# Problem Statement A \(.4~\mathrm{kg}\) puck is shot at \(80~\mathrm{km/h}\) towards a stationary goalie (mass \(90~\mathrm{kg}\)) on his knees on the ice. The goalie catches the puck. The coefficient of kinetic friction (\(\mu_k\)) between the goalie and ice is \(.05\). Find how far the goalie slides before coming to rest. --- ## Step 1: Convert Units - Puck mass, \(m_p = .4~\mathrm{kg}\) - Goalie mass, \(m_g = 90~\mathrm{kg}\) - Puck velocity, \(v_p = 80~\mathrm{km/h}\) Convert \(v_p\) to m/s: \[ v_p = 80~\mathrm{km/h} \times \frac{100~\mathrm{m}}{1~\mathrm{km}} \times \frac{1~\mathrm{h}}{360~\mathrm{s}} = 22.22~\mathrm{m/s} \] --- ## Step 2: Conservation of Momentum Initial momentum (goalie is stationary): \[ p_{\text{initial}} = m_p v_p + m_g \cdot = m_p v_p \] After catching, both move together: Let final speed = \(V\) \[ (m_g + m_p)V = m_p v_p \] \[ V = \frac{m_p v_p}{m_g + m_p} \] Plug in values: \[ V = \frac{.4 \times 22.22}{90 + .4} = \frac{8.888}{90.4} \approx .098~\mathrm{m/s} \] --- ## Step 3: Friction Force and Deceleration Friction force: \[ f_k = \mu_k (m_g + m_p) g \] Deceleration (\(a\)): \[ f_k = (m_g + m_p)a \implies a = \frac{f_k}{m_g + m_p} = \mu_k g \] \[ a = .05 \times 9.8 = .49~\mathrm{m/s^2} \] --- ## Step 4: Distance Slid Before Stopping Use kinematics (\(v_f^2 = v_i^2 + 2ad\), with \(v_f = \), \(a\) is negative): \[ = V^2 - 2ad \] \[ d = \frac{V^2}{2a} \] Plug in values: \[ d = \frac{(.098)^2}{2 \times .49} \approx \frac{.0096}{.98} \approx .0098~\mathrm{m} \] --- ## Final Answer **The goalie and puck slide approximately \(.01~\mathrm{m}\) (1 cm) before coming to rest.** --- ### Summary Table | Quantity | Value | |-----------------|------------------| | Puck Mass | \(.4~\mathrm{kg}\) | | Goalie Mass | \(90~\mathrm{kg}\) | | Initial Speed | \(22.22~\mathrm{m/s}\) | | Final Speed | \(.098~\mathrm{m/s}\) | | Friction | \(.05\) | | Distance Slid | \(\approx .01~\mathrm{m}\) |