Two unequal masses P and Q moving along straight lines are brought to rest by applying equal retarding forces. If P moves twice the time as Q, but goes only `(1//3)^(rd)` of the distance covered by Q before coming to rest, the ratio of their initial velocities is

Options

A. `1:2`
B. `1:6`
C. `2:3`
D. `2:1`

Answer

To solve the problem, we need to find the ratio of the initial velocities of two unequal masses P and Q, given certain conditions about their motion under equal retarding forces. Let's break down the solution step by step.

### Step 1: Understand the given information
- Mass P moves for a time \( T_P = 2T \) (twice the time of mass Q).
- Mass Q moves for a time \( T_Q = T \).
- The distance covered by mass P is \( S_P = \frac{1}{3} S_Q \) (one-third of the distance covered by mass Q).

### Step 2: Use the equations of motion
For both masses, we can use the first equation of motion:
\[ V = U + AT \]
where:
- \( V \) is the final velocity (0 for both masses since they come to rest),
- \( U \) is the initial velocity,
- \( A \) is the acceleration (negative for retarding force),
- \( T \) is the time of motion.

#### For mass P:
Since \( V_P = 0 \):
\[ 0 = U_1 - A_1 (2T) \]
Rearranging gives:
\[ A_1 = \frac{U_1}{2T} \]

#### For mass Q:
Since \( V_Q = 0 \):
\[ 0 = U_2 - A_2 (T) \]
Rearranging gives:
\[ A_2 = \frac{U_2}{T} \]

### Step 3: Relate the accelerations
Since the retarding forces are equal, we have:
\[ F = m_1 A_1 = m_2 A_2 \]
This implies:
\[ A_1 = \frac{F}{m_1} \quad \text{and} \quad A_2 = \frac{F}{m_2} \]
Thus:
\[ \frac{A_1}{A_2} = \frac{m_2}{m_1} \]

### Step 4: Use the second equation of motion for distance
Using the second equation of motion:
\[ S = UT + \frac{1}{2} AT^2 \]

#### For mass P:
\[ S_P = U_1 (2T) - \frac{1}{2} A_1 (2T)^2 \]
Substituting \( A_1 \):
\[ S_P = U_1 (2T) - \frac{1}{2} \left(\frac{U_1}{2T}\right) (4T^2) \]
\[ S_P = 2U_1 T - U_1 T = U_1 T \]

#### For mass Q:
\[ S_Q = U_2 (T) - \frac{