### Brain Teasers

# Two Bullets

A piece of firewood is standing on end. Which is more likely to knock the piece of wood over - a lead bullet fired from a gun or a rubber bullet of the same mass travelling at the same speed (and hitting the same spot)?

Assume that the rubber bullet does not disintegrate on impact.

Assume that the rubber bullet does not disintegrate on impact.

### Hint

The rubber bullet will ricochet off of the wood - what difference does this make?### Answer

The rubber bullet will be more likely to knock the wood over. The lead bullet will embed itself in the wood, so the wood will have to absorb enough energy to stop the bullet's momentum. The rubber bullet will ricochet off of the wood, so the wood will have to absorb enough energy to stop the bullet PLUS send the bullet back in the opposite direction. The rubber bullet will supply more total force to the wood and will therefore be more likely to knock it over.Hide Hint Show Hint Hide Answer Show Answer

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## Comments

Good one..

I knew the answer but i didn;t know how to explain it scientifically.

I knew the answer but i didn;t know how to explain it scientifically.

Very good description!!

Is that Newton's 3rd law? For every action there is an equal and opposite re-action?

I though that was Newton's first law?

I'm with AbdulAziz, I knew the answer, just not how to explain it!

Newton's first law is INERTIA

This is good teaser. I've tried to explain the concept to my dad but he didn't believe me.

You learn something new every day

Good one

Good one

for this to be true the bullet would have to hit straight on and not at an angle.

here's an example that is fairly common knowledge. you take a bat or other large object and try to break a wooden board is the board easier to break hitting straight on or at an angle?

of course its easier to break hitting straight on. when it hits at an angle not all of the energy is transfered straight into the board its deflected instead of absorbed.

here's an example that is fairly common knowledge. you take a bat or other large object and try to break a wooden board is the board easier to break hitting straight on or at an angle?

of course its easier to break hitting straight on. when it hits at an angle not all of the energy is transfered straight into the board its deflected instead of absorbed.

The problem with your reasoning is that the rubber bullet ricochets off of the wood not because the wood has to absorb more energy, but because energy is REFLECTED by the wood. Newton's 3rd law still illustrates why the wood is more likely to be knocked over, but the rubber bullet, if it has the same mass and momentum, hits the wood with the same amount of energy as the lead bullet. The difference is that more of the kinetic energy is transformed into potential energy and back.

I agree with tangled brain. Your reasoning is basically sound, but the terminology is incorrect.

The just-ricocheted rubber bullet will have have MORE kinetic energy than the just-embedded lead bullet. However, the rubber bullet will impart more momentum to the log.

How does this work? It's because kinetic energy is a scalar quantity and momentum is a vector. Vectors have direction. The momentum of the just-struck log will have more momentum than the rubber bullet started with, because the richoeted bullet has negative momentum.

Mathematically, it looks like this:

Let's say the bullet weighs .007 kg and strikes a 10kg log at 300 meters/second.

The initial kinetic energy of the bullet is 1/2 (mass * velocity^2)... 315 joules in this case.

The initial momentum of the bullet is mass * velocity... 2.1 kg-m/s in this case.

The collision of the rubber bullet, according to the riddle, is elastic. BOTH momentum and energy must be convserved. This leaves only ONE possible solution for the resulting bullet and log velocities:

Rubber bullet after collision:

. . . . . Mass = .007 kg

. . . Velocity = -299.58 m/s (yep, that's a negative velocity. Velocity is a vector)

Kinetic energy = 314.119 joules (it kept almost all of it)

. . . Momentum = -2.0971 kg-m/s (note, once again, a negative number)

Log after collision:

. . . . . Mass = 10 kg

. . . Velocity = .4197 m/s

Kinetic energy = .88074 joules

. . . Momentum = 4.1971 kg-m/s

Add the two momentums (to double-check) and you get the initial momentum: 4.1971 + (-2.0971) = 2.1

By comparison, the lead bullet does NOT experience an elastic collision. It couldn't have, because it's been embedded in the log. So we need calculate only the velocity of the bullet-log system.

