Applying the Law of Conservation of Momentum

Oct 13, 2025 by miodeya.com 13 views

Introduction

In this discussion, we will explore the physics behind a bullet hitting a soft wooden block attached to a wire. The bullet becomes stuck in the wooden block, causing it to swing upwards. We will analyze the situation and show that the bullet's kinetic energy is transferred to the wooden block, resulting in the block's upward motion.

The Scenario

A bullet of mass 12 g moves at a speed of $450 \, \text{m/s}$ . The bullet hits a soft wooden block of mass 2.5 kg, which is attached to a wire. The bullet becomes stuck in the wooden block, causing it to swing upwards.

Conservation of Momentum

When the bullet hits the wooden block, the two objects stick together and move as a single system. We can apply the law of conservation of momentum to this system. The momentum of the bullet before the collision is equal to the momentum of the bullet and the wooden block after the collision.

Let's denote the mass of the bullet as $m_b = 12 \, \text{g} = 0.012 \, \text{kg}$ and the mass of the wooden block as $m_w = 2.5 \, \text{kg}$ . The initial momentum of the bullet is given by:

p_i = m_b v_b = 0.012 \, \text{kg} \times 450 \, \text{m/s} = 5.4 \, \text{kg m/s}

The final momentum of the bullet and the wooden block is given by:

p_f = (m_b + m_w) v_f

where $v_f$ is the final velocity of the bullet and the wooden block.

Applying the Law of Conservation of Momentum

We can now apply the law of conservation of momentum to the system:

p_i = p_f

Substituting the expressions for $p_i$ and $p_f$ , we get:

5.4 \, \text{kg m/s} = (0.012 \, \text{kg} + 2.5 \, \text{kg}) v_f

Simplifying the equation, we get:

v_f = \frac{5.4 \, \text{kg m/s}}{2.512 \, \text{kg}} = 2.15 \, \text{m/s}

Kinetic Energy

The kinetic energy of the bullet before the collision is given by:

K_i = \frac{1}{2} m_b v_b^2 = \frac{1}{2} \times 0.012 \, \text{kg} \times (450 \, \text{m/s})^2 = 606.75 \, \text{J}

The kinetic energy of the bullet and the wooden block after the collision is given by:

K_f = \frac{1}{2} (m_b + m_w) v_f^2 = \frac{1}{2} \times 2.512 \, \text{kg} \times (2.15 \, \text{m/s})^2 = 11.43 \, \text{J}

Energy Transfer

The kinetic energy of the bullet is transferred to the wooden block, resulting in the block's upward motion. The energy transfer can be calculated as:

\Delta K = K_i - K_f = 606.75 \, \text{J} - 11.43 \, \text{J} = 595.32 \, \text{J}

Conclusion

In this discussion, we have shown that the bullet's kinetic energy is transferred to the wooden block, resulting in the block's upward motion. The law of conservation of momentum was applied to the system, and the kinetic energy of the bullet and the wooden block was calculated before and after the collision. The energy transfer was also calculated, showing that the kinetic energy of the bullet is transferred to the wooden block.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics. John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage Learning.

Additional Resources

[1] Khan Academy. (n.d.). Conservation of momentum. Retrieved from https://www.khanacademy.org/science/physics/one-dimensional-motion/conservation-of-momentum/v/conservation-of-momentum
[2] Physics Classroom. (n.d.). Conservation of momentum. Retrieved from https://www.physicsclassroom.com/class/momentum/Lesson-1/Conservation-of-Momentum
A Bullet of Mass 12 g Moves at a Speed of $450 \, \text{m/s}$ : A Physics Discussion ====================================================================================

Q: What is the momentum of the bullet before the collision?

A: The momentum of the bullet before the collision is given by:

p_i = m_b v_b = 0.012 \, \text{kg} \times 450 \, \text{m/s} = 5.4 \, \text{kg m/s}

Q: What is the final momentum of the bullet and the wooden block?

A: The final momentum of the bullet and the wooden block is given by:

p_f = (m_b + m_w) v_f

where $v_f$ is the final velocity of the bullet and the wooden block.

Q: How do we apply the law of conservation of momentum to the system?

A: We can apply the law of conservation of momentum to the system by setting the initial momentum equal to the final momentum:

p_i = p_f

Substituting the expressions for $p_i$ and $p_f$ , we get:

5.4 \, \text{kg m/s} = (0.012 \, \text{kg} + 2.5 \, \text{kg}) v_f

Simplifying the equation, we get:

v_f = \frac{5.4 \, \text{kg m/s}}{2.512 \, \text{kg}} = 2.15 \, \text{m/s}

Q: What is the kinetic energy of the bullet before the collision?

A: The kinetic energy of the bullet before the collision is given by:

K_i = \frac{1}{2} m_b v_b^2 = \frac{1}{2} \times 0.012 \, \text{kg} \times (450 \, \text{m/s})^2 = 606.75 \, \text{J}

Q: What is the kinetic energy of the bullet and the wooden block after the collision?

A: The kinetic energy of the bullet and the wooden block after the collision is given by:

K_f = \frac{1}{2} (m_b + m_w) v_f^2 = \frac{1}{2} \times 2.512 \, \text{kg} \times (2.15 \, \text{m/s})^2 = 11.43 \, \text{J}

Q: What is the energy transfer between the bullet and the wooden block?

A: The energy transfer between the bullet and the wooden block is given by:

\Delta K = K_i - K_f = 606.75 \, \text{J} - 11.43 \, \text{J} = 595.32 \, \text{J}

Q: What is the significance of the law of conservation of momentum in this scenario?

A: The law of conservation of momentum is significant in this scenario because it shows that the momentum of the bullet is transferred to the wooden block, resulting in the block's upward motion. This demonstrates the principle of conservation of momentum, which is a fundamental concept in physics.

Q: What are some real-world applications of the law of conservation of momentum?

A: Some real-world applications of the law of conservation of momentum include:

Designing safety features for vehicles, such as airbags and crumple zones, to absorb the impact of a collision and reduce the risk of injury.
Developing strategies for preventing accidents, such as warning systems and emergency braking systems.
Understanding the behavior of complex systems, such as the motion of objects in a collision or the behavior of fluids in a pipe.

Conclusion

In this Q&A article, we have explored the physics behind a bullet hitting a soft wooden block attached to a wire. We have applied the law of conservation of momentum to the system and calculated the kinetic energy of the bullet and the wooden block before and after the collision. The energy transfer between the bullet and the wooden block was also calculated, showing that the kinetic energy of the bullet is transferred to the wooden block.