Here are the solutions to the questions: 4.1 Write down the definition of momentum in words. Momentum is the product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction. 4.2 The net force acting on object A is zero between $t = 10 \text{ s}$ and $t = 20 \text{ s}$. Use the graph and a relevant equation to explain why this statement is TRUE. Step 1: Observe the graph between $t = 10 \text{ s}$ and $t = 20 \text{ s}$. From the graph, the momentum of object A is constant at $50 \text{ kg} \cdot \text{m/s}$ during this time interval. Step 2: State the relevant equation relating net force and momentum. The net force acting on an object is equal to the rate of change of its momentum. $$ F_{net} = \frac{\Delta p}{\Delta t} $$ Step 3: Explain why the statement is true. Since the momentum ($p$) is constant between $t = 10 \text{ s}$ and $t = 20 \text{ s}$, the change in momentum ($\Delta p$) is zero. Therefore, according to the equation $F_{net} = \frac{\Delta p}{\Delta t}$, the net force acting on object A is zero. 4.3 Calculate the magnitude of the impulse that object A experiences between $t = 20 \text{ s}$ and $t = 50 \text{ s}$. Step 1: Determine the initial and final momentum from the graph. At $t = 20 \text{ s}$, the initial momentum $p_i = 50 \text{ kg} \cdot \text{m/s}$. At $t = 50 \text{ s}$, the final momentum $p_f = -120 \text{ kg} \cdot \text{m/s}$. Step 2: Calculate the impulse using the impulse-momentum theorem. Impulse ($J$) is the change in momentum: $$ J = \Delta p = p_f - p_i $$ $$ J = (-120 \text{ kg} \cdot \text{m/s}) - (50 \text{ kg} \cdot \text{m/s}) $$ $$ J = -170 \text{ kg} \cdot \text{m/s} $$ Step 3: State the magnitude of the impulse. The magnitude of the impulse is the absolute value of $J$. $$ |J| = |-170 \text{ kg} \cdot \text{m/s}| = 170 \text{ kg} \cdot \text{m/s} $$ The magnitude of the impulse is $\boxed{\text{170 kg} \cdot \text{m/s}}$. 4.4 At $t = 50 \text{ s}$, object A collides with another object, B, which has a momentum of $70 \text{ kg} \cdot \text{m/s}$ EAST. Use the information from the graph and the relevant principle to calculate the momentum of object B after the collision. Step 1: Define a positive direction and identify the momentum values before and after the collision for object A from the graph. Let EAST be the positive direction. Momentum of object A just before collision ($p_{A,before}$) at $t = 50 \text{ s}$ is $-120 \text{ kg} \cdot \text{m/s}$ (meaning $120 \text{ kg} \cdot \text{m/s}$ WEST). Momentum of object A just after collision ($p_{A,after}$) at $t = 50 \text{ s}$ is $50 \text{ kg} \cdot \text{m/s}$ (meaning $50 \text{ kg} \cdot \text{m/s}$ EAST). Step 2: Identify the momentum of object B before the collision. Momentum of object B before collision ($p_{B,before}$) is $70 \text{ kg} \cdot \text{m/s}$ EAST, so $p_{B,before} = +70 \text{ kg} \cdot \text{m/s}$. Step 3: Apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. $$ p_{A,before} + p_{B,before} = p_{A,after} + p_{B,after} $$ Step 4: Substitute the known values and solve for $p_{B,after}$. $$ (-120 \text{ kg} \cdot \text{m/s}) + (70 \text{ kg} \cdot \text{m/s}) = (50 \text{ kg} \cdot \text{m/s}) + p_{B,after} $$ $$ -50 \text{ kg} \cdot \text{m/s} = 50 \text{ kg} \cdot \text{m/s} + p_{B,after} $$ $$ p_{B,after} = -50 \text{ kg} \cdot \text{m/s} - 50 \text{ kg} \cdot \text{m/s} $$ $$ p_{B,after} = -100 \text{ kg} \cdot \text{m/s} $$ Step 5: State the momentum of object B after the collision, including direction. The negative sign indicates the direction is opposite to EAST, which is WEST. The momentum of object B after the collision is $\boxed{\text{100 kg} \cdot \text{m/s WEST}}$. That's 2 down. 3 left today — send the next one.