Familiar? But confirmed correct one

[CartoonKid]

4. A 5.00 g bullet moving with an initial speed of 400 m/s is fired into and passes through
a 1.00 kg block (See figure).
400 m/s
v 5.00 cm
The block, initially at rest on a frictionless, horizontal surface is connected to a spring
with force constant 900 N/m. If the block moves 5.00 cm to the right after impact, find:
(a) the speed at which the bullet emerges from the block,
[Ans: 100 m/s]
(b) the mechanical energy converted into internal energy in the collision.
[Ans: 374 J]

http://server2.uploadit.org/files/ultimacarlos-untitled.JPG

This question is almost similar to the one asked by someone here. The different is that a spring has been added to the block. I have lost the solution given by my lecturer. I tried to do it with conservation of energy but it failed. I forgot about how to solve this question. Can somebody please help me?

 

[ehild]

4. A 5.00 g bullet moving with an initial speed of 400 m/s is fired into and passes through
a 1.00 kg block (See figure).
400 m/s
v 5.00 cm
The block, initially at rest on a frictionless, horizontal surface is connected to a spring
with force constant 900 N/m. If the block moves 5.00 cm to the right after impact, find:
(a) the speed at which the bullet emerges from the block,
[Ans: 100 m/s]
(b) the mechanical energy converted into internal energy in the collision.
[Ans: 374 J]

http://server2.uploadit.org/files/ultimacarlos-untitled.JPG

The problem can be solved by assuming that the bullet stays inside the block for a very short time so the spring can not deform and exert force on the block during the impact. In this case, the momentum is conserved.
Let be m the mass of the bullet, M that of the block, vo the speed of the bullet at the beginning, vf the speed of the bullet after inpact and V the speed of the block after impact.

mv_0=mv_f+MV \rightarrow V_f=v_0-M/m*V

You can apply conservation of energy for the motion of the block after the impact:

1/2 MV^2=1/2kx^2 \rightarrow V=\sqrt{k/M}*x

So

v_f=v_0-M/m*V=M/m*\sqrt{k/M}*x


ehild

 

[CartoonKid]

Thank you very much. This is the solution in my lost solution. I have a question here. What makes us to make such assumption? I mean by looking at the question at the first time, how do we know we have to make such an assumption. I really dislike the question which requires assumption, however, it's inevitable. Fortunately my tutor said our final exam wouldn't involve question which needs assumption. Anyway, I would like to know. Thanks.

 

[ehild]

What makes us to make such assumption? I mean by looking at the question at the first time, how do we know we have to make such an assumption. I really dislike the question which requires assumption, however, it's inevitable. Fortunately my tutor said our final exam wouldn't involve question which needs assumption. Anyway, I would like to know. Thanks.

The problem would be undetermined without this assumption. The momentum would not be conserved during the interaction of the block and bullet, as there is an external force exerted by the spring. Energy is not conserved, either, as this force of interaction certainly depends on the relative speed between bullet and block when the bullet travels inside the block. Newton's second law would yield one equation for the bullet and one for the block, but to solve them we would need the force of interaction.

You can find out from the text what you can suppose. There is no indication about the size of the block. Also, there is nothing about the time it stays inside.
Imagine that the bullet loses most of its momentum at entering the block. The block starts to move, the spring starts to get compressed, and decelerates the block, with the bullet still in. It is completely unsure if it gets out or not without knowing how long that block is.

Usually problems concerning collision of two bodies are solved in this way. We do not worry about the external forces during the interaction, assuming it happens in very short time, and we happily apply the law of conservation of momentum. Only after we got the new velocities we consider the effect of external forces. It is the same with problems about exploding rockets and asking where the pieces would fall. We use conservation of momentum as if gravity would not exist. But it does, even during the explosion, not only afterwards.

So this is only an approximation, but we never can solve a real problem without simplifying assumptions.

ehild

 

[arildno]

"So this is only an approximation, but we never can solve a real problem without simplifying assumptions."

Not only is it an approximation, but usually, a damned good one as well..
(Not to mention how simplifying it is..)

 

[roam]

And can anyone please explain how to solve part (b)? How do we find the mechanical energy converted into internal energy in the collision? Do we need to minus the final kenetic energy from the initial kenetic energy?

 

[ehild]

And can anyone please explain how to solve part (b)? How do we find the mechanical energy converted into internal energy in the collision? Do we need to minus the final kenetic energy from the initial kenetic energy?

Yes. The mechanical energy that was converted to heat is the difference between the initial KE of the bullet and the sum of the KE of both
the bullet and block just after the bullet emerged from the block.

ehild