Question

a gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. (figure 1)
part a
if the bullet has a mass of 27.7 g and a speed of 230 m/s, how high will the block rise into the air after the bullet becomes embedded in it?
express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Convert bullet mass to kg

$m_b = 27.7\ \text{g} = 0.0277\ \text{kg}$

Step2: Apply conservation of momentum

Let $m_b$ = bullet mass, $v_b$ = bullet speed, $m_w$ = wood mass, $v_f$ = combined speed.
$$m_b v_b = (m_b + m_w) v_f$$
$$v_f = \frac{m_b v_b}{m_b + m_w} = \frac{0.0277 \times 230}{0.0277 + 1.40} = \frac{6.371}{1.4277} \approx 4.462\ \text{m/s}$$

Step3: Apply conservation of energy

Kinetic energy converts to gravitational potential energy:
$$\frac{1}{2}(m_b + m_w)v_f^2 = (m_b + m_w)gh_{max}$$
Cancel $(m_b + m_w)$, solve for $h_{max}$:
$$h_{max} = \frac{v_f^2}{2g}$$
Use $g = 9.81\ \text{m/s}^2$:
$$h_{max} = \frac{(4.462)^2}{2 \times 9.81} = \frac{19.91}{19.62} \approx 1.015\ \text{m}$$

Step4: Round to 3 significant figures

$h_{max} \approx 1.02\ \text{m}$

Answer:

$1.02$ meters