A lead bullet of mass 0.05kg is fired with a velocity of 200ms ^ -1 into a block of mass 0.95kg. Given that…

Analysis

This is a classic Inelastic Collision problem. In this scenario, the bullet embeds itself into the block, and they move together as a single unit after the impact. We need to use the Law of Conservation of Linear Momentum to find the final velocity, and then calculate the kinetic energy of the combined mass.

Concept

1. Conservation of Momentum: The total momentum before impact equals the total momentum after impact. $$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$ Where: • $m_1 = 0.05\text{ kg}$ (mass of bullet) • $u_1 = 200\text{ ms}^{-1}$ (velocity of bullet) • $m_2 = 0.95\text{ kg}$ (mass of block) • $u_2 = 0\text{ ms}^{-1}$ (block is initially at rest) • $v = \text{final velocity of the combined mass}$

2. Kinetic Energy Formula: $K.E. = \frac{1}{2}mv^2$

Step-by-Step Solution

Step 1: Find the final velocity ($v$) Using the conservation of momentum:

Step 2: Calculate the final kinetic energy Now, use the combined mass ($M = m_1 + m_2 = 1.0\text{ kg}$) and the final velocity ($v = 10\text{ ms}^{-1}$):

Conclusion

The final kinetic energy of the system after the impact is 50 Joules.

Exam Tip

In collision problems, always remember that while momentum is conserved, kinetic energy is usually lost (converted into heat or sound) during an inelastic collision. If you calculate the initial kinetic energy ($0.5 \times 0.05 \times 200^2 = 1000\text{ J}$) and compare it to the final kinetic energy ($50\text{ J}$), you will see a significant loss, which confirms the collision was inelastic.