4.1 Write down the definition of momentum in words.

From the graph, the momentum of object A is constant at $50 \text{ kg} \cdot \text{m/s}$ during this time interval.

The net force acting on an object is equal to the rate of change of its momentum.

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.

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 $p_f = -120 \text{ kg} \cdot \text{m/s}$.

Impulse ($J$) is the change in momentum:

The magnitude of the impulse is the absolute value of $J$.

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.

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).

Momentum of object B before collision ($p_{B,before}$) is $70 \text{ kg} \cdot \text{m/s}$ EAST, so $p_{B,before} = +70 \text{ kg} \cdot \text{m/s}$.

The total momentum before the collision is equal to the total momentum after the collision.

The negative sign indicates the direction is opposite to EAST, which is WEST.
The momentum of object B after the collision is $\boxed{\text{100 kg} \cdot \text{m/s WEST}}$.