Problem 27 - Applied Mathematics

Question

A bullet of mass \(m\) is fired with speed \(u\) into a fixed block of wood and emerges with speed \(2 u / 3 .\) When the experiment is repeated with the block free to move the bullet emerges with speed \(u / 2\) relative to the block. Assuming the same constant resistance to penetration in both cases, find the mass and the final speed of the block in the second case. (Neglect the effect of gravity throughout)

Step-by-Step Solution

Answer

The mass of the block is 2m and the final speed of the block is \( \frac{u}{6} \).

Step 1 - Understand the given information

Identify and represent all given variables and conditions. Assigned variables: Mass of bullet = m, Initial speed of bullet = u, Emerged speed when block is fixed = \( \frac{2u}{3} \), Emerged speed when block is free = \( \frac{u}{2} \).

Step 2 - Apply conservation of momentum (fixed block)

In the first case where the block is fixed, apply the principle of conservation of momentum. Initial momentum is \( m \cdot u \). Final momentum is \( m \cdot \frac{2u}{3} \). Using the initial and final momentum, set up the momentum equation: \[ mu = m \cdot \frac{2u}{3} + F t \] where F is the resistance force, and t is the time of penetration.

Step 3 - Determine the resistance force

Recognize that resistance force is the same in both cases. Calculate the force F using the momentum equation from the fixed block scenario: \[ F t = mu - m \cdot \frac{2u}{3} = m \cdot \frac{u}{3} \] Thus, \( F t = m \cdot \frac{u}{3} \).

Step 4 - Apply conservation of momentum (free block)

Now consider the second case: initial momentum is \( mu \), final momentum when the block is free is \( m \cdot \frac{u}{2} + M V \), where M is the mass of the block and V is its final speed. The resistance force, \( m \cdot \frac{u}{3} = F t \), is the same.

Step 5 - Solve for the block's speed and mass

Set up the momentum equation based on the second scenario: \[ mu = m \cdot \frac{u}{2} + M V \] and the work-energy principle relating to the resistance force yields: \( \frac{1}{2}mu^{2} - \frac{1}{2}m\left( \frac{u}{2}\right)^{2} = \frac{1}{2}MV^{2} \). Solve both equations simultaneously to find M and V.

Key Concepts

Physics Problems

Solving physics problems often requires breaking down the information you have and figuring out how it connects. In our exercise, a bullet and a block interact through two scenarios. To solve it, you need to model the motion of the objects involved. You should:

• Identify given values and unknowns.
• Set up and solve relevant equations.
• Apply core concepts like the conservation of momentum and forces.

These steps form a structured problem-solving approach, ensuring you can tackle other complex physics problems effectively.

Momentum

Momentum is a measure of an object's motion and is calculated using the formula: \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In our exercise, the bullet's initial and final momenta vary depending on whether the block is fixed or moving.

• For a fixed block: The bullet's momentum changes as it emerges.
• For a moving block: Both bullet and block momenta must be considered.

If total momentum before and after the collision is equal, it validates the Principle of the Conservation of Momentum.

Resistance Force

The resistance force acts against the bullet as it penetrates the block. Despite the two different scenarios, this force remains constant. We calculate it using the change in momentum over time: \( F = \frac{\text{Change in Momentum}}{t} \).

• In the fixed block case, the force is derived from the bullet's speed reduction.
• In the moving block case, the resistance force remains \( F = m \frac{u}{3} \).

Understanding this constant force helps you connect the different scenarios and solve for unknowns effectively.

Kinematics

Kinematics is the study of motion, describing objects' velocity, acceleration, and displacement over time. In this problem, kinematic principles guide the bullet’s behavior as it slows down. Moreover, using the work-energy principle, we relate the resistance force to changes in kinetic energy: \[ \frac{1}{2}mu^2 - \frac{1}{2}m\big( \frac{u}{2} \big)^2 = \frac{1}{2}MV^2 \].

• Calculate initial and final velocities.
• Relate these velocities to work done by forces.

Understanding these concepts leads to finding the block's final speed and mass.