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

AI Thread Summary
A circus clown is launched vertically at 18.0 m/s and collides with a performer at 4.5 m height, leading to an inelastic collision. The discussion focuses on calculating the speed of the clown and performer duo when they return to the original launch height. Energy conservation principles are applied, with potential energy (PE) and kinetic energy (KE) equations being critical for solving the problem. The speed of the clown before grabbing the performer is calculated to be approximately 15.34 m/s, which is essential for determining their combined speed post-collision. The final calculations emphasize the importance of momentum conservation during the collision to find their speed at the original height.
Sneakatone
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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.
 
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The equation you used is for momentum, which shouldn't have to come into play here.

Try thinking about it in terms of energy
 
1/2m1v1^2=1/2m1v'1^2+1/2m2v2'^2
 
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?
 
should the second equation be m1v1+m2v2=m1v'1+m2v'1+m2v'2?
 
That is still momentum.

What is the equation for the kinetic energy of an object?
What about potential energy?
 
KE=1/2mv^2
PE=mv
 
almost. PE = mgh

try now
 
I would assume mass is 1 since they have the same mass so
1*9.81*4.5=44.145J
 
  • #10
That's a terrible idea!

use m for mass and see where you get
 
  • #11
PE=44.14m
 
  • #12
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?
 
  • #13
so can I set PE=KE to find mass(m)?
 
  • #14
You can set PE = KE, but look at what happens to the masses
 
  • #15
the masses cancel out,
 
  • #16
Exactly. Now you can find the height they are at when they start to fall down, and in turn the velocity
 
  • #17
Careful. When the clown grabs the other performer, energy is not conserved. (Treat that as an inelastic collision.)
 
  • #18
I did not understand what you said in the last part .@dinospamoni
 
  • #19
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
 
  • #20
V_2=sqrt[mgh/.5(2m]

H_2=(0.5*2m*v_3^3)/2mg
 
  • #21
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
 
Last edited:
  • #22
So we know what the mass is?
 
  • #23
Whoops forgot to include mass in a term on that post. Fixed now.

They still cancel though
 
  • #24
If the masses cancel then v2= 0.46
 
  • #25
That's not what I'm getting. Don't forget to subtract the g*h_1 from both sides
 
  • #26
I accidentally divided instead,
my new answer is 15.35 m/s=v2
 
  • #27
Nailed it. Now use that velocity in .5*2m*(v_2)^2 to find the maximum height they reach using 2mgh
 
  • #28
dinospamoni said:
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? :confused:
 
  • #29
coconut62 said:
Why is the mass of the clown before he grabs the person 2m? :confused:
It isn't. That was an error.
 
  • #30
Whoops sorry about that
 
  • #31
my max height will be 6m
 
  • #32
hm i got twice that. Not sure what happened. Anyway, now you can use the potential energy at that height to find the velocity at the original height
 
  • #33
Sneakatone said:
I accidentally divided instead,
my new answer is 15.35 m/s=v2
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?

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
1/2mv^2=mgh

v=sqrt(gh2)
v=15.34
is this the right method?
 
  • #35
Sneakatone said:
1/2mv^2=mgh

v=sqrt(gh2)
v=15.34
is this the right method?
No.

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 ?
 
  • #36
the speed after is 15.34m/s
how would I find KE without knowing what mass is?
 
  • #37
Sneakatone said:
1/2mv^2=mgh

v=sqrt(gh2)
v=15.34
is this the right method?
How did you get that speed? What does that speed mean?
 
  • #38
I got speed from using the height 12m in the equation 1/2mv^2=mgh to get v.

the speed means is that it is the velocity where from the original launch height.
 
  • #39
SammyS said:
No.

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 ?

Sneakatone said:
the speed after is 15.34m/s
how would I find KE without knowing what mass is?

Doc pointed out that 15.34m/s is the speed of the clown just before she grabs the performer.



All you know about mass is that the mass of the clown is the same as the mass of the performer. -- but that's enough.
 
  • #40
with that speed I found a height of 12m from using 1/2mv^2=mgh,
what am I suppose to do with that?
 
  • #41
Sneakatone said:
with that speed I found a height of 12m from using 1/2mv^2=mgh,
what am I suppose to do with that?
What speed are you referring to?

Please use the "Quote" feature of the Forum, or state the value you are referring to, directly in your post.
 
  • #42
the speed I am talking about is the one I solved v=15.34 (speed of the clown just as she's about to grab the performer)
 
  • #43
Sneakatone said:
the speed I am talking about is the one I solved v=15.34 (speed of the clown just as she's about to grab the performer)
If the clown had not grabbed the performer, but just continued to rise unimpeded, then yes, the maximum height the clown would have achieved would be 12 meters.

You still haven't computed the speed of the clown - performer combination immediately after the clown grabs the performer.

Here's a hint:

The momentum of the clown plus the performer is the same just before the grabbing and just after the grabbing.
 

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