Ballistic Pendulum: Videos & Practice Problems

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. The maximum height reached by the pendulum can be calculated using the equation mi>yc=v22g. Additionally, the angle with the vertical can be determined using the pendulum equation, leading to a comprehensive understanding of motion and energy transformations in collisions.

Ballistic Pendulum Problems

The ballistic pendulum problem involves a collision between a moving object, such as a bullet, and a stationary block that swings upward after the collision. This scenario combines principles of momentum and energy conservation to analyze the motion and height reached by the pendulum.

To begin solving a ballistic pendulum problem, it is essential to identify key points in the motion: point A (before the collision), point B (immediately after the collision), and point C (at the maximum height). The goal is to determine the maximum height (denoted as \( y_c \)) that the pendulum reaches after the collision.

The first step involves applying the conservation of momentum during the collision phase (from point A to point B). The equation for this is:

\[ m_1 v_{1,\text{initial}} + m_2 v_{2,\text{initial}} = m_1 v_{1,\text{final}} + m_2 v_{2,\text{final}} \]

Here, \( m_1 \) is the mass of the bullet, \( v_{1,\text{initial}} \) is its initial velocity, \( m_2 \) is the mass of the block, and \( v_{2,\text{initial}} \) is zero since the block is at rest. After the collision, both the bullet and block move together, making it a completely inelastic collision.

Next, the conservation of energy principle is applied from point B to point C. The energy conservation equation is:

\[ K_B + U_B + W_{\text{non-conservative}} = K_C + U_C \]

In this case, the kinetic energy at point B is given by:

\[ K_B = \frac{1}{2} m v_B^2 \]

At point C, the kinetic energy is zero because the pendulum momentarily stops, and the potential energy is at its maximum:

\[ U_C = m g y_c \]

By setting the lowest point of the pendulum (point B) as the reference height (where potential energy is zero), the energy equation simplifies to:

\[ \frac{1}{2} m v_B^2 = m g y_c \]

From this, we can derive the expression for the maximum height:

\[ y_c = \frac{v_B^2}{2g} \]

To find \( v_B \), we substitute the values from the momentum equation. For example, if \( m_1 = 0.2 \, \text{kg} \) (bullet) and \( m_2 = 40 \, \text{kg} \) (block), the equation becomes:

\[ 0.2 \times 700 + 40 \times 0 = (0.2 + 40) v_B \]

Solving for \( v_B \) yields a value that can then be plugged back into the height equation to find \( y_c \). For instance, if \( v_B \) is calculated to be \( 3.48 \, \text{m/s} \), substituting this into the height equation gives:

\[ y_c = \frac{(3.48)^2}{2 \times 9.8} \approx 0.62 \, \text{m} \]

After determining the height, the next step is to calculate the angle \( \theta_y \) that the pendulum makes with the vertical. This can be done using the relationship between the length of the pendulum \( l \), the height \( y_c \), and the angle:

\[ l - y_c = l \cos(\theta_y) \]

Rearranging gives:

\[ \cos(\theta_y) = \frac{l - y_c}{l} \]

Taking the inverse cosine allows for the calculation of the angle. For example, if \( l = 2 \, \text{m} \) and \( y_c = 0.62 \, \text{m} \), the angle can be found as:

\[ \theta_y = \cos^{-1}\left(\frac{2 - 0.62}{2}\right) \approx 46.4^\circ \]

In summary, the ballistic pendulum problem illustrates the application of conservation laws in a dynamic system, allowing for the calculation of maximum height and angle through systematic analysis of momentum and energy transformations.

Final Speed of Ballistic Pendulum Projectile

In this scenario, we analyze a ballistic pendulum problem where a bullet is fired into a wooden block. The bullet has a mass of \( m_1 = 0.005 \, \text{kg} \) and the block has a mass of \( m_2 = 1 \, \text{kg} \). The length of the pendulum is \( 2 \, \text{m} \), and after the bullet strikes the block, the center of mass of the system rises by \( y = 0.11 \, \text{m} \) (11 centimeters).

To find the speed of the bullet as it exits the block, we apply the principles of conservation of momentum and conservation of energy. The process can be divided into three key events: (a) before the bullet hits the block, (b) during the collision, and (c) after the block has swung to its highest point.

