Question

item 7 agent arlene devised the following method of measuring the muzzle velocity of a rifle (the figure below). she fires a bullet into a 4.162 - kg wooden block resting on a smooth surface, and attached to a spring of spring constant k = 163.5 n/m. the bullet, whose mass is 7.870 g, remains embedded in the wooden block. she measures the maximum distance that the block compresses the spring to be 9.460 cm. (figure 1) figure < 1 of 1 > part a what is the speed v of the bullet? express your answer to four significant figures and include the appropriate units.

Explanation:

Step1: Calculate the kinetic - energy of the block - bullet system just after the collision

The elastic - potential energy of the spring at maximum compression is given by $U = \frac{1}{2}kx^{2}$, where $k = 163.5\ N/m$ and $x=9.460\ cm = 0.09460\ m$.
$U=\frac{1}{2}\times163.5\times(0.09460)^{2}$
$U=\frac{1}{2}\times163.5\times0.00894916$
$U = 0.7301\ J$
By conservation of mechanical energy, the kinetic energy of the block - bullet system just after the collision $K=\frac{1}{2}(M + m)v_{f}^{2}$ is equal to the elastic - potential energy of the spring at maximum compression, so $\frac{1}{2}(M + m)v_{f}^{2}=\frac{1}{2}kx^{2}$, where $M = 4.162\ kg$ and $m=7.870\ g=0.007870\ kg$.

Step2: Solve for the velocity of the block - bullet system just after the collision $v_{f}$

From $\frac{1}{2}(M + m)v_{f}^{2}=\frac{1}{2}kx^{2}$, we can cancel out $\frac{1}{2}$ on both sides and get $(M + m)v_{f}^{2}=kx^{2}$. Then $v_{f}=\sqrt{\frac{kx^{2}}{M + m}}$.
Substitute $k = 163.5\ N/m$, $x = 0.09460\ m$, $M = 4.162\ kg$, and $m = 0.007870\ kg$ into the formula:
$v_{f}=\sqrt{\frac{163.5\times(0.09460)^{2}}{4.162 + 0.007870}}$
$v_{f}=\sqrt{\frac{163.5\times0.00894916}{4.16987}}$
$v_{f}=\sqrt{\frac{1.46018866}{4.16987}}$
$v_{f}=\sqrt{0.349196}$
$v_{f}=0.5909\ m/s$

Step3: Use conservation of momentum to find the initial velocity of the bullet

By conservation of momentum, $mv=(M + m)v_{f}$, where $v$ is the initial velocity of the bullet. Then $v=\frac{(M + m)v_{f}}{m}$.
Substitute $M = 4.162\ kg$, $m = 0.007870\ kg$, and $v_{f}=0.5909\ m/s$ into the formula:
$v=\frac{(4.162+0.007870)\times0.5909}{0.007870}$
$v=\frac{4.16987\times0.5909}{0.007870}$
$v=\frac{2.46497428}{0.007870}$
$v = 313.2\ m/s$

Answer:

$v = 313.2\ m/s$