Questions: In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement theta as shown (Figure 1). Part A Find an expression for v0, the initial speed of the fired object. Express your answer in terms of some or all of the variables m, M, L, and theta and the acceleration due to gravity, g.

In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement theta as shown (Figure 1).

Part A

Find an expression for v0, the initial speed of the fired object. Express your answer in terms of some or all of the variables m, M, L, and theta and the acceleration due to gravity, g.

Transcript text: In a ballistic pendulum an object of mass $m$ is fired with an initial speed $v_{0}$ at a pendulum bob. The bob has a mass $M$, which is suspended by a rod of length $L$ and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement $\theta$ as shown (Figure 1). Part A Find an expression for $v_{0}$, the initial speed of the fired object. Express your answer in terms of some or all of the variables $m, M, L$, and $\theta$ and the acceleration due to gravity, $g$. View Available Hint(s) $v_{0}=$ $\square$

Solution

Solution Steps

Step 1: Conservation of Momentum During Collision

When the object of mass $ m $ collides with the pendulum bob of mass $ M $ and they stick together, the total momentum before and after the collision must be conserved. The initial momentum is given by: \[ p_{\text{initial}} = m v_0 \]

After the collision, the combined mass $ (m + M) $ moves with a velocity $ v $: \[ p_{\text{final}} = (m + M) v \]

By conservation of momentum: \[ m v_0 = (m + M) v \] \[ v = \frac{m v_0}{m + M} \]

Step 2: Conservation of Energy During the Swing

After the collision, the combined mass $ (m + M) $ swings to a maximum height. At the maximum angular displacement $ \theta $, the kinetic energy is converted into potential energy. The height $ h $ can be found using the length of the pendulum $ L $ and the angle $ \theta $: \[ h = L (1 - \cos \theta) \]

The potential energy at the maximum height is: \[ U = (m + M) g h = (m + M) g L (1 - \cos \theta) \]

The kinetic energy just after the collision is: \[ K = \frac{1}{2} (m + M) v^2 \]

By conservation of energy: \[ \frac{1}{2} (m + M) v^2 = (m + M) g L (1 - \cos \theta) \]

Step 3: Solving for $ v $

Canceling $ (m + M) $ from both sides: \[ \frac{1}{2} v^2 = g L (1 - \cos \theta) \] \[ v^2 = 2 g L (1 - \cos \theta) \] \[ v = \sqrt{2 g L (1 - \cos \theta)} \]

Step 4: Substituting $ v $ Back into the Momentum Equation

From Step 1, we have: \[ v = \frac{m v_0}{m + M} \]

Substitute $ v $ from Step 3: \[ \frac{m v_0}{m + M} = \sqrt{2 g L (1 - \cos \theta)} \]

Step 5: Solving for $ v_0 $

\[ v_0 = \frac{(m + M)}{m} \sqrt{2 g L (1 - \cos \theta)} \]