# Homework Help: Momentum and energy disagreement

1. Mar 14, 2004

### ShawnD

A 30g bullet is fired horizontally into a 12kg block which is suspended on a long cord. What speed of bullet would cause the center of gravity of the block to rise 8cm?

energy approach

change of kinetic energy for bullet = change of potential for bullet and block

$$\frac{1}{2}mv^2 = mgh$$

$$\frac{1}{2}(0.03)v^2 = (12.03)(9.8)(0.08)$$

$$v = 25.08 \frac{m}{s}$$ <-------------------------------

momentum approach

change in kinetic energy (block and bullet) = change in potential (block and bullet)

$$\frac{1}{2}mv^2 = mgh$$

$$\frac{1}{2}(12.03)v^2 = (12.03)(9.8)(0.08)$$

$$v = 1.252$$

momentum of bullet = momentum of bullet and block

$$mv = mv$$

$$(0.03)v = (12.03)(1.252)$$

$$v = 502 \frac{m}{s}$$ <-------------------------------

Half the class thinks the answer is 25.08, the other half thinks it's 502. The answer in the back of the book is 502.

Why would the answer be 502? All of the energy in the system comes from the bullet, thus, the bullet's initial kinetic energy should equal the final potential energy of the block and the bullet. Where would all the energy go if this system is 100% efficient.

Look at the energy difference, assuming the answer is 502m/s.
Here is the energy of the bullet initially.

$$E = \frac{1}{2}(0.03)(502)^2$$

$$E = 3780 J$$

Here is the final energy of the block and bullet

$$E = (12.03)(9.8)(0.08)$$

$$E = 9.432 J$$

Where did the energy go???

2. Mar 15, 2004

### cookiemonster

The system isn't 100% efficient. Or else the bullet wouldn't stick.

cookiemonster

3. Mar 15, 2004

### ShawnD

Inefficiencies cannot account for a 99.75% loss of energy.

4. Mar 15, 2004

### cookiemonster

Sorry, "inefficient" isn't the correct word. "Inelastic" is a much better one.

A ball falling to the ground has some energy. Once it smacks into the ground and stops, it loses a lot more than 99.75% of its energy, even if you do consider the movement of the Earth.

cookiemonster

5. Mar 15, 2004

### outandbeyond2004

The block does not move vertically upwards. It moves like a pendulum bob. The equations in the energy approach look like they are for the vertical straight line rather than the pendulum bob.

The momentum approach works because you find out how fast the block is moving shortly after the bullet hits it. At that point it is almost all kinetic energy. When the block has risen 8 cm, it is all potential energy.

6. Mar 15, 2004

### HallsofIvy

outandbeyond2004: The only change in potential energy IS from vertical motion. That part was correct.

ShawnD: cookiemonster is correct: this was not an elastic collision so "energy" in the sense of only potential and kinetic energy is not conserved. There will be an increase in temperature in the block due to the energy the block absorbs stopping the bullet. And, yes, that can be a significant portion of the energy.

7. Mar 15, 2004

### Chi Meson

MOre noticeable than the block's increase in temperature would be the sound created by the bullet hitting the block. The heat and the sound are forms of energy and they must come from somewhere; they come from the initial KE of the bullet. The bullet's KE cannot both produce sound and heat and still be conserved to produce the equivalent PE.

Remember, there is no such thing as the "Law of Conservation of Kinetic Energy."

Rule of Thiumb: To find the velocity outcomes of any collision, you analyze conservation of momentum. In order to see what sort of change in speed or height or temperature an object will undergo, analyze the work and energies.

8. Mar 15, 2004

### outandbeyond2004

It is true that the only change in PE is from the vertical part of the block's motion. So?

If you slowly move the block vertically, there is almost no KE but a change in PE. That is not what happens in the problem -- the block has a lot of KE shortly after the bullet hits it.

The following alternative problem may help understand what is going on.

This problem is the same except the bullet is fired vertically upwards in line with the cord and the block's center of mass. The energy approach as given earlier will fail, again, whereas the momentum approach will work correctly. You have to find the velocity for the KE of the block shortly after the bullet hits it.

9. Mar 15, 2004

### outandbeyond2004

I forgot to add this. We are clearly expected to make some simplifying assumptions, such as to ignore the heat&sound generated in the bullet's collision with the block and the time needed to accelerate the block to its max KE.

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