Physics Problem 3

Question

A 4.30 g bullet moving at 717 m/s strikes a 630 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 472 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Answer 1

Momentum is mass times velocity. 

 

The initial momentum of the bullet is 4.3 g * 717 m/s = 3083.1 g-m/s.  Since the block is not moving, that's the total momentum of the system.

 

After the collision, the bullet's momentum is 4.3 * 472 = 2029.6 g-m/s.  So it has given 1053.5 g-m/s of momentum to the block.

 

Since the block's mass is 630 g, it's mass must be 1053.5 / 630 = 1.67 m/s.

 

To find the velocity of the center of mass of the system, you divide the total momentum by the total mass.  The total mass of the system is 630 + 4.3 = 634.3 m/s.  And its total momentum before and after the collision is 3083.1 g-m/s.  So the center of mass of the system is traveling at 3083.1 / 634.3 = 4.86 m/s.