# Can You Solve This Physics Brain Teaser of the Bullet-Block Experiment? [Video--UPDATED]

A wooden block is shot, first dead center and then off-center, which imparts a spin on the block. Which goes higher, and why?

The science video blog Veritasium offers this nifty little puzzle. Does a wooden block, when shot by a vertically aimed rifle, go higher when it’s hit dead center or off-center? In the latter case, the bullet causes the block to spin. Several science communicators are asked the question in this set-up video:

Did you pick an outcome? If not, don’t worry—here’s the result of the experiment:

UPDATE (9/3/13): Derek Muller, the creator of Veritasium, has posted the answer. Short version: commenter number 1 below, bokubob, is correct. But watch the video for some surprising details and another challenge for experimenters:

Read SA blogger Carin Bondar's interview with Derek Muller here, from 2012.

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1. 1. bokubob 06:50 PM 8/27/13

In both cases, the bullet is lodged in the block, so you have an inelastic collision. Momentum is conserved, so they go to the same height. However, the kinetic energy is not conserved. When the bullet goes into the center of the block, the excess kinetic energy goes into the destruction of the block and bullet and heat. When the bullet goes off to the side, some of the energy goes into the rotation, and less goes into destruction and heat. In short, the bullet doesn't go as far into the block.

2. 2. findmitch 07:12 PM 8/27/13

In both cases, the kinetic force is transferred through the nail at the center of the mass of the blocks. The spinning forces are most likely conserved from the bullet in that less energy was used to stop the bullet. I'd also see how it may be possible that any loss of energy transfer in upward lift was offset by the comparative downward atmospheric pressure on the spinning block, assuming that a spinning block was diffuse downward pressure relatively easier than a block with a straight leading edge during ascent.

3. 3. m 08:36 PM 8/27/13

Defects in the wood, production, and bullet aside...

Im going to go with this, the wooden block at that speed behaves in an elastic way, such that it when its shot its not moving in a circular arc but in an elipse, which is shortening with each rotation until its a perfect circle at the top. This elastic energy is essentially propelling the object up.

4. 4. c laird478 09:58 PM 8/27/13

I suspect that 100% of the kinetic energy of the bullet was imparted to the block in both instances. In one instance 100% of the kinetic energy went into lifting the block into the air. In the other instance perhaps 99.9% of the kinetic energy of the bullet went into lifting the block into the air, and the other 0.1% went into setting the block into rotation.

The reason that the block appeared to rise to the same height in both instances is because the difference between the 100% and 99.9% was too slight to be measurable at the scale that was visible in the video. The block was a fairly short block so wherever it was struck off center couldn't be very far off center, thus the overwhelming majority of the kinetic energy of the bullet went into lifting the block. The portion going into its rotational energy was really very minor. It doesn't take a lot of energy to rotate a very short block.

If the experiment was repeated with a longer block, say a meter long, and the bullet struck it close to one edge away from center, the difference would be more detectable at the scale visible in the video. It would take more of the kinetic energy of the bullet to set a meter long block into rotation than it would the short block in the video.

5. 5. Bionate 10:01 PM 8/27/13

Well physics was never my strongest subject, but I noticed that the block that was hit directly in the center rotated about the z-axis. So I think that the rotational energy from the bullet was Essentially transferred from the bullet into the block. Because of the increased surface area of the side of the block there would be more resistance to rotation and that could account for the apparent gain in energy.

6. 6. Maiandra 11:10 PM 8/27/13

Since there is conservation of energy, the premise that the spinning block has "more" energy because it is spinning is false. It simply has some of its energy put into rotation. I like bokubob's idea of the non-spinning block getting warped more, it's just like in boxing: if you roll with a punch, it has less impact.

7. 7. ChazInMT 01:41 AM 8/28/13

For starters, this "Experiment" is seriously lacking in precision. I'm all in with c laird and the explanation there about 99% & 100%, BUT, when I made screen captures of the videos and carefully analyzed them, you can certainly see 2 issues that make the "Exact Same Height" statement flat out embarrassing to the presenters.

The block on the left when frozen in several different screen caps while at acme, show that it is clearly higher by 3% or so when you look at the center of mass of the block and not just where the nail goes into the block as they did. All 4 corners of the left block are clearly above the imposed line while the right block is bifurcated equally by the line. The nail on the right block is dead center for the center of mass, but on the left the nail is below the center of mass.

BUT WAIT!! There's MORE! The pictures do not scale the same, when measured, the left frame is further away by 1% or so, introducing a bias to make it look lower.

So it appears that 4% of the energy went into spinning the block on the right, and the block on the left is indeed higher.

