Right, let's go. 41. The momentum of an object at a given instant is defined as the product of its mass and velocity ($p = mv$). While acceleration changes velocity over time, at any given instant, momentum depends only on mass (inertia) and velocity (which includes speed and direction). Therefore, it is independent of acceleration at that instant. Answer: (B) acceleration 42. This scenario is a direct application of Newton's Third Law of Motion. When you kick a stone, you exert a force on it (action). In response, the stone exerts an equal and opposite force back on your foot (reaction), which is why you get hurt. Answer: (D) Reaction 43. Step 1: Relate kinetic energy and momentum. Momentum is $p = mv$. Kinetic energy is $KE = \frac{1}{2}mv^2$. We can express kinetic energy in terms of momentum: $$KE = \frac{1}{2}m v^2 = \frac{1}{2m} (mv)^2 = \frac{p^2}{2m}$$ Step 2: Apply the given conditions. We are given that $p_P = p_Q = p$ and $KE_Q > KE_P$. Using the formula $KE = \frac{p^2}{2m}$: $$KE_Q = \frac{p^2}{2m_Q}$$ $$KE_P = \frac{p^2}{2m_P}$$ Since $KE_Q > KE_P$, we have: $$\frac{p^2}{2m_Q} > \frac{p^2}{2m_P}$$ Step 3: Compare masses. Since $p^2$ and $2$ are positive, we can cancel them from the inequality: $$\frac{1}{m_Q} > \frac{1}{m_P}$$ This implies $m_P > m_Q$. So, Q weighs less than P. Step 4: Compare velocities. Since $p_P = p_Q$, we have $m_P v_P = m_Q v_Q$. Because $m_P > m_Q$, for the momentum to be equal, it must be that $v_Q > v_P$. Therefore, Q is moving faster than P. Answer: (B) is moving faster than P 44. The speed of the center of mass of a system remains constant if no external forces act on the system. A collision involves only internal forces, so the speed of the center of mass is conserved. We can calculate it using the initial conditions. Step 1: Identify the given values. Mass of cart A: $m_A = 0.20 \, \text{kg}$ Initial velocity of cart A: $v_A = 3.0 \, \text{m/s}$ Mass of cart B: $m_B = 0.40 \, \text{kg}$ Initial velocity of cart B: $v_B = 2.0 \, \text{m/s}$ (traveling away from A, so in the same direction) Step 2: Calculate the total momentum of the system. $$P_{total} = m_A v_A + m_B v_B$$ $$P_{total} = (0.20 \, \text{kg})(3.0 \, \text{m/s}) + (0.40 \, \text{kg})(2.0 \, \text{m/s})$$ $$P_{total} = 0.60 \, \text{kg} \cdot \text{m/s} + 0.80 \, \text{kg} \cdot \text{m/s}$$ $$P_{total} = 1.40 \, \text{kg} \cdot \text{m/s}$$ Step 3: Calculate the total mass of the system. $$M_{total} = m_A + m_B = 0.20 \, \text{kg} + 0.40 \, \text{kg} = 0.60 \, \text{kg}$$ Step 4: Calculate the speed of the center of mass. The speed of the center of mass $V_{CM}$ is given by: $$V_{CM} = \frac{P_{total}}{M_{total}}$$ $$V_{CM} = \frac{1.40 \, \text{kg} \cdot \text{m/s}}{0.60 \, \text{kg}}$$ $$V_{CM} = \frac{14}{6} \, \text{m/s} = \frac{7}{3} \, \text{m/s}$$ $$V_{CM} \approx 2.33 \, \text{m/s}$$ Answer: (C) $\boxed{\text{2.3 m/s}}$ 45. This problem involves the conservation of momentum during a collision. Step 1: Identify the given values and convert units. Mass of ball A: $m_A = 200 \, \text{g} = 0.200 \, \text{kg}$ Mass of ball B: $m_B = 400 \, \text{g} = 0.400 \, \text{kg}$ Initial velocity of ball A: $v_{A,initial} = +0.3 \, \text{m/s}$ (assuming positive direction) Initial velocity of ball B: $v_{B,initial} = ?$ (moving in opposite direction, so it will be negative) Final velocity of ball A: $v_{A,final} = 0 \, \text{m/s}$ (comes to rest) Final velocity of ball B: $v_{B,final} = 0 \, \text{m/s}$ (comes to rest) Step 2: Apply the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision: $$m_A v_{A,initial} + m_B v_{B,initial} = m_A v_{A,final} + m_B v_{B,final}$$ Step 3: Substitute the known values into the equation. $$(0.200 \, \text{kg})(0.3 \, \text{m/s}) + (0.400 \, \text{kg})(v_{B,initial}) = (0.200 \, \text{kg})(0 \, \text{m/s}) + (0.400 \, \text{kg})(0 \, \text{m/s})$$ Step 4: Simplify and solve for $v_{B,initial}$. $$0.06 \, \text{kg} \cdot \text{m/s} + 0.400 \, \text{kg} \cdot v_{B,initial} = 0$$ $$0.400 \, \text{kg} \cdot v_{B,initial} = -0.06 \, \text{kg} \cdot \text{m/s}$$ $$v_{B,initial} = \frac{-0.06 \, \text{kg} \cdot \text{m/s}}{0.400 \, \text{kg}}$$ $$v_{B,initial} = -0.15 \, \text{m/s}$$ The negative sign indicates that ball B was initially moving in the opposite direction to ball A, as stated in the problem. Answer: (C) $\boxed{\text{-0.15 m/s}}$ Got more? Send 'em 📸