Ballistic Pendulum - Online Tutor, Practice Problems & Exam Prep

The ballistic pendulum problem involves a bullet colliding with a block, causing the block to swing upward. This scenario combines conservation of momentum and energy principles. Initially, momentum is conserved during the collision, leading to a final velocity for the combined mass. Then, energy conservation is applied to find the maximum height reached by the pendulum. The height \( y_c \) can be expressed as $v^2/2g$. The angle with the vertical can be calculated using the pendulum equation, relating height and length.

Ballistic Pendulum Problems

Hey, guys. So for this video, I'm going to introduce you to a type of problem or a type of collision problem called a ballistic pendulum. Let's check this out. The basic setup of a ballistic pendulum is you have an object that collides with a block and then that block swings up on a pendulum like this to some height. So, what we have here in this problem is we have a collision in the first part, and then once these things collide, the block moves upwards and starts changing speeds and heights. So, you have a motion or energy part to this problem as well. What I'm going to show you in this video is that when we solve this problem, it is just like any other problem where we have collisions and then changing speeds and heights or motion and energy parts. We're really just going to combine both of our conservation equations, momentum, and energy. We've seen this thing before. It's nothing new. Remember, this is just one of the special variations that you might see when you have a collision. You have a collision and then you have this pendulum here. So, because we're going to talk about a pendulum, we'll also need this pendulum equation as well, which just relates the different variables in pendulums. So let's take a look at our problem here. First thing we want to do is just draw diagrams and then label the points of interest. So let's take a look here. Remember there are 3 parts to this problem. The first part is before the collision. Before the collision, I'm going to call that point a. The bullet is still firing inside of the blocker. It's still moving towards the block. Afterwards, when it embeds itself, that's going to be after the collision. That's point b. That's where the motion or the swinging pendulum part of our problem starts. And then finally, when it reaches the maximum height over here, that's going to be where the motion ends. I'm going to call that point c. What we're looking for in this problem is the maximum heights that the pendulum will reach. So what ends up happening is that this pendulum rises to some heights and I'm going to call that yc. So that's really my target variable. So how do we solve for it? Well, remember, we're just going to write out both of our equations, our conservation of momentum and energy equations. So let's go ahead and do that.

So for this part a here, I'm going to calculate, I'm going to start writing the equations, so I'm going to use conservation of momentum in the interval from a to b. So I have m1v1, initial, which is a + m2v2, initial equals m1v1, final + m2v2, final. And then for the interval from b to c, that's going to be a conservation of energy equation. This is going to be kb + ub + work done by non-conservative forces equals kc + uc. That's initial energy equals final energy. So which equation do we start off with? Well, hopefully, you guys realize if we're trying to solve for yc, that's actually going to come from the potential energy at point c. So, we're going to use the conservation of energy equation first. So let's take a look at each of our terms and, you know, plug in values and start solving. So do we have any kinetic energy at point b? That's right here. Well, after the bullet hits the block, the bullet is going to move upwards. That's why it swings. So there's definitely some kinetic energy here. What about the gravitational potential? Remember, there are no springs here in this problem. Well, what happens is we actually don't know what this height is even at the lowest point of the problem, we have the lowest point of the pendulum. But what happens in these situations if we don't know the heights? We're just going to actually set the lowest point of the pendulum to be where y equals 0 because then our potential energy equals 0 and it makes our problems simpler. There's also no work done by non-conservative forces. There's no work done by u or friction. What about k final? What happens here is that the maximum height when the block just stops swinging, it stops, and its velocity is equal to 0, so there's no kinetic energy, but there is some gravitational potential because it's risen some height here. So that's really what our conservation of energy equation boils down to. So now we just expand the terms. So this is going to be 1/2 and this is going to be mvb squared, except what happens is our little m actually becomes big m. The bullet is going to embed itself inside of the block, so this is a completely inelastic collision. So you have this bullet here that's now embedded itself inside of the block like this, and they're both going to travel together. So what happens is your big m is equal to m1 + m2. So this is going to be 1/2 of big m vb squared equals big m g yc. So, we're looking for this yc. We're going to cancel out the masses and then write an expression for yc. We're going to divide the g over to the other side and then yc equals vb squared over 2g. We've come up with this kind of expression before. So, before I can solve for this yc, I actually first need to figure out what this vb is. What's the velocity at the start of the motion? Now remember, what happens is this is the start of the motion, but it's also the end of the collision. So, if I get stuck here and I don't know what this vb is, I can always go to my conservation of momentum equation and solve for it there because I have vb here. So that's all we have to do here. So we're going to go ahead and write, this m1. So this is going to be the bullets, m1 equals 0.2, and then my, block m2 is equal to 40. The initial speed of the block, v2, is equal to 0. So what basically happens is this is going to be 0.2 times 700 + 40 times 0 because it's at rest, and this is equal to Now the... (The response was too long, and I have to cut it short. If you would like me to continue or focus on particular sections, let me know!)

Final Speed of Ballistic Pendulum Projectile

Hey, everybody. So let's take a look at our problem here. We've got a bullet that's being fired into a wooden block. What we're told is that the length of this pendulum here is 2. The mass of the bullet, I'm going to call this \( m_1 \), is 0.005. The mass of the block, \( m_2 \), is 1. This is a ballistic pendulum type problem. You have a bullet fired into a block. What happens is when it hits, it travels in a curved path until it rises some height. We're told that when this pendulum ends its swing, it's going to be a distance of 11 centimeters higher. The center of mass rises by a distance of 11 centimeters. If I draw this line here from where it ends, then this distance between where it started and ended is going to be 11 centimeters. I'm going to call that \( y \) and that's going to be 0.11. We want to calculate the speed of the bullet as it travels and emerges from the block. This is different from normal ballistic pendulum problems because the bullet actually keeps on going once it's passed through the block.

In this problem, there are 3 events: Part A, where the bullet is still traveling before hitting the block; Part B, where the collision happens and the motion starts; and Part C, where the block reaches the end of its swing upwards. We use conservation of momentum from A to B, and conservation of energy from B to C.

For momentum conservation: m_1v_1a+m_2v_2a = m_1v_1b+m_2v_2b where \( v_1a \) is the speed of the bullet before the collision (450 m/s) and \( v_2a \) is 0 since the block is initially at rest.

For energy conservation: KE B = PE C where potential energy at C is \( m_2 \cdot g \cdot 0.11 \), assuming no kinetic energy at point C as the block stops moving.

When simplifying, we get for \( v_2b \) that approximately equals \( \sqrt{2 \cdot 9.8 \cdot 0.11} = 1.47 \) m/s. Plugging this into the momentum equation: 0.005v_1b+1.47=2.25 , we find \( v_1b \) which equals 156 m/s. The bullet impacts the block and loses a lot of its initial speed as it transfers energy to the block.

Let me know if you have any questions.