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

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.

The net force acting on an object is given by the rate of change of momentum:

Since $\Delta p = 0$ between $t = 10$ s and $t = 20$ s:

Therefore, the net force acting on object A is zero between $t = 10$ s and $t = 20$ s. The statement is TRUE.

4.3:

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 $p_{final} = -120 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$.

Impulse ($J$) is equal to the change in momentum ($\Delta p$):

The magnitude of the impulse is the absolute value:

4.4:

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.

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, $p_{B,final} = ?$

The negative sign indicates the direction is WEST.
The momentum of object B after the collision is $\boxed{\text{100 kg}\cdot\text{m}\cdot\text{s}^{-1} \text{ WEST}}$.