# An explosion breaks an object into two pieces

1. May 31, 2004

### pupatel

Can someone help me with this...what is the answer to this and how do i get it?????????

An explosion breaks an object into two pieces, one of which has 1.5 times the mass of the other. If 7800 J were released in the explosion, how much kinetic energy did each piece acquire?

2. May 31, 2004

### Janitor

Note that since there are only two pieces, you can get by without explicit use of vectors for the momenta.

From the point of view of the center of mass, conservation of momentum implies

m1 v1 = m2 v2

or, substituting what you know about masses and taking m1 to be the lesser,

m1 v1 = 1.5 m1 v2

where I am letting each speed be a positive number, since I am not using vector language. You can divide that last equation through by m1 to get that v2 = (2/3) v1.

Kinetic energy, again from the center of mass point of view, is

(1/2) m1 v1^2 + (1/2) m2 v2^2 = 7800.

Subtituting again,

(1/2) m1 v1^2 + (1/2)(3/2) m1 [(2/3) v1]^2 = 7800,

which means

KE1 + (1/3) KE1 = 7800.

So KE1 = (3/4) 7800, and that means KE2 = (1/4) 7800. Take it from here with your calculator.

Last edited: May 31, 2004
3. Jun 1, 2004

### wisky40

I agree with Janitor in almost everything; but I get: KE1+(2/3)KE1=Total energy => KE1=(3/5) of total energy.

4. Jun 1, 2004

### Janitor

Wisky40, I looked back through what I wrote and I can't find an error--but that doesn't mean I am correct! I would advise pupatel to go through each step on his/her own and not take anything for granted.

5. Jun 1, 2004

### wisky40

I think EK1(1+(3/2)(2/3)^2)=EK1(1+(2/3))=EK1(5/3).

6. Jun 1, 2004

### Chen

No, it means KE1 + (2/3) KE1 = 7800. I suspect you multiplied 3/2 by (2/3)2 but then divided it by 2, which you shouldn't do because 1/2 is part of the KE expression.

7. Jun 1, 2004

### Janitor

Yep, that's what I did. You guys are completely correct. My mistake!