Good morning. A ballistic pendulum is a classic tool for calculating the speed of a projectile.
The ballistic pendulum is a block suspended from strings at which you fire the projectile.
In my demonstration, I use a blow dart gun to fire a blow dart at the hanging block of wood,
the blow dart collides with the wood block and embeds itself in the wood block.
(That is a perfectly inelastic collision.) Yes, Billy, This is a perfectly inelastic
collision. The wood block and blow dart swing together up to a maximum height and that maximum
height can be used to calculate the velocity of the dart before the collision. Of course,
this happens pretty quickly, so this is what it looks like at 40 times slower than real speed.
We can just measure the speed of the dart using the slow-motion video, right? So,
we don’t really need ballistic pendulums anymore. Yeah Bo, that may be true, however, this still
serves as a great physics problem. Bo, please read and translate the problem.
♪ Flipping Physics ♪ A dart moving horizontally at a constant speed
embeds itself in a hanging, stationary wood block. The wood block and dart rise to a maximum vertical
height. Solve for the speed of the dart before the collision in terms of the mass of the dart,
the mass of the wood block, the maximum vertical height of the dart and wood block,
and known constants. This is called a Ballistic Pendulum. Okay. Let’s label some variables.
The mass of the blow dart is m sub d for dart, the mass of the wood block is m sub
let’s not use b, that’s too similar to d for dart, let’s use m sub w for wood block. The initial
velocity of the wood block is zero. The maximum vertical height of the dart and wood block is
let’s use h final for that. And we are solving for the initial velocity of the dart,
so v sub d i equals question mark. Thank you Bo. Bobby, please identify
the different events in this problem and the general physics principles during those events.
“Identify the different events in this problem and the general physics principles
during those events”? What does that mean? (What things happen in this problem and what
physics will we use to solve the problem?) Okay. Uh, there is a collision …
(Right. And then what happens?) It swings on the strings?
(And what is that called?) A pendulum?
(Okay, so, there is a collision and a pendulum, right?)
Yeah. (So those are the two
“events”. What general physics principles are we going to use to analyze those “events”?)
Okay. I think I get it. First there is a collision. During that collision, all the forces
in the x-direction are internal to the system, the force the dart applies to the wood block is equal
and opposite to the force the wood block applies to the dart, those two forces form a Newton’s
third law force pair. Therefore, the net force in the x-direction equals zero. And because net force
equals the derivative of momentum with respect to time, the momentum of the dart and wood block
system does not change. In other words, during the collision linear momentum in the x-direction is
conserved. So, one event is the collision where the general physics principle is conservation of
linear momentum in the x-direction. Actually, we could just say that,
in the x-direction, net force equals the change in momentum over the change in time
which equals zero and also get to conservation of linear momentum in the x-direction that way
and therefore we do not actually need derivatives to understand this, right mr.p?
(Yes Billy, that is correct. Bobby, please continue?)
Sure. The other event is where the two objects swing up on the strings as a pendulum.
And There is no work done by a force of friction or work done by a force applied on the system,
therefore, mechanical energy is conserved. So, after the collision, the next event is a
pendulum where mechanical energy is conserved. Absolutely Bobby, thanks. Now, rather than
looking at this as initial and final and parts 1 and 2, I am instead going to have us identify
three different moments during this event and then identify what happens from moment 1 to 2 and then
from moment 2 to 3. Moment 1 is right before the collision. Moment 2 is right after the collision.
And moment 3 is when the dart and wood block are at their maximum height. We can now say that from
moment 1 to moment 2, linear momentum in the x-direction is conserved. And from moment 2
to moment 3, mechanical energy is conserved. This does mean we need to give some of our known values
different subscripts which will actually clarify exactly where these initial and final points are.
Velocity of the wood block initial is now velocity of the wood block at moment 1. Height final is now
height at moment 3 and velocity of the dart initial is now velocity of the dart at moment
1. Billy, please analyze the collision to solve for the velocity of the dart at moment 1.
Certainly. … Well, Bobby already showed us that linear momentum in the x-direction is
conserved from 1 to 2 during the collision and momentum equals mass times velocity. So,
the mass of the dart times the velocity of the dart at 1 plus the mass of the wood block times
the velocity of the wood block at 1 equals the mass of the dart times the velocity of
the dart at 2 plus the mass of the wood block times the velocity of the wood block at 2.
The velocity of the wood block at 1 is zero, so that term cancels out.
The dart and wood block are stuck together at 2, so those two velocities are equal to one another,
and we can factor out velocity at 2. And we can solve for the velocity of the dart at 1,
it equals the quantity mass of the dart plus the mass of the wood block all divided by the mass
of the dart all multiplied by the velocity of both the dart and the wood block at 2.
But, we do not know the velocity at 2 So Oh, we put that in our equation holster! Yeah,
that equation goes in our equation holster! Very nice Billy. Yes, we can put that equation
into our equation holster, and we know we need to solve for the velocity at 2 in terms of the
known variables. Bo, please do that. Sure. From moment 2 to moment 3,
Bobby already explained that mechanical energy is conserved.
Let’s set the horizontal zero line at the center of mass of the dart and wood block at moment 2.
At moment 2 the only type of mechanical energy is kinetic energy. At moment 3,
the only type of mechanical energy is gravitational potential energy. And
Everybody brought mass to the party! Hold up. Which mass is that? The mass of the
dart or the mass of the wood block? They have to be the same mass in order to have a mass party.
