is not conserved. (Treat that as an inelastic collision.)

 

[Sneakatone]

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

 

[dinospamoni]

Right. I completely forgot about that Doc Al.

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

 

[Sneakatone]

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

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

 

[dinospamoni]

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

 

[Sneakatone]

So we know what the mass is?

 

[dinospamoni]

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

They still cancel though

 

[Sneakatone]

If the masses cancel then v2= 0.46

 

[dinospamoni]

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

 

[Sneakatone]

I accidentally divided instead,
my new answer is 15.35 m/s=v2

 

[dinospamoni]

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

 

[coconut62]

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?

 

[Doc Al]

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

It isn't. That was an error.

 

[dinospamoni]

Whoops sorry about that