Let's break down the problem step by step. ## **Given Data** - Mass of bullet, \( m_{\text{bullet}} = 30 \) g \( = 0.03 \) kg - Mass of block B, \( m_B = 3 \) kg - Mass of carrier C, \( m_C = 30 \) kg - Initial velocity of bullet, \( v_{\text{bullet}} = 450 \) m/s (horizontal) - Initial velocity of B and C = 0 (at rest) - Coefficient of kinetic friction between B and C, \( \mu_k = 0.2 \) - All motions are along the same straight line (horizontal). --- ## **(a) Velocity of bullet + block B after the first impact** This is a **perfectly plastic collision** (bullet embeds in B), so they move together after collision. **Conservation of Momentum:** \[ \text{Initial momentum} = \text{Final momentum} \] Let \( v_1 \) be the velocity just after impact: \[ m_{\text{bullet}} v_{\text{bullet}} + m_B (0) = (m_{\text{bullet}} + m_B) v_1 \] \[ (0.03)(450) + 3(0) = (0.03 + 3) v_1 \] \[ 13.5 = 3.03 v_1 \] \[ v_1 = \frac{13.5}{3.03} \approx 4.46 \text{ m/s} \] **Final Answer (a):** \[ \boxed{v_1 = 4.46 \text{ m/s}} \] This is the velocity of the bullet+B block immediately after the bullet embeds in B. --- ## **(b) Final velocity of the carrier C** **Step 1:** After the first impact, block B (with embedded bullet) slides on carrier C (\( v_1 = 4.46 \) m/s), while C is at rest. **Step 2:** Block B slides **until it hits the end of C**. At this moment, there's a **perfectly plastic collision** between B and C. They stick together and move with a final velocity. **But before this collision, friction acts between B and C**, reducing the velocity of B relative to C. Let's analyze these steps in detail. --- ### **Step 1: Friction slows B before impact with C (relative motion)** - Initial velocity of B relative to C: \( v_{\text{rel,0}} = 4.46 - 0 = 4.46 \) m/s - Friction force acts on B (and equal and opposite on C). - Let the time to impact be \( t \), and assume B slides distance \( d \) before hitting the end of C. #### **Let’s denote:** - \( v_{B0} = 4.46 \) m/s (initial velocity of B relative to ground) - \( v_{C0} = 0 \) (initial velocity of C relative to ground) Let \( v_B \), \( v_C \) be their velocities at time \( t \), before the impact at the end. #### **Friction Analysis** - **Friction force,** \( f = \mu_k N = \mu_k m_B g = 0.2 \times 3 \times 9.81 = 5.886 \) N - **Acceleration on B (relative to ground):** \[ a_B = -\frac{f}{m_B} = -\frac{5.886}{3} = -1.962 \text{ m/s}^2 \] - **Acceleration on C (relative to ground):** \[ a_C = +\frac{f}{m_C} = +\frac{5.886}{30} = +0.1962 \text{ m/s}^2 \] #### **Relative velocity (B with respect to C):** \[ v_{\text{rel}} = v_B - v_C \] The **relative acceleration:** \[ a_{\text{rel}} = a_B - a_C = (-1.962) - (0.1962) = -2.1582 \text{ m/s}^2 \] The **initial relative velocity** is \( v_{\text{rel},0} = 4.46 \) m/s. When B hits the end of C, the relative velocity is still 4.46 m/s (since they started at the same point; the time to impact depends on the length of C, but the key is the relative velocity at the moment just before the impact). But what really matters is **their velocities right before the impact** (not just their relative velocity). #### **Let’s express their velocities after time \( t \):** \[ v_B = v_{B0} + a_B t = 4.46 - 1.962 t \] \[ v_C = v_{C0} + a_C t = 0 + 0.1962 t \] **Relative displacement until B hits the end (length \( L \) of C):** \[ L = (v_{B0} - v_{C0}) t + \frac{1}{2} (a_B - a_C) t^2 = v_{\text{rel},0} t + \frac{1}{2} a_{\text{rel}} t^2 \] **But, since the length is not given, we can proceed by considering the velocities just before the collision:** - B: \( v_B = 4.46 - 1.962 t \) - C: \( v_C = 0.1962 t \) --- ### **Step 2: Perfectly plastic collision between B and C** Now, **just before impact:** - Block B: \( v_B \) (found above) - Carrier C: \( v_C \) (found above) - Masses: \( m_B = 3 \) kg, \( m_C = 30 \) kg They stick together and move with common velocity \( v_f \). **Conservation of momentum (system B+C):** \[ m_B v_B + m_C v_C = (m_B + m_C) v_f \] \[ v_f = \frac{m_B v_B + m_C v_C}{m_B + m_C} \] Now, substitute the expressions for \( v_B \) and \( v_C \): \[ v_f = \frac{3 (4.46 - 1.962 t) + 30 (0.1962 t)}{33} \] \[ = \frac{3 \times 4.46 - 3 \times 1.962 t + 30 \times 0.1962 t}{33} \] \[ = \frac{13.38 - 5.886 t + 5.886 t}{33} \] \[ = \frac{13.38}{33} \] \[ = 0.4055 \text{ m/s} \] **Notice:** The \( t \) terms cancel! This means, **no matter how long B slides on C, the final velocity after the perfectly plastic collision is determined solely by the initial momentum after the first collision.** --- ## **Final Answers** ### **(a) Velocity of bullet + B after first impact** \[ \boxed{4.46 \text{ m/s}} \] ### **(b) Final velocity of the carrier** \[ \boxed{0.406 \text{ m/s}} \] --- ### **Detailed Explanation for Part (b):** - After the bullet embeds in B, the combined mass (3.03 kg) moves at 4.46 m/s and slides on C (which is at rest). - While sliding, friction acts between B and C: it slows down B and speeds up C, but the **force is internal to the system (B+C)**, so **total momentum of B+C is conserved until the impact at the end**. - When B hits the end, B and C collide and stick together (perfectly plastic collision). - Momentum is again conserved in this collision, and, as shown, the final velocity depends only on the initial momentum of B after the bullet embedded, divided by the total mass. - **Friction changes the individual velocities of B and C before the collision, but not the sum of their momenta.** ##### **Key Point:** The final velocity of the carrier (and the stuck-together B) is: \[ v_f = \frac{\text{initial momentum after bullet embeds}}{\text{total mass}} = \frac{(m_{\text{bullet}} + m_B) \times v_1}{m_{\text{bullet}} + m_B + m_C} = \frac{3.03 \times 4.46}{33.03} = 0.409 \text{ m/s} \] (Using more significant digits for accuracy) ## **Final boxed answer:** \[ \boxed{0.406 \text{ m/s}} \] --- **Summary Table:** | Step | Calculation | Answer | |---------------------|--------------------------------------------------|--------------| | (a) Bullet+B velocity | \( v_1 = \frac{0.03 \times 450}{3.03} \) | 4.46 m/s | | (b) Final carrier velocity | \( v_f = \frac{3.03 \times 4.46}{33.03} \) | 0.406 m/s | --- ### **Key Explanation for (b):** - Friction is an internal force within the B+C system, so it redistributes momentum between B and C but does not change the total system momentum. - The final velocity after the plastic collision is determined by applying conservation of momentum to the entire system at the instant after the bullet embeds and before friction has changed the total system's momentum. If you have further questions or want to explore the time/distance of sliding, let me know!