Final answer:
Using the conservation of momentum principle, the velocity of the 12 kg block after a 0.005 kg bullet bounces off of it with a final velocity of 150 m/s west is calculated to be 0.229 m/s east.
Step-by-step explanation:
To find the velocity of the block after the collision, we need to use the formula for the conservation of linear momentum: pinitial = pfinal. Initial momentum of the system (bullet + block) is the momentum of the bullet since the block is at rest: mbullet × vbullet initial. 0.005 kg × 400 m/s = 2 kg·m/s east. The final momentum is the sum of the bullet's momentum (after bouncing back) and the block's momentum (which was at rest initially): (mbullet × vbullet final) + (mblock × vblock final). (0.005 kg × -150 m/s) + (12 kg × vblock final). Note that the bullet's final velocity is negative as it is now traveling westward. By setting the initial and final momentum equal, we can solve for the block's velocity: 2 kg·m/s = (-0.75 kg·m/s) + (12 kg × vblock final). 12 kg × vblock final = 2.75 kg·m/s, vblock final = (2.75 kg·m/s) / 12 kg = 0.229 m/s east. The velocity of the block is 0.229 m/s east after the bullet bounces off.