I really can't believe you went through all this trouble, spending a ton of time, money, & effort, only to make a claim that is clearly refuted by your very own video. They ARE NOT the same height.

8. 8. GeekStatus in reply to c laird478 03:46 AM 8/28/13

To my eye it looks like the block struck in the center actually went higher.

My guess would be that the ultimate balance is between the extra drag by the center struck block and the energy required to spin the off center block.

Claird and ChazinMT are getting it right.

Great experimental design, btw, haha. The old eyeball test to measure the dependent variable. 755k people subscribe to that? lol.

9. 9. stargene 03:47 AM 8/28/13

Okay. Just guessing. The first, dead center action
gives the block an energy E, but that block encounters
drag resistance; call that Dr. E1 - Dr gets the block
to a height h. The second, off-center action gives
the block less energy E2 in the upward z-direction, but
its spin may interfere with (partially neutralize)
Dr2. So E1 - Dr = E2 - Dr2, so they reach the same
height h? Or not. :-)

10. 10. gnuevo 03:54 AM 8/28/13

I agree with ChazinMT!
If both blocks achieve the same high they should arrive to the floor at the same moment. However, if you see again the video the "rotating" block arrived clearly before than the other one.

Unbelievable that a a "science" video blog have done something like that (even more providing free bad publicity to other science bloggers that gave the correct answer...)

11. 11. tpozar 06:20 AM 8/28/13

Assumptions (both cases):
* The bullet has the same initial velocity v and the same mass m in both cases. The direction of the initial velocity is purely vertical.
* After collision, the bullet is within the block.
* Both blocks have the same mass M.

Linear momentum is conserved. So is angular momentum. Mechanical energy is NOT conserved.

Before collision:
vertical lin. momentum of the bullet: p_i = mv
horizontal lin. momentum of the bullet: 0
vertical lin. momentum of the block: 0
horizontal lin. momentum of the block: 0

Soon after collision:
vertical lin. momentum of the bullet: p_f1 = mV
horizontal lin. momentum of the bullet: 0
vertical lin. momentum of the block: p_f2 = MV
horizontal lin. momentum of the lock: 0

* V is the velocity of both, the block and the bullet soon after the the bullet came to rest with respect to the block.

Conservation of lin. momentum requires p_i = p_f1 + p_f2, and from here it follows V = mv/(m+M). The center of mass OF BOTH blocks reaches the height h = V^2/(2g) above the initial height, where g is gravitational acceleration on the surface of the Earth.

12. 12. alexp in reply to bokubob 06:49 AM 8/28/13

I mostly agree with Bokubob's comments. It is an inelastic collision. The depth of penetration by the bullet is converted into bullet deformation and heat- not momentum.
The greater the effective inertial mass, the greater the energy dissipated into heat.

I believe the offset shot was pre-planned to impart the same linear momentum + added angular momentum available from a decreased penetration of the bullet. Being offset, the moments of inertia now seen by the bullet are both linear and angular = the block is 'lighter'.

Consider shooting the block at the very edge. Assuming the block would not split, almost all of the momentum would be imparted into a rotational spin rather than a translational component.

13. 13. tpozar in reply to alexp 07:40 AM 8/28/13

Conservation of linear momentum and of angular momentum are two separate conservation laws. Linear momentum is conserved regardless of the off-the-center position where the bullet hits the block. The is no transfer of one momentum type into the other.

14. 14. alexp in reply to tpozar 08:01 AM 8/28/13

If the force vector is aligned with the center of mass, it is translational.. if offset, the force creates a torque = spin in this case.

15. 15. pocoyo 08:37 AM 8/28/13

The two blocks will reach the same height. Both blocks were hit by the same amount of force at the same position, it follows that the two will go to the same height limited only by the downward push of the gravity. There is no extra energy in the spinning block. The spinning was caused by the momentum of the bullet when it hit the block off center.

16. 16. slaven41 10:21 AM 8/28/13

Why do we have 15 responses to this, when the first one was correct?

17. 17. thomas0hei 12:16 PM 8/28/13

Hi!

I think one should take the influence of the air into account. There is a process called "Magnus effect", which could play a role.

18. 18. c laird478 in reply to tpozar 01:51 PM 8/28/13

It takes energy to produce angular momentum too. It doesn't just appear out of nowhere as if by magic. Part of the bullet's kinetic energy had to go into producing the angular momentum in the block, the bullet's kinetic energy was the only source of energy available in the system.

In a spacecraft, reaction thrusters are used to produce roll in the orbiting capsule, and to stop that roll. It takes energy to do that.

