Calculating Kinetic Energy of an Exploding Baseball

[Robertoalva]

exploding baseball

1. A 150 g trick baseball is thrown at 59 km/h. It explodes in flight into two pieces, with a 41 g piece continuing straight ahead at 85 km/h. How much energy do the pieces gain in the explosion?

Homework Equations

The Attempt at a Solution

is it asking for kinetic energy? or am I wrong?

 

[ChiralWaltz]

What indicates it is kinetic energy?

 

[Robertoalva]

the gain in motion ? it was accelerating at 59 km/h and the at 85 km/h

 

[ChiralWaltz]

Lets just say kinetic energy is defined by "motion" at this point. Do you have a formula for kinetic energy?

 

[jedishrfu]

okay so it looks like one piece is traveling in the original direction with an increased velocity and the second piece must be traveling in the oposite direction. The 41g piece has a delta velocity of 85-59=16 km/h and from that you can get its KE.

For the 150-41g piece you could assume it got the same amount of energy and from that compute its delta velocity to subtract from the 59km/h original velocity.

I'd also convert everything to MKS units for consistency ie no kilometers use meters, no hours use seconds and no grams use kilograms.

 

[Robertoalva]

so basically, i have to get the KE of the whole thing and then get the KE of the 59km/h and subtract it from the whole thing? It should give me the same for both pieces right?

 

[Robertoalva]

the formula for kinetic energy is KE= 1/2 m v^2 right?

 

[Robertoalva]

1. A 150 g trick baseball is thrown at 56 km/h. It explodes in flight into two pieces, with a 38 g piece continuing straight ahead at 85 km/h. How much energy do the pieces gain in the explosion?

Homework Equations

KE= 1/2 m v^2

The Attempt at a Solution

converted everything to meters kg and seconds then i got the kinetic energy of the 38g piece and got the velocity of the other piece, this is where i got lost.

 

[ChiralWaltz]

Something that helps me work physics problems is to write down all of the given information before I start working on the problem. For your example, I would write:

Initial Values
m_i=150 g
v_i=59 km/hr

Final Values (2 pieces)
m_f=41 g
v_f=85 km/hr

m_f= 109 g
v_f= ?


Then I would look at my "cheat sheet", where I have written all these formulas down to reference quickly while I learn the material. Notice kinetic energy use m and v in the formula and the ball is in motion and energy is the desired answer. Another one I see is the conservation of momentum.

I write formulas I'm thinking about:
KE = (1/2)mv²
ρ = mv (ρ=momentum)

Do you understand the concept of conservation of momentum?

 

[TSny]

Can you show how you got the velocity of the other piece?

 

[Robertoalva]

v=sqrt( (1/2 m)/KE)

 

[TSny]

How did you get a value for KE to plug into this equation? Can you state what physics principle you are using? (Edit:The equation v = sqrt((1/2 m)/KE) isn't correct.)

 

[Robertoalva]

first I did the 38g piece and it gave me 1.23 or something near that, and I assume it is the same for the other piece of the ball.

 

[TSny]

How did you get the number 1.23? What does the number 1.23 represent? It will be very helpful if you can show the individual mathematical steps.

 

[Robertoalva]

KE = 1/2(.038g)(8.05m/s)^2 = 1.23J

 

[TSny]

KE = 1/2(.038g)(8.05m/s)^2 = 1.23J

OK. The units for the .038 should be kg, right? The speed of 8.05 m/s is not correct. Can you show how you converted 85 km/h to m/s?

 

[Robertoalva]

sorry, i got confused hahah it should be like this

KE= 1/2(.038kg)(23.611)^2 = 10.59J

 

[Robertoalva]

now I think what is my mistake, 85000m/3600s= 23.611m/s

 

[TSny]

KE= 1/2(.038kg)(23.611)^2 = 10.59J

Right. So, that's the KE of the 38 g piece. If you could figure out the KE of the other piece, then you could find the gain in KE of the system due to the explosion. You're going to need the velocity of the other piece. Can you think of a way to get it?

 

[Robertoalva]

the concept of conservation of momentum, is the sum of the momentum on both objects?

 

[Robertoalva]

v= sqrt( (1/2m)/KE)

 

[TSny]

No, this is not a correct equation. But, also, even if it were correct, you don't know the KE of the second piece. If you did, you wouldn't need to find the velocity in order to answer the question about the KE.

You'll be able to find the KE of the second piece using KE = (1/2)mv². But first you must find the velocity v of the second piece. To do this, you will need to realize that something is conserved in the explosion. (It is not KE that is conserved, because we know that the question is stating that the KE of the system increases due to the explosion.)

 

[Robertoalva]

oh! momentum! p = m v ?

 

[TSny]

Good. So, see if you an use that idea to find v for the second piece.

 

[ChiralWaltz]

Could you clarify with an equation? I'm not sure I understand what you are saying. Please go check out hyperphysics webpage http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/hph.html

conservation of momentum is under mechanics
the laws of thermodynamics are under heat and thermodynamics

 

[Robertoalva]

so i have to get the momentum on the whole thing and then in the 38 g piece and after that subtract it?

 

[TSny]

Yes, What do you get?

 

[Robertoalva]

to get the velocity of the big piece do i have to use this formula?

vf1= ((m1-m2)/(m1+m2))(v1i)+ (2m2/(mi+m2)vi2

 

[Robertoalva]

also, is mass in grams or in kilograms?

 

[TSny]

to get the velocity of the big piece do i have to use this formula?

vf1= ((m1-m2)/(m1+m2))(v1i)+ (2m2/(mi+m2)vi2

No, that equation is not relevant to this problem. Try to set up the conservation of momentum relation on your own.

How much momentum is there before the explosion? What direction is this momentum?

How much momentum does the small piece have after the explosion? What direction is this momentum?