To solve the problem step by step, we will analyze the situation using the principles of conservation of momentum and relative motion.
### Step 1: Understand the System
The man weighs 80 kg and is standing on a trolley that weighs 320 kg. The trolley is on a frictionless horizontal surface, meaning there are no external horizontal forces acting on the system.
**Hint:** Remember that in a frictionless system, the total momentum before and after any event must remain constant.
### Step 2: Initial Momentum
Initially, both the man and the trolley are at rest. Therefore, the initial momentum of the system is:
\[
\text{Initial Momentum} = (80 \, \text{kg} + 320 \, \text{kg}) \times 0 \, \text{m/s} = 0 \, \text{kg m/s}
\]
**Hint:** The total momentum of a system at rest is zero.
### Step 3: Define Velocities
Let \( V_p \) be the velocity of the trolley after the man starts walking. The man walks at a speed of 1 m/s relative to the trolley. Thus, the velocity of the man relative to the ground will be:
\[
V_m = V_p + 1 \, \text{m/s}
\]
**Hint:** Use the concept of relative velocity to express the man's speed in terms of the trolley's speed.
### Step 4: Final Momentum
According to the conservation of momentum:
\[
\text{Final Momentum} = \text{Initial Momentum}
\]
This gives us:
\[
80 \, \text{kg} \cdot (V_p + 1) + 320 \, \text{kg} \cdot V_p = 0
\]
**Hint:** Set up the equation based on the final momentum being equal to the initial momentum.
### Step 5: Solve for \( V_p \)
Expanding the equation:
\[
80V_p + 80 + 320V_p = 0
\]
Combining like terms:
\[
400V_p + 80 = 0
\]
Solving for \( V_p \):
\[
400V_p = -80 \implies V_p = -\frac{80}{400} = -\frac{1}{5} \, \text{m/s}
\]
**Hint:** Isolate \( V_p \) to find the velocity of the trolley.
### Step 6: Calculate the Man's Velocity
Now substituting \( V_p \) back into the equation for the man's velocity:
\[
V_m = V_p + 1 = -\frac{1}{5} + 1 = \frac{4}{5} \, \text{m/s}
\]
**Hint:** Ensure you correctly add the velocities to find the man's speed relative to the ground.
### Step 7: Calculate Displacement
To find the displacement of the man relative to the ground after 4 seconds:
\[
\text{Displacement} = V_m \times t = \left(\frac{4}{5} \, \text{m/s}\right) \times 4 \, \text{s} = \frac{16}{5} \, \text{m} = 3.2 \, \text{m}
\]
**Hint:** Use the formula for displacement, which is velocity multiplied by time.
### Final Answer
The displacement of the man relative to the ground after 4 seconds is **3.2 meters**.
