Question
A bullet of mass 10 g moving at 400 m/s embeds itself in a wooden block of mass 990 g initially at rest on a frictionless surface. Find the speed of the bullet-block system after the collision.
Solution — Step by Step
When the bullet embeds in the block, the two objects move together after the collision. This is a perfectly inelastic collision — the objects stick together and share a common final velocity.
The surface is frictionless, so there is no external horizontal force. Linear momentum is conserved.
- Mass of bullet:
- Initial velocity of bullet:
- Mass of block:
- Initial velocity of block: (at rest)
- Final common velocity:
The final speed (4 m/s) is much less than the bullet’s initial speed (400 m/s), which makes physical sense. The heavy block (99× heavier) absorbs most of the momentum, slowing the system dramatically.
The momentum before = kg·m/s. The momentum after = kg·m/s. Momentum is conserved. ✓
Why This Works
Conservation of momentum applies whenever the net external force on the system is zero. On a frictionless surface, the only significant horizontal forces are the internal forces (bullet pushing block forward, block pushing bullet backward). These internal forces don’t change the total momentum of the system.
Kinetic energy is not conserved here. Initial KE = J. Final KE = J. The remaining 792 J is converted to heat, sound, and deformation — that’s the energy that buries the bullet in the wood.
Common Mistake
A very common error is forgetting to convert grams to kilograms before substituting. If you plug in and without converting, you get m/s by coincidence (since the ratio is the same), but in problems where units matter (energy calculations), wrong units give wrong answers. Always convert to SI units first.
The formula for perfectly inelastic collision simplifies nicely: when the second body is initially at rest. Memorise this form for JEE MCQs — it saves time.