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.