Bullet and log together:

Mass = 10kg + .007kg = 10.007kg

Momentum = 2.1 kg-m/s (exactly the same as the bullet's prior to embedding)

Velocity = 2.1 / 10.007 = .2099 m/s

Q.E.D., the lead bullet imparts only about half as much momentum as the rubber bullet.

You might be wondering...

The kinetic energy of the log-bullet system is very low: 1/2 [10.007kg * (2.099m/s)^2] = .22044 joules. Where's the missing energy? Lead bullets are designed to deliver a large amount of energy to the interior of a target, and that's exactly what it did. It tore a whole in the log, sending dust and tiny splinters in all directions (net momentum = zero, but lots of kinetic energy); these splinters in turn ripped apart other sections of the log, and after a second or two it all decayed to the microscopic level where it is now measured as heat energy.

[Note for the curious: A large amount of destructive kinetic energy translates to a very small amount of heat energy. It would take about 15 bullets to raise the average temperature of a 10kg log by only one degree centigrade.]

The just-ricocheted rubber bullet will have have MORE kinetic energy than the just-embedded lead bullet. However, the rubber bullet will impart more momentum to the log.

How does this work? It's because kinetic energy is a scalar quantity and momentum is a vector. Vectors have direction. The momentum of the just-struck log will have more momentum than the rubber bullet started with, because the richoeted bullet has negative momentum.

Mathematically, it looks like this:

Let's say the bullet weighs .007 kg and strikes a 10kg log at 300 meters/second.

The initial kinetic energy of the bullet is 1/2 (mass * velocity^2)... 315 joules in this case.

The initial momentum of the bullet is mass * velocity... 2.1 kg-m/s in this case.

The collision of the rubber bullet, according to the riddle, is elastic. BOTH momentum and energy must be convserved. This leaves only ONE possible solution for the resulting bullet and log velocities:

Rubber bullet after collision:

. . . . . Mass = .007 kg

. . . Velocity = -299.58 m/s (yep, that's a negative velocity. Velocity is a vector)

Kinetic energy = 314.119 joules (it kept almost all of it)

. . . Momentum = -2.0971 kg-m/s (note, once again, a negative number)

Log after collision:

. . . . . Mass = 10 kg

. . . Velocity = .4197 m/s

Kinetic energy = .88074 joules

. . . Momentum = 4.1971 kg-m/s

Add the two momentums (to double-check) and you get the initial momentum: 4.1971 + (-2.0971) = 2.1

By comparison, the lead bullet does NOT experience an elastic collision. It couldn't have, because it's been embedded in the log. So we need calculate only the velocity of the bullet-log system.

Bullet and log together:

Mass = 10kg + .007kg = 10.007kg

Momentum = 2.1 kg-m/s (exactly the same as the bullet's prior to embedding)

Velocity = 2.1 / 10.007 = .2099 m/s

Q.E.D., the lead bullet imparts only about half as much momentum as the rubber bullet.

You might be wondering...

The kinetic energy of the log-bullet system is very low: 1/2 [10.007kg * (2.099m/s)^2] = .22044 joules. Where's the missing energy? Lead bullets are designed to deliver a large amount of energy to the interior of a target, and that's exactly what it did. It tore a whole in the log, sending dust and tiny splinters in all directions (net momentum = zero, but lots of kinetic energy); these splinters in turn ripped apart other sections of the log, and after a second or two it all decayed to the microscopic level where it is now measured as heat energy.

[Note for the curious: A large amount of destructive kinetic energy translates to a very small amount of heat energy. It would take about 15 bullets to raise the average temperature of a 10kg log by only one degree centigrade.]

The author sort of slides over the point that the "bullets" are the same mass. Of course, this means that the rubber bullet it 3 or 4 times the size of the lead bullet.

"Bullet" is too generic a term, as not every bullet will "embed" itself in the firewood. A hollow-point bullet, designed for expansion, will simply mushroom on impact and knock the wood over without penetration. A frangible bullet with enough mass will disintegrate upon impact, but will still impart enough inertia on the wood to knock it over.

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