Instruction says the momentum versus time graph of object A ,originally moving horizontally EAST, is shown below

Question

Instruction says the momentum versus time graph of object A ,originally moving horizontally EAST, is shown below

Asked on March 28, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-28T14:53:09.244Z

4.1: Momentum is the product of an object's mass and its velocity. 4.2: Step 1: Analyze the graph between $t = 10$ s and $t = 20$ s. From the graph, the momentum of object A is constant at $50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ between $t = 10$ s and $t = 20$ s. This means the change in momentum ($\Delta p$) during this interval is zero. Step 2: Use the relevant equation. The net force acting on an object is given by the rate of change of momentum: $$F_{net} = \frac{\Delta p}{\Delta t}$$ Since $\Delta p = 0$ between $t = 10$ s and $t = 20$ s: $$F_{net} = \frac{0}{\Delta t} = 0$$ Therefore, the net force acting on object A is zero between $t = 10$ s and $t = 20$ s. The statement is TRUE. 4.3: Step 1: Identify the initial and final momentum values. At $t = 20$ s, the initial momentum of object A is $p_{initial} = 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$. At $t = 50$ s, the final momentum of object A is $p_{final} = -120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$. Step 2: Calculate the impulse. Impulse ($J$) is equal to the change in momentum ($\Delta p$): $$J = \Delta p = p_{final} - p_{initial}$$ $$J = (-120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}) - (50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1})$$ $$J = -170 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$ Step 3: Determine the magnitude of the impulse. The magnitude of the impulse is the absolute value: $$\text{Magnitude of impulse} = |-170 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}| = \boxed{\text{170 kg}\cdot\text{m}\cdot\text{s}^{-1}}$$ 4.4: Step 1: State the relevant principle. The relevant principle is the Principle of Conservation of Linear Momentum. It states that in an isolated system, the total linear momentum before a collision is equal to the total linear momentum after the collision. Step 2: Identify the initial and final momentum values for objects A and B. Let EAST be the positive direction. From the graph, just before the collision (at $t = 50$ s): Momentum of object A, $p_{A,initial} = -120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ (120 kg·m·s⁻¹ WEST). Momentum of object B, $p_{B,initial} = +70 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ (70 kg·m·s⁻¹ EAST). From the graph, just after the collision (at $t = 50$ s, after the vertical jump): Momentum of object A, $p_{A,final} = +50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ (50 kg·m·s⁻¹ EAST). Momentum of object B after the collision, $p_{B,final} = ?$ Step 3: Apply the principle of conservation of linear momentum. $$p_{A,initial} + p_{B,initial} = p_{A,final} + p_{B,final}$$ $$-120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} + 70 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} = 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} + p_{B,final}$$ $$-50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} = 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} + p_{B,final}$$ $$p_{B,final} = -50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} - 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$ $$p_{B,final} = -100 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$ The negative sign indicates the direction is WEST. The momentum of object B after the collision is $\boxed{\text{100 kg}\cdot\text{m}\cdot\text{s}^{-1} \text{ WEST}}$.

Accepted Answer · 2026-03-28T14:53:09.244Z

4.1: Momentum is the product of an object's mass and its velocity. 4.2: Step 1: Analyze the graph between $t = 10$ s and $t = 20$ s. From the graph, the momentum of object A is constant at $50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ between $t = 10$ s and $t = 20$ s. This means the change in momentum ($\Delta p$) during this interval is zero. Step 2: Use the relevant equation. The net force acting on an object is given by the rate of change of momentum: $$F_{net} = \frac{\Delta p}{\Delta t}$$ Since $\Delta p = 0$ between $t = 10$ s and $t = 20$ s: $$F_{net} = \frac{0}{\Delta t} = 0$$ Therefore, the net force acting on object A is zero between $t = 10$ s and $t = 20$ s. The statement is TRUE. 4.3: Step 1: Identify the initial and final momentum values. At $t = 20$ s, the initial momentum of object A is $p_{initial} = 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$. At $t = 50$ s, the final momentum of object A is $p_{final} = -120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$. Step 2: Calculate the impulse. Impulse ($J$) is equal to the change in momentum ($\Delta p$): $$J = \Delta p = p_{final} - p_{initial}$$ $$J = (-120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}) - (50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1})$$ $$J = -170 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$ Step 3: Determine the magnitude of the impulse. The magnitude of the impulse is the absolute value: $$\text{Magnitude of impulse} = |-170 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}| = \boxed{\text{170 kg}\cdot\text{m}\cdot\text{s}^{-1}}$$ 4.4: Step 1: State the relevant principle. The relevant principle is the Principle of Conservation of Linear Momentum. It states that in an isolated system, the total linear momentum before a collision is equal to the total linear momentum after the collision. Step 2: Identify the initial and final momentum values for objects A and B. Let EAST be the positive direction. From the graph, just before the collision (at $t = 50$ s): Momentum of object A, $p_{A,initial} = -120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ (120 kg·m·s⁻¹ WEST). Momentum of object B, $p_{B,initial} = +70 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ (70 kg·m·s⁻¹ EAST). From the graph, just after the collision (at $t = 50$ s, after the vertical jump): Momentum of object A, $p_{A,final} = +50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ (50 kg·m·s⁻¹ EAST). Momentum of object B after the collision, $p_{B,final} = ?$ Step 3: Apply the principle of conservation of linear momentum. $$p_{A,initial} + p_{B,initial} = p_{A,final} + p_{B,final}$$ $$-120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} + 70 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} = 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} + p_{B,final}$$ $$-50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} = 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} + p_{B,final}$$ $$p_{B,final} = -50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} - 50 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$ $$p_{B,final} = -100 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$ The negative sign indicates the direction is WEST. The momentum of object B after the collision is $\boxed{\text{100 kg}\cdot\text{m}\cdot\text{s}^{-1} \text{ WEST}}$.