Bullet embedding in a block: common speed

A bullet of mass 0.02 kg is fired horizontally at a speed of 400 m/s into a block of mass 1.98 kg moving in the same direction at a speed of 100 m/s, and remains embedded in it. Find the common velocity of the bullet and the block after collision.

Step-by-step

1. m₁v₁ + m₂v₂ = (m₁ + m₂)v
   Momentum is conserved in the collision (no external horizontal force)
2. v = (0.02×400 + 1.98×100) ÷ (0.02+1.98) = 103
   Perfectly inelastic collision: the bodies share one speed afterwards

Why it works

The characteristic of a perfectly inelastic collision is that after the collision, both objects move together with the same velocity. In this case, momentum is conserved but kinetic energy is not. The method is to first calculate the sum of their momenta before the collision, then divide by the total mass after the collision. When moving in the same direction, the two momenta are added rather than subtracted; they are only subtracted when moving in opposite directions. This is where mistakes are most easily made.

Common mistakes

• 1) Treating same-direction motion as opposite direction, so subtracting the two momenta.
• 2) Using conservation of energy to calculate; kinetic energy is not conserved in a perfectly inelastic collision.
• 3) Omitting one object when calculating the total mass after the collision.

Related

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