Abullet is fired vertically into a 1.40 kg block of wood at rest directly above it. if the bullet has a mass of 29.0 g and a speed of 510 m/s, how high will the block rise after the bullet becomes embedded in it?

5.47 m  
The bullet undergoes a non-elastic collision with the block of wood and momentum is conserved. The initial momentum is 0.029 kg * 510 m/s = 14.79 kg*m/s. The combined mass of the block and bullet is 1.40 kg * 0.029 kg = 1.429 kg. Since momentum is conserved, the velocity of both combined will then be 14.79 kg*m/s / 1.429 kg = 10.34989503 m/s. 
 With a local gravitational acceleration of 9.8 m/s^2, it will take 10.34989503 m/s / 9.8 m/s^2 = 1.056111738 s for their upward velocity to drop to 0, just prior to descending. 
 The equation for distance under constant acceleration is
 d = 0.5 A T^2
 so
 d = 0.5 * 9.8 m/s^2 * (1.056111738 s)^2
 d = 4.9 m/s^2 * 1.115372003 s^2
 d = 5.465322814 m
  Rounding to 3 significant figures gives a height of 5.47 meters.

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