A 0.05 kg bullet is fired with a velocity of 200 m/s into a stationary 0.95 kg block. The bullet embeds itself in the block. Calculate the final velocity of the combined mass and its kinetic energy.

• Mass of the lead bullet ($m_1$) = $0.05 \, \text{kg}$
• Initial velocity of the lead bullet ($v_1$) = $200 \, \text{m/s}$
• Mass of the lead block ($m_2$) = $0.95 \, \text{kg}$
• Initial velocity of the lead block ($v_2$) = $0 \, \text{m/s}$ (since it is at rest)

When the bullet is fired into the block and the block can move freely, it implies that the bullet embeds itself in the block, and they move together as a single combined mass after the impact. In such a collision, momentum is conserved, but kinetic energy is not.

Let $V_f$ be the final velocity of the combined bullet-block system.
The total momentum before the collision equals the total momentum after the collision:

Substitute the given values:

The final kinetic energy ($KE_f$) is given by the formula $KE = \frac{1}{2} M V^2$, where $M$ is the total mass and $V$ is the final velocity.

Substitute the total mass and the final velocity:

The final kinetic energy after impact is $\text{50 J}$.