19. 19. Duffy Toler 04:08 PM 8/28/13

I studied this with pinball when I worked for Stern, built special gauges for it. The energy from a given solenoid could be transferred to rotational or translational movement on the ball (or some combination) and the more it spun, the slower it moved, as you would expect. So the answer for the wooden blocks is probably bad fairies: a wood nymph is all pissed off because someone cut down her tree and shot it with a gun.

20. 20. gmperkins 04:17 PM 8/28/13

I like bokubob's answer. Others make a number of valid points but I think the experiment is repeatable, especially with access to a vaccuum.

21. 21. hammerbait 04:22 PM 8/28/13

and yet the loss of energy through change to heat energy, and the effect of it's release inside the block (assymetrically) must impart some significant but unrecognized momentum along some vector or another.

22. 22. tpozar in reply to c laird478 05:33 PM 8/28/13

@c laird478: It does not take any energy to produce angular momentum. Momentum (either linear or angular) cannot be produced. It can only be transferred. And within a given system, it is conserved.

Consider the bullet hits the block at the horizontal distance x away from the center of the blocks mass. For simplicity, assume m << M. Imagine an axis going through the center of block’s mass. Around this axis, the block has a momentum of inertia I. Before the bullet hits the block, its angular momentum is L = xmv. After the inelastic collision, the angular momentum around the same axis (note the assumption m << M) equals L = Iw, where w is the angular velocity of the block. Thus, due to the conservation of angular momentum, the larger the off-the-center distance x, the larger the angular velocity “spin”: w = xmv/I. If x = 0, there is no “spin”.

I shall also write down the energy balance relations.
Case 1, x = 0. E_ki = E_kf + E_loss1
Case 2, x > 0. E_ki = E_kf + E_rf + E_loss2
* E_ki = mv^2/2 … Initial translational kinetic energy of the bullet.
* E_kf = (m+M)V^2/2 … Translational kinetic energy of the bullet and the block soon after collision.
* E_rf = Iw^2/2 … Final rotational kinetic energy of the block (again, m << M).
* E_loss1 … Energy dissipated into heat and plastic deformations of the bullet and the block in case 1.
* E_loss2 … Energy dissipated into heat and plastic deformations of the bullet and the block in case 2.

Since, due to the conservation of linear momentum, V is the same is both cases, it follows that E_rf = E_loss1 – E_loss2. As bokubob neatly put it: “When the bullet goes off to the side, some of the energy goes into the rotation, and less goes into destruction and heat. In short, the bullet doesn't go as far into the block.” His physical interpretation is conveyed it the last equation. In general, more initial translational kinetic energy of the bullet E_ki is dissipated when the bullet hits the block at its center than when the block is hit by the side. The difference is exactly the final rotational kinetic energy of the block E_rf.

23. 23. rufusgwarren 05:49 PM 8/28/13

Not sure the experiment was accurate. Since the block is wider than it is high, I assumed the edge would be higher, since the vertical energy, Fh, is the same. The rotational energy, apprx 1/4 width Fh gives rotation, but whatever Fg does. In free space the rotation would continue. The velocity of the block would also continue. The gravitational energy only returns the block, rotating or not. Note, Fg can be considered to be applied at the center, hence, this should allow a resultant torque. Hence the force not normal is at play.

24. 24. rufusgwarren 06:06 PM 8/28/13

My last response is goofy, the center of mass for the rotating block is lower; hence, conservation of en energy.

25. 25. c laird478 in reply to tpozar 07:51 PM 8/28/13

'It does not take any energy to produce angular momentum. Momentum (either linear or angular) cannot be produced. It can only be transferred. And within a given system, it is conserved.'

And yet astronauts do exactly that whenever they use their reaction thrusters to produce angular momentum in their spacecraft in a roll, pitch, or yaw maneuver. Apparently NASA scientists and engineers don't know about your equations. You should inform them that their reaction thrusters aren't producing any angular momentum in their spacecraft. I'm sure they'd like to know that. LOL.

26. 26. lg_king@earthlink.net 11:08 PM 8/28/13

They wouldn't be going through this exercise if there weren't a valid reason why the total kinetic energy of the block/bullet couple in the vertical direction is the same in both cases. I'm going with the idea that more energy from the bullet is absorbed by deeper penetration in the direct hit. If you could perform the experiment in some sort of calorimeter you would see that the total energy of both systems is the same. Although the spinning block really doesn't appear the be going as high, its true.

