Let's break down and solve the problem step by step: ## **Given Data** - **Mass of puck, \( m_p \) = .4 kg** - **Initial velocity of puck, \( v_p \) = 80 m/h = \(\frac{80}{3.6} = 22.22 \) m/s** - **Mass of goalie (Ullmark), \( m_g \) = 90 kg** - **Coefficient of friction, \( \mu \) = .05** - **Initial velocity of goalie = (stationary)** ## **Step 1: Find the velocity after catching the puck** By **conservation of momentum**: \[ m_p v_p + m_g v_g = (m_p + m_g)V \] \[ (.4)(22.22) + (90)() = (.4 + 90)V \] \[ 8.888 = 90.4V \] \[ V = \frac{8.888}{90.4} \approx .0983 \text{ m/s} \] --- ## **Step 2: Find the Distance Slid Before Stopping** The only force stopping Ullmark is friction: \[ F_{\text{friction}} = \mu (m_p + m_g)g \] \[ a = \frac{F_{\text{friction}}}{m_p + m_g} = \mu g \] \[ a = .05 \times 9.8 = .49 \text{ m/s}^2 \] This is the **deceleration**. Use the kinematic equation: \[ v^2 = u^2 + 2as \] Final velocity = , initial velocity = \(V\), \(a = -.49\): \[ = (.0983)^2 + 2(-.49)s \] \[ = .00966 - .98s \] \[ .98s = .00966 \] \[ s = \frac{.00966}{.98} \approx .00986 \text{ m} \] \[ s \approx .99 \text{ cm} \] --- ## **Conclusion** **Ullmark (including the puck) slides about .99 cm before coming to a stop.** --- ### **Note** - The answer in the image is 13 cm, which suggests a possible miscalculation or typo in the velocity conversion. - Double-check the units for puck speed: If **80 km/h** (not m/h), then \( v_p = \frac{80 \times 100}{360} = 22.22 \) m/s (which we used). - If any value is different, please clarify for recalculation.

# Problem Breakdown and Solution ## Given Data - **Mass of puck, \( m_p \)**: 0.4 kg - **Initial velocity of puck, \( v_p \)**: 80 km/h (converted to m/s) \[ v_p = \frac{80 \times 1000}{3600} \approx 22.22 \text{ m/s} \] - **Mass of goalie (Ullmark), \( m_g \)**: 90 kg - **Coefficient of friction, \( \mu \)**: 0.05 - **Initial velocity of goalie**: 0 m/s (stationary) ## Step 1: Calculate the Velocity After Catching the Puck Using **conservation of momentum**: \[ m_p v_p + m_g v_g = (m_p + m_g)V \] Substituting the values: \[ (0.4)(22.22) + (90)(0) = (0.4 + 90)V \] \[ 8.888 = 90.4V \] Solving for \( V \): \[ V = \frac{8.888}{90.4} \approx 0.0983 \text{ m/s} \] ## Step 2: Calculate the Distance Slid Before Stopping The force stopping Ullmark is friction: \[ F_{\text{friction}} = \mu (m_p + m_g)g \] Where \( g \approx 9.8 \text{ m/s}^2 \): \[ a = \frac{F_{\text{friction}}}{m_p + m_g} = \mu g \] Calculating deceleration \( a \): \[ a = 0.05 \times 9.8 \approx 0.49 \text{ m/s}^2 \] Using the kinematic equation: \[ v^2 = u^2 + 2as \] Where: - Final velocity \( v = 0 \) - Initial velocity \( u = V \approx 0.0983 \text{ m/s} \) - Deceleration \( a = -0.49 \text{ m/s}^2 \) Substituting into the equation: \[ 0 = (0.0983)^2 + 2(-0.49)s \] \[ 0 = 0.00966 - 0.98s \] Rearranging to solve for \( s \): \[ 0.98s = 0.00966 \] \[ s = \frac{0.00966}{0.98} \approx 0.00986 \text{ m} \approx 0.986 \text{ cm} \] ## Conclusion **Ullmark (including the puck) slides approximately 0.99 cm before coming to a stop.** ### Note The discrepancy with the answer of 13 cm may suggest a need to verify the puck's initial speed. If the puck's speed was meant to be in a different context (like km/h), please provide clarification for a recalculation.