Finding the energy lost in a collision as the bullet collides to the block

[Lolagoeslala]

Homework Statement

a 50 g bullet hits and becomes embedded in a wooden block of as 450 g which is at rest on a surface (μ = 0.41) and is attached to a spring bumper. The block and bullet compress the spring bumper a distance of 20 cm. Is the spring constant is 250 N/m
find initial speed of bullet? how much energy has been lost in the collision?

The Attempt at a Solution

a)
mbvb +mwvw=mtvc
(0.05 kg)vb=(0.5kg)(4.47m/s)
vb=44.7m/s

b)
ET2= 1/2(0.5 kg)(4.47 m/s)^2
ET2= 4.995 J

ET1= 1/2(0.05 kg)(44.7m/s)^2
ET1= 49.95 J

ET1-W=ET2
W= 44.957 J

How would i represent this in a percentage?

 

[frogjg2003]

Where did you get 4.47m/s?

 

[Lolagoeslala]

Where did you get 4.47m/s?

i calculated it...

1/2mvc^2=1/2kx^2
1/2(0.5kg)vc^2=1/2(250N/m)(0.2m)^2
0.25vc^2=5J
4.47m/s=vc^2

 

[frogjg2003]

You forgot about friction.

 

[Lolagoeslala]

You forgot about friction.

so umm...

1/2mvc^2 - Wf =1/2kx^2
1/2(0.5kg)vc^2 - (0.5kgx9.8m/s^2x0.41)=1/2(250N/m)(0.2m)^2
0.25vc^2 - (2.009J) =5J
5.29m/s=vc

IS THIS CORRECT??
but the speed has increased?

 

[frogjg2003]

That's the force of friction, you need to multiply by the distance traveled.
Yes, the speed has increased. Your original calculation only accounted for the energy in the spring, and not the energy lost to friction. More energy means a higher speed.

 

[Lolagoeslala]

That's the force of friction, you need to multiply by the distance traveled.
Yes, the speed has increased. Your original calculation only accounted for the energy in the spring, and not the energy lost to friction. More energy means a higher speed.

1/2mvc^2 - Wf =1/2kx^2
1/2(0.5kg)vc^2 - (0.5kgx9.8m/s^2x0.41x0.2m)=1/2(250N/m)(0.2m)^2
0.25vc^2 - (0.4018J) =5J
4.648m/s=vc

This is the right answer right?

 

[frogjg2003]

Yes.
Now, what is the momentum of the block and bullet together?

 

[Lolagoeslala]

Yes.
Now, what is the momentum of the block and bullet together?

i think you are trying to ask what is the final momentum well

mtvc
= (0.5kg)(4.648m/s)
=2.324 kgm/s

now when finding the speed of the bullet i am assuming you do this
mbvb +mwvw=mtvc
(0.05 kg)vb=(0.5kg)(4.648m/s)
vb= 46.48 m/s

 

[frogjg2003]

Right, and now you can determine the energy of the bullet before it stuck the block.

 

[Lolagoeslala]

Right, and now you can determine the energy of the bullet before it stuck the block.

so would you just do this??

ET2= 1/2(0.5 kg)(4.648 m/s)^2
ET2= 5.400976 kgm/s

ET1= 1/2(0.05 kg)(46.48m/s)^2
ET1= 54.00976 kgm/s

ET1-W=ET2
W= 48.608784 kgm/s

is this right??
hwow ould i represent this in percentage?

 

[frogjg2003]

Yes, that's right, but why would you need to represent it as a percentage?

 

[Lolagoeslala]

Yes, that's right, but why would you need to represent it as a percentage?

Oh wait i don't need to represent it as a percentage..
so this is right OMG! this is?? THANK YOU!