A bullet is fired from a gun with an initial horizontal velocity of 425 m/sec. It passes through a 2-kg block of wood, A, and becomes embedded in a second, 1.5-kg block of wood, B. The bullet causes Blocks A and B to begin moving to the right with velocities of 2 m/s and 3 m/s, respectively. The mass (g) of the bullet is most nearly:

We know from the conservation of momentum principle that in an isolated system, the total momentum before a collision is equal to the total momentum after the collision. Our equation for momentum is equal to:

$$\begin{gathered} p = mv \\ p = momentum \\ m = mass \\ v = velocity \end{gathered}$$

Let's first focus on what happens at the very end of the situation described in the problem. When the bullet enters block B, it stops moving. Therefore:

$$\begin{gathered} p_{initial}=p_{final} \\ {p}_{bullet-initial}+p_{block-b-intiial}=p_{bulet-final}+p_{block-b-final} \\ \left(m\right)\left(v_{after-block-a}\right)+(1.5kg)\left(0\right)=\left(m\right)\left(0\right)+(1.5kg)(3m/s) \\ \left(m\right)\left(v_{after-block-a}\right)=4.5 \end{gathered}$$

Now, let's set up an equation that represents the conservation of momentum that occurs between the bullet and block A as it passes through:

$$\begin{gathered} p_{initial}=p_{final} \\ {p}_{bullet-initial}+p_{block-a-intiial}=p_{bullet-final}+p_{block-a-final} \\ \left(m\right)(425m/s)+\left(2kg\right)\left(0\right)=\left(m\right)\left(v_{after-block-a}\right)+\left(2kg\right)(2m/s) \\ 425m-4=\left(m\right)\left(v_{after-block-a}\right) \end{gathered}$$

Since we found a numerical equivalent to $\left(m\right)\left(v_{after-block-a}\right)$, we can simply plug it in to solve for the mass of the bullet:

$$\begin{gathered} 425m-4=\left(m\right)\left(v_{after-block-a}\right) \\ 425m-4=4.5 \\ 425m=8.5 \\ m = 0.02 \end{gathered}$$

Therefore, the mass of our bullet is approximately .02kg, or 20g.