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Hey all--

I had a test yestrerday dealing with momentum, energy convservation, collision, etc, and there was one question that was marked incorrect, however I feel that I may have done it correctly. Please check over my work against my teacher's solution to see what you guys think...

anyway here's the question:

EDIT: I made a visual for this (dont criticize i did it in MsPaint)

http://eric.segonline.net/physicsexample.jpg [Broken]

A 2KG block and an 8-KG block are both attached to an ideal spring (k=200 N/m) and are both initially at rest on a horizontal firctionless surface. A 100-gram ball of clay is thrown at the 2KG block, and the velocity of the clay-block system is 3.9 m/s immediately after contact, and before the spring begins to compress.

What is the final velocity of the 8kg block once the spring compresses and then regains its original length.

My method:

Energy and Momentum are conserved:

.5mv^2=.5mv^2 and mv(f) = mv(i)

combining the 2 gives the equation:

V(1F) = (M1-M2)/(M1+M2) * V(1I)

and

V(2F) = 2M1/(M1+M2) * V(1I)

Note-- this derivation works for elastic collisions, combining both kinetic energy AND momentum conservation.

From this, I find V(2)= 1.62 m/s <--- final velocity of 8kg block

in addition, V(1) = -2.28 m/s <--- final velocity of 2kg block/clay

If you stick these back into the original equations, you will find that momentum is conserved from the original clay/2KG block system, and the initial kinetic energy is the same.

However, my teacher provided this response:

K(i)=V(f)+(K(f)

When spring returns to original length:

V(8KG block) = V(2kg block/clay)

.5(M(clay/2kg block))v^2 = 0 + .5(M(clay/2kgblock) + M(8kgblock))v^2

from this, he finds the final velocity of the 8kg block to be 1.78 m/s. In addition, he claims that when the spring is not compressed, it acts the same way as a rope, meaning velocities are the same.

Which one is more accurate?

Thanks for any input.

Eric

I had a test yestrerday dealing with momentum, energy convservation, collision, etc, and there was one question that was marked incorrect, however I feel that I may have done it correctly. Please check over my work against my teacher's solution to see what you guys think...

anyway here's the question:

EDIT: I made a visual for this (dont criticize i did it in MsPaint)

http://eric.segonline.net/physicsexample.jpg [Broken]

A 2KG block and an 8-KG block are both attached to an ideal spring (k=200 N/m) and are both initially at rest on a horizontal firctionless surface. A 100-gram ball of clay is thrown at the 2KG block, and the velocity of the clay-block system is 3.9 m/s immediately after contact, and before the spring begins to compress.

What is the final velocity of the 8kg block once the spring compresses and then regains its original length.

My method:

Energy and Momentum are conserved:

.5mv^2=.5mv^2 and mv(f) = mv(i)

combining the 2 gives the equation:

V(1F) = (M1-M2)/(M1+M2) * V(1I)

and

V(2F) = 2M1/(M1+M2) * V(1I)

Note-- this derivation works for elastic collisions, combining both kinetic energy AND momentum conservation.

From this, I find V(2)= 1.62 m/s <--- final velocity of 8kg block

in addition, V(1) = -2.28 m/s <--- final velocity of 2kg block/clay

If you stick these back into the original equations, you will find that momentum is conserved from the original clay/2KG block system, and the initial kinetic energy is the same.

However, my teacher provided this response:

K(i)=V(f)+(K(f)

When spring returns to original length:

V(8KG block) = V(2kg block/clay)

.5(M(clay/2kg block))v^2 = 0 + .5(M(clay/2kgblock) + M(8kgblock))v^2

from this, he finds the final velocity of the 8kg block to be 1.78 m/s. In addition, he claims that when the spring is not compressed, it acts the same way as a rope, meaning velocities are the same.

Which one is more accurate?

Thanks for any input.

Eric

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