- #1

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What I was initially using was:

Momentumi = Momentumf..

0 = m1v1-1.4m2v2 (x-comp)

v1 = 1.4v2

The thing is I'm just not sure what to do next..am I on the right path?

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- Thread starter MD2000
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- #1

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What I was initially using was:

Momentumi = Momentumf..

0 = m1v1-1.4m2v2 (x-comp)

v1 = 1.4v2

The thing is I'm just not sure what to do next..am I on the right path?

- #2

Doc Al

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- #3

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8200 = 1/2 m1v1^2 + 1/2(1.4m1)(1.4v1)^2

The thing is we would still have two unknowns..I dunno..theres soemthing I'm obviously not doing right..or can we use KEi = KEf?

- #4

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MD2000 said:

8200 = 1/2 m1v1^2 + 1/2(1.4m1)(1.4v1)^2

The thing is we would still have two unknowns..I dunno..theres soemthing I'm obviously not doing right..or can we use KEi = KEf?

You can't use [itex]KE_{i} = KE_{f}[/itex] here becuse this is essentially an inelastic collision (energy is not conserved) but what you are doing is [itex]Q = KE_{f} - KE_{i}[/itex] where Q is the explosion energy. Looks like you are missing something, the equations are correct.

- #5

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- #6

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MD2000 said:

In your case [itex]KE_{i} = 0[/itex], but I have written it to make sure you don't use conservation of kinetic energy which is not even applicable here.

[itex]KE[/itex] refers to the kinetic energy of both masses (before the explosion, the kinetic energy of the combined mass is zero). The system consists of the combined unexploded mass and hence (after the explosion has taken place) two two resulting masses which are born at the instant the explosion takes place.

- #7

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Q = KEf in this case..

8200J = 1/2m1v1^2 (block 1)

8200J = 1/2 (1.4m1)(1.4v1)^2 (block 2)

Am I looking at this right?

Or would it be

8200 = 1/2 (m1 + 1.4m1)(v1 + 1.4v1) ^ 2

Regardless..wouldn't we still be left with two unknowns?

- #8

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You would still have two unknowns, but thats because this is an ill-posed problem. You should still apply the correct equations though.

- #9

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Written seperately..

8200 J = 1/2m1v1^2 + 1/2m2v2^2

8200 J = 1/2m1v1^2 + 1/2(1.4m1)(1.4v2)^2

Wouldn't that be the sum of the two?

- #10

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2190.17 = 1/2m1v1^2

So 8200 - 2190.17 = 6009.83 for the larger block? I dunno..I think I jus made that all up..lol

- #11

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any help guys?

- #12

Doc Al

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This is correct.MD2000 said:Wouldnt the second be the sum?

Written seperately..

8200 J = 1/2m1v1^2 + 1/2m2v2^2

This is not. (You misapplied your result from the momentum equation.)8200 J = 1/2m1v1^2 + 1/2(1.4m1)(1.4v2)^2

Better yet is to find the relationship between the two KEs:

KE1 = 1/2m1v1^2

KE2 = 1/2m2v2^2 = 1/2(1.4m1)(v1/1.4)^2 = (1/1.4) KE1

- #13

arildno

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Just PRIOR to the explosion, the object contains 8200J of CHEMICAL energy [itex]E_{i.c}[/itex]that is going to be converted into KINETIC energy through the explosion. Thus, we have:

[tex]E_{i.c}=\frac{1}{2}(m_{1}\vec{v}_{1}^{2}+m_{2}\vec{v}_{2}^{2})[/tex]

Furthermore, we have conservation of momrentum:

[tex]m_{1}\vec{v_{1}}=-m_{2}\vec{v}_{2}\to\vec{v}_{1}=-\frac{m_{2}}{m_{1}}\vec{v}_{2}[/tex]

The RATIO of their aquired energies are therefore:

[tex]\frac{E_{1}}{E_{2}}=\frac{m_{1}}{m_{2}}\frac{\vec{v}_{1}^{2}}{\vec{v}_{2}^{2}}=\frac{m_{2}}{m_{1}}=1.4[/tex]

Thus, we get:

[tex]8200=E_{1}+E_{2}=2.4E_{2}[/tex]

- #14

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Or is it 2.4E2 = 8200 which is 3416.67..not exactly sure how u got the 2.4 tho

I really appreciate your help guys..srry for being an idiot..lol

- #15

arildno

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[tex]E_{1}=\frac{1}{2}m_{1}\vec{v}_{1}^{2},E_{2}=\frac{1}{2}m_{2}\vec{v}_{2}^{2}[/tex]

Thus, we may rewrite our original equation as:

[tex]E_{i.c}=E_{1}+E_{2}[/tex]

Get that?

As for "why doing the ratio?" remember that if we know:

[tex]\frac{E_{1}}{E_{2}}=1.4[/tex]

then we must have:

[tex]E_{1}=1.4E_{2}[/tex]

- #16

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Actually Im srry I got that part..I wasnt sure about the 2.4E2 tho?

- #17

arildno

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Look at my edit, and figure that last bit out for yourself.

- #18

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so..

2.4E2 = 8200

E2 = 3416.67

- #19

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EDIT: Oh I just saw Arildno said just what I came to say here :rofl:

- #20

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MD2000 said:An explosion breaks an object into two pieces, one of which has 1.4 times the mass of the other. If 8200 J were released in the explosion, how much kinetic energy did the heavier piece acquire?

QUOTE]

The problem is, is the velocity of 2 objects same or different? If it is the same then it is easier because the ratio of the 2 energies is related to the ratio of the 2 masses. Somehow I feel this question is incomplete in the first place.

If we assume both objects have equal final velocity, then:

0.5m(v^2) + 0.5(1.4m)(v^2) = 8200

- #21

Doc Al

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Then you've missed the point of the problem, which is to realize that momentum is conserved in an explosion.thiotimoline said:If we assume both objects have equal final velocity, then:

- #22

Tom Mattson

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