Tabletop Fire

Physics Conservation of Momentum. Bullet Problem.?

A 5.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.630 m/s and 1.40 m/s, respectively, are thereby given to the blocks (Fig. 7-28b). Neglect the mass removed from the first block by the bullet.

(a) Find the speed of the bullet immediately after it emerges from the first block.

(b) Find the bullet's original speed.

Break the problem into pieces. In this case, it's easiest to tackle the second block first.

The bullet embeds itself in the second block. After this, the momentum of the system is
mv = 1.8055 * 1.40 m/s = 2.5277 kgm/s
Before impact, all this momentum was carried by the bullet: v = 2.5277/ 0.0055 = 459.6 m/s.
Which happens to be the speed of the bullet when it emerged from the first block (a)

The bullet passes through the first block, transferring some of its momentum to the block.
After the impact, the bullet has a momentum of 2.5277 kgm/s, and the block has a momentum of:
1.20 * 0.63 = 0.756 kgm/s. Before this impact, the block had a momentum of zero, and all the momentum in the system was in the bullet.
mv = 0.756 + 2.5277 = 3.2837.

v = 3.2837/0.0055 = 597 m/s. (b)

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