Momentum: explosion of a block

[huynhtn2]

Homework Statement

A block of mass 8.45 kg in outerspace is moving at 2.47 m/s with no external forces acting on it. After an explosion, the block is split into two parts both having mass equal to half the mass of the original block. The explosion supplies the two masses with an additional 19.3 J of kinetic energy. Neither mass leaves the line of original motion. Calculate the magnitude of the velocity of the mass that is moving at a greater velocity.

Homework Equations

ke=(1/2)mv^2
p=mv

The Attempt at a Solution

(1/2)(8.45)(2.47)^2 =ke
9.28 joules = ke

What should i do?

 

[zhermes]

How about:
conservation of energy, i.e. the initial energy = the final energy + 19.3 J; how can that help you?

 

[huynhtn2]

How about:
conservation of energy, i.e. the initial energy = the final energy + 19.3 J; how can that help you?

9.28 =(1/4)mv^2 +(1/4)mv^2 +19.3

 

[zhermes]

9.28 =(1/4)mv^2 +(1/4)mv^2 +19.3

Good, but remember that the velocities are different, i.e. they should be [tex]v_1[/tex] and [tex]v_2[/tex].

Now, you also know conservation of momentum: initial momentum = final momentum. This gives you another equation.
At this point you should have two equations with two unknowns (the two velocities), so you can solve them.

 

[huynhtn2]

Good, but remember that the velocities are different, i.e. they should be [tex]v_1[/tex] and [tex]v_2[/tex].

Now, you also know conservation of momentum: initial momentum = final momentum. This gives you another equation.
At this point you should have two equations with two unknowns (the two velocities), so you can solve them.

can you please show me how to do this. i would be very grateful.

 

[zhermes]

Set up the two equations:
You have conservation of energy, [tex]E_{initial} + 19.3 = E_{final}[/tex] and conservation of momentum, [tex]p_{initial} = p_{final}[/tex].

When you write them out, you'll see you have only two unknowns: the velocities. If you have trouble solving them, post them here and I'll give you some pointers.

 

[huynhtn2]

Set up the two equations:
You have conservation of energy, [tex]E_{initial} + 19.3 = E_{final}[/tex] and conservation of momentum, [tex]p_{initial} = p_{final}[/tex].

When you write them out, you'll see you have only two unknowns: the velocities. If you have trouble solving them, post them here and I'll give you some pointers.

energy
(1/2)(8.45)(2.47)^2+19.3= (1/4)mv1^2+(1/4)mv2^2

momentum
(8.45)(2.47)=(4.225)v1+(4.225)v2

 

[zhermes]

Awesome. In the momentum equation, solve the equation for either of the velocities---then plug that result into the energy equation. This will eliminate one of the variables, and you can find the other (remember to plug in the block mass for m). You can then plug back into either of the original equations to find the other velocity.

 

[huynhtn2]

Awesome. In the momentum equation, solve the equation for either of the velocities---then plug that result into the energy equation. This will eliminate one of the variables, and you can find the other (remember to plug in the block mass for m). You can then plug back into either of the original equations to find the other velocity.

from momentum
4.95-v1=v2

energy
21.35 = v1^2 +v2^2
4.62 = v1+v2
4.62=v1+(4.94-v1)

i did something wrong

 

[zhermes]

21.35 = v1^2 +v2^2
4.62 = v1+v2

NO!
[tex] (a+b)^2 \neq a^2 + b^2[/tex]
i.e.
[tex] \sqrt{a^2 + b^2} \neq a+b [/tex]

 

[huynhtn2]

NO!
[tex] (a+b)^2 \neq a^2 + b^2[/tex]
i.e.
[tex] \sqrt{a^2 + b^2} \neq a+b [/tex]

err... 4.62=v1^2+(4.9-v1)
0 = v1^2-v1+0.32

there is a negative under the square root when i use quad formula

 

[huynhtn2]

err... 4.62=v1^2+(4.9-v1)
0 = v1^2-v1+0.32

there is a negative under the square root when i use quad formula

Thank you su much for you help, i got the answer! you are good at helping!

 

[zhermes]

:D thanks, happy to help