A man standing on a trolley pushes another identical a trolley (both trolleys are at rest on a rough surface), are set in motion and stop alter some time so that they If the ratio of mass of first trolley with man to mass of second trolley is `3`, then find the ratio of the stopping distances of the second trolley to that of the first trolley. (Assume coefficient of friction to be the same for both the trolleys)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the masses
Let the mass of the second trolley be \( M \). According to the problem, the mass of the first trolley with the man is \( 3M \).

### Step 2: Apply the conservation of momentum
When the man pushes the second trolley, the momentum before and after must be conserved. Initially, both trolleys are at rest, so the total initial momentum is zero. After the push, let \( v_1 \) be the velocity of the first trolley (with the man) and \( v_2 \) be the velocity of the second trolley.

Using conservation of momentum:
\[ (3M)v_1 + (M)v_2 = 0 \]
This implies:
\[ 3Mv_1 + Mv_2 = 0 \implies v_2 = -3v_1 \]
(Note: The negative sign indicates that the trolleys move in opposite directions.)

### Step 3: Calculate the stopping distances
The stopping distance for an object under uniform deceleration can be calculated using the equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 when stopping),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (deceleration in this case),
- \( s \) is the stopping distance.

Rearranging gives:
\[ s = \frac{u^2}{2a} \]

For the first trolley (with the man):
\[ s_1 = \frac{v_1^2}{2a} \]

For the second trolley:
\[ s_2 = \frac{v_2^2}{2a} \]

### Step 4: Substitute for \( v_2 \)
From our earlier finding, \( v_2 = -3v_1 \). Thus:
\[ s_2 = \frac{(-3v_1)^2}{2a} = \frac{9v_1^2}{2a} \]

### Step 5: Find the ratio of stopping distances
Now we can find the ratio of the stopping distances:
\[ \frac{s_2}{s_1} = \frac{\frac{9v_1^2}{2a}}{\frac{v_1^2}{2a}} = \frac{9v_1^2}{v_1^2} = 9 \]

### Conclusion
The ratio of the stopping distances of the second trolley to that of the first trolley is:
\[ \frac{s_2}{s_1} = 9 \]