Uh Both masses are the total mass of the dart and the wood block.
So, the two masses are the same and we can say … Everybody brought total mass to the party!
Now we can rearrange the equation to solve for velocity at 2. And substitute that back
into the equation we previously holstered to get
the velocity of the dart at 1 equals the quantity mass of the dart plus the mass of the wood block
all divided by the mass of the dart all multiplied by the square root of 2 times
acceleration due to gravity times height at 3. Thank you Bo. Now, I know this answer is a bunch
of variables, which, for many of you, is less satisfying than ending with a number. We will
substitute in real numbers to this equation to check its veracity in a bit, however, first,
let’s see if this solution for the velocity of the dart at moment 1 makes sense. More specifically,
Bobby, according to our equation, if we increase the velocity of the dart at
moment 1, keeping all other variables constant, what happens to height 3?
Okay, so the mass of the dart and wood block are unchanged and the acceleration due to gravity
is also unchanged. However, we have increased the velocity of the dart before the collision
and the question is, what happens to the height of the dart and wood block at the end. Well,
according to our equation, height 3 increases. (I agree Bobby. But does that make sense?)
Yes? (I’m sorry, what I mean is,
use physics to explain if that makes sense.)
Okay. Well, if the velocity of the dart at moment 1 is increased, that increases the momentum of
the system at 1. According to conservation of momentum, that increases the momentum at 2.
That means the velocity at 2 is increased. And therefore kinetic energy at 2 is increased,
and according to conservation of energy, gravitational potential energy at 3 is increased,
that means the height at 3 must be increased. So, yes. Increasing just the velocity of the dart at 1
should increase the height at 3. Thanks Bobby. Billy, if the velocity
of the dart at 1 is again increased, however, we want the height at 3 to be the same as before,
and if the only variable we change is the mass of the wood block, how would we have to change the
mass of the wood block to make that happen? Please look at it in terms of the equation
and also explain if the physics makes sense. Okay. The velocity of the dart at 1 is increased.
Everything but the mass of the wood block remains the same. According to our equation, the mass of
the wood block would have to be increased. But does that make sense with respect to physics?
Well, again, increasing the velocity of the dart at 1 increases the momentum of the dart and wood
block system at 1 which, according to conservation of momentum, increases the momentum of the system
at 2. However, according to conservation of energy, in order for the height at 3 to remain
the same, we need the velocity at 2 to also remain the same. In order for that to occur,
we need the mass of the wood block to be increased to make the momentum at 2 larger. So, yes,
this makes sense according to physics as well. I agree Billy. Bo, same question, only for
the mass of the dart. In other words, if the velocity of the dart at 1 is again increased,
however, we want the height at 3 to be the same as before, if the only variable we change is the
mass of the dart, how would we have to change the mass of the dart to make that happen?
Please look at it in terms of the equation and also explain if the physics makes sense.
Alright. According to our equation, if we increase the velocity of the dart at 1,
in order for the height at 3 to stay the same, the mass of the dart needs to be decreased.
Hold up, really? How can you tell from the equation?
Oh, when estimating this, because the mass of the dart is so small relative to the mass of the wood
block, we can consider the change in the total mass of the dart and wood block to be negligible.
And because the mass of the dart is in the denominator and we need the value of the equation
to increase, we need to decrease the mass of the dart, again, because it is in the denominator.
Ohhh. Thanks. You’re welcome. Okay, but
does this make sense according to physics. Well, again, increasing the velocity of the dart at 1
would increase the momentum at 1, however, we need the momentum at 1 to remain constant,
so we need to decrease the mass of the dart to keep the momentum at 1 constant.
So, yeah, the physics makes sense as well. Cool. Very nice. Alright, I think it is time to actually
put some numbers in this. In my demonstration, the mass of the dart is 0.0051 kilograms, the mass of
the wood block is 0.1329 kilograms and the maximum vertical height above the horizontal zero line,
which is height 3 of the center of mass of the dart and wood block system, is 0.056 meters.
Plugging all those numbers into our equation gives us 28.363 or 28 meters per second with 2 sig figs
for the velocity of the dart at moment 1 or while it is moving at a constant velocity
before colliding with the wood block. And we can measure the velocity of the
dart before the collision using this high frame rate video, right mr.p?
Exactly Bo. The dart moves a horizontal distance of 0.297 meters in 10 frames
right before it collides with the wood block. Considering the frame rate of this video is
960 frames per second, we can multiply 10 frames by 1 second over 960 frames to get
a change in time of 0.010416 repeating seconds. Therefore, we measured the
constant velocity of the dart before the collision to be 28.512 meters per second.
Considering relative error equals the observed value minus the accepted value all divided by
the accepted value and then multiplied by 100, and using the measured velocity as the observed value,
and the calculated velocity as the accepted value, we get a relative error between the measured and
predicted velocities of the dart before the collision to be 0.53 percent with 2 sig figs.
I think that is definitely close enough to say The physics works! The physics works!
Uh huh, uh huh! The physics works!
The physics works! Uh huh, uh huh! The physics works! The physics
works! Uh huh, uh huh! The physics works! The
physics works! Uh huh, uh huh! And thank you very much for
learning with me today, I enjoyed learning with you.