For the momentum conservation from event (a) to (b), we use the equation:

$$ m_1 v_{1a} + m_2 v_{2a} = m_1 v_{1b} + m_2 v_{2b} $$

Here, \( v_{1a} \) is the initial speed of the bullet, which is given as \( 450 \, \text{m/s} \), and \( v_{2a} = 0 \, \text{m/s} \) since the block is initially at rest. The equation simplifies to:

$$ 0.005 \times 450 + 1 \times 0 = 0.005 v_{1b} + 1 v_{2b} $$

Next, we apply conservation of energy from event (b) to (c). The kinetic energy of the block and bullet after the collision is converted into potential energy at the highest point of the swing:

$$ \frac{1}{2} m_2 v_{2b}^2 = m_2 g y $$

Here, \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)) and \( y \) is the height the block rises, which is \( 0.11 \, \text{m} \). This simplifies to:

$$ \frac{1}{2} v_{2b}^2 = g y $$

Solving for \( v_{2b} \), we find:

$$ v_{2b} = \sqrt{2gy} = \sqrt{2 \times 9.8 \times 0.11} \approx 1.47 \, \text{m/s} $$

Now, substituting \( v_{2b} \) back into the momentum equation gives:

$$ 2.25 = 0.005 v_{1b} + 1.47 $$

Rearranging this yields:

$$ 0.005 v_{1b} = 2.25 - 1.47 = 0.78 $$

Finally, solving for \( v_{1b} \) results in:

$$ v_{1b} = \frac{0.78}{0.005} = 156 \, \text{m/s} $$

This indicates that the bullet retains a significant portion of its speed after passing through the block, losing some energy to the block in the process. Understanding these principles of momentum and energy conservation is crucial in solving similar physics problems.

Here’s what students ask on this topic:

What is a ballistic pendulum and how does it work?

A ballistic pendulum is a device used to measure the velocity of a projectile, such as a bullet. It consists of a block suspended as a pendulum. When the projectile collides with the block, it embeds itself, causing the block to swing upward. The problem involves two main parts: the collision and the pendulum motion. Initially, the conservation of momentum is used to determine the velocity of the block after the collision. Subsequently, the conservation of energy is applied to find the maximum height the pendulum reaches. This height can be used to calculate the initial velocity of the projectile. The ballistic pendulum is a practical application of physics principles, combining momentum and energy conservation.

How do you solve a ballistic pendulum problem using conservation of momentum and energy?

To solve a ballistic pendulum problem, start by applying the conservation of momentum to the collision phase. Use the equation: $m_1{v}_{1}i+m_2{v}_{2}i=\; (m_1+m_2)v_{\mathrm{b}}$. This equation helps find the velocity of the block and projectile after the collision. Next, apply the conservation of energy to the pendulum motion: $\frac{{v_{\mathrm{b}}}^2}{2}=g{y}_{\mathrm{c}}$. This equation calculates the maximum height the pendulum reaches. By combining these principles, you can determine the initial velocity of the projectile.

What is the significance of the angle in a ballistic pendulum problem?

The angle in a ballistic pendulum problem is significant because it represents the maximum displacement of the pendulum from the vertical. This angle can be calculated using trigonometric relationships, specifically the pendulum equation: $l-y_{\mathrm{c}}=lcos({\theta }_{\mathrm{y}})$. Solving for the angle involves isolating the cosine term and using the inverse cosine function. The angle provides insight into the pendulum's motion and helps verify the calculations of height and velocity. Understanding this angle is crucial for accurately analyzing the pendulum's behavior and the energy transformations involved.

How is the velocity of the block calculated in a ballistic pendulum problem?

The velocity of the block in a ballistic pendulum problem is calculated using the conservation of momentum. Initially, the momentum of the system is given by the equation: $m_1{v}_{1}i+m_2{v}_{2}i$. After the collision, the combined mass of the block and projectile moves with a common velocity $v_{\mathrm{b}}$. The equation becomes: $(m_1+m_2)v_{\mathrm{b}}$. Solving for $v_{\mathrm{b}}$ involves rearranging the equation and substituting known values. This velocity is crucial for further calculations involving energy conservation and determining the pendulum's maximum height.

What is the formula for calculating the maximum height in a ballistic pendulum problem?

The formula for calculating the maximum height in a ballistic pendulum problem is derived from the conservation of energy principle. It is given by: $y_{\mathrm{c}}=\frac{{v_{\mathrm{b}}}^2}{2g}$. Here, $y_{\mathrm{c}}$ is the maximum height, $v_{\mathrm{b}}$ is the velocity of the block after the collision, and $g$ is the acceleration due to gravity. This formula is used to determine how high the pendulum rises after the collision, providing insight into the energy transformation from kinetic to potential energy.