bullet that has a mass of m_(bullet)=4.00g is moving horizontally with a velocity of overrightarrow (v)_(bullet)=+380m/s vhere the + sign Indicates that it is moving to the right.[See the part (a)of the drawing.]The mass of he first block is m_(block1)=1300g and its velocity is overrightarrow (v)_(block1)=0.480m/s after the bullet passes hrough it. [See part (b) of the drawing .] The mass of the second block is m_(block2)=1530g (c) What is the velocity overrightarrow (v)_(block)2 of the second block after the bullet imbeds itself? Include a plus or minus sign in your answer to denote direction.

To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision.<br /><br />Let's denote the velocity of the second block after the bullet imbeds itself as $\overrightarrow{v}_{block2}$.<br /><br />The total momentum before the collision is given by:<br /><br />$P_{initial} = m_{bullet} \cdot \overrightarrow{v}_{bullet} + m_{block1} \cdot \overrightarrow{v}_{block1} + m_{block2} \cdot \overrightarrow{v}_{block2}$<br /><br />The total momentum after the collision is given by:<br /><br />$P_{final} = m_{bullet} \cdot \overrightarrow{v}_{bullet} + m_{block1} \cdot \overrightarrow{v}_{block1} + m_{block2} \cdot \overrightarrow{v}_{block2}$<br /><br />Since the total momentum is conserved, we can set $P_{initial}$ equal to $P_{final}$ and solve for $\overrightarrow{v}_{block2}$.<br /><br />$P_{initial} = P_{final}$<br /><br />$m_{bullet} \cdot \overrightarrow{v} + m_{block1} \cdot \overrightarrow{v}_{block1} + m_{block2} \cdot \overrightarrow{v}_{block2} = m_{bullet} \cdot \overrightarrow{v}_{bullet} + m_{block1} \cdot \overrightarrow{v}_{block1} + m_{block2} \cdot \overrightarrow{v}_{block2}$<br /><br />Simplifying the equation, we get:<br /><br />$m_{bullet} \cdot \overrightarrow{v}_{bullet} + m_{block1} \cdot \overrightarrow{v}_{block1} + m_{block2} \cdot \overrightarrow{v}_{block2} = m_{bullet} \cdot \overrightarrow{v}_{bullet} + m_{block1} \cdot \overrightarrow{v}_{block1} + m_{block2} \cdot \overrightarrow{v}_{block2}$<br /><br />$0 = 0$<br /><br />This equation is always true, which means that the velocity of the second block after the bullet imbeds itself is the same as the velocity of the first block after the bullet passes through it.<br /><br />Therefore, the velocity of the second block after the bullet imbeds itself is $\overrightarrow{v}_{block2} = 0.480 \, \text{m/s}$, in the same direction as the velocity of the first block.