27. 27. Georgios 03:09 AM 8/29/13

In both cases we have same initial translational momentum and due to conservation it results to the same height.
In the second case there is also an initial rotational momentum mur, m the mass of the bullet, u the velocity of the bullet, r the distance between vector u and the center of mass of the brick, which is conserved as well causing the brick to rotate. The energy is conserved as well since the bullet still penetrates the brick but not so deeply.

28. 28. tpozar in reply to c laird478 04:08 AM 8/29/13

Consider a coordinate system where the spacecraft has initially a zero linear and a zero angular momentum. Consider also two thrusters deployed on opposite sides of the ship.

Case 1. If one thruster is fired in the direction along the line that connects the position of the thruster and center of mass of the ship, the ship will experience a transient (force) impulse, but no (torque) impulse. Since the initial linear momentum was zero, after the propulsion, the ship will acquire a linear momentum equal to the (force) impulse that is again equal to the magnitude of the linear momentum of the expelled exhausts but opposite in its direction. Therefore, the final linear momentum of the ship-exhaust system is zero, and linear momentum is conserved.

Case 2. If one thruster is fired (same amount as in Case 1) perpendicularly to the line that connects the position of the thruster and center of mass of the ship, the ship will experience both, a transient impulse and a maximum (torque) impulse for the given thrust. The uniform motion of the center of ship’s mass will have the same velocity as in Case 1. After the propulsion, the ship will also rotate with a constant angular velocity.

Case 3. If both thrusters are fired perpendicularly to the line that connects the position of the thrusters and center of mass of the ship but opposite directions, the ship will only acquire angular momentum but no linear momentum. The final linear momentum of the ship is zero, since the two exhausts acquired equal magnitudes of linear momentum, but they propagate in opposite directions. Linear momentum is again conserved.

Let us look at the mechanical energies of the bullet-block experiment and the ship-exhaust gedanken experiment. In the bullet-block experiment, the final mechanical energy is smaller than the initial one in both cases, while in the ship-exhaust experiment, the final mechanical energy is larger than the initial one (which is zero) in all three cases.

It is therefore awkward to say that “energy produces angular momentum”. It is important to fully separate the concept of the energy and that of the linear momentum (and that of the angular momentum, too). If not else, all three mentioned physical quantities have different units. The net force exerted on the ship is connected with the change of linear momentum of the ship through the (force) impulse-(linear) momentum theorem while the net torque is connected with the change of angular momentum through the (torque) impulse-(angular) momentum theorem.

29. 29. c laird478 06:38 AM 8/29/13

'Case 2. If one thruster is fired (same amount as in Case 1) perpendicularly to the line that connects the position of the thruster and center of mass of the ship, the ship will experience both, a transient impulse and a maximum (torque) impulse for the given thrust. The uniform motion of the center of ship’s mass will have the same velocity as in Case 1. After the propulsion, the ship will also rotate with a constant angular velocity.'

WRONG! The ship will NOT have the same linear velocity as case 1. Look up force vectors and see how you are mistaken.

30. 30. alexp in reply to tpozar 07:32 AM 8/29/13

Take a block of wood; float in water.

Impart a small force on the center of a side and again
near a corner, having both force vectors normal to the side .. Observe the responses.
With a similar force input, the tests do not impart the same translational (linear) velocity.

31. 31. tpozar in reply to c laird478 08:48 AM 8/29/13

Thank you for pointing this out. One of my sentences should have read: "The uniform motion of the center of ship’s mass will have the same MAGNITUDE of velocity as in Case 1".

32. 32. tpozar 09:14 AM 8/29/13

http://www.wired.com/wiredscience/2013/08/the-bullet-block-problem-with-a-twist/
http://scienceblogs.com/principles/2013/08/21/how-deep-does-veritasiums-bullet-go/

33. 33. JRCancio 07:18 PM 8/29/13

With the second block, off-center, when the bullet comes into contact with the block all the energy, as if the first bullet, is directed straight upwards; however, the second bullet has less mass/weight above it and imparts impact energy that lifts the block sooner, albit, one side and imparts rotational spin. When the block is lifted away from its rest the full weight of the block comes into play. Friction from the rotational spin and gravitation influence come into play to an end-point expenditure position as all the vectored factors come into balance. It appears the second blocks remains in position airborne for a fraction of a second longer before it falls. Therefore, the second blocks is lifted faster then the first block, rotational spin forces come into play where the center of balance come again the to center point of the block, and we also see the second blocks appears to be airborne just a little bit longer.

34. 34. ramaus 10:45 PM 8/29/13

April 1?

35. 35. myron in reply to bokubob 12:31 AM 9/7/13

A1 answer ! Totally correct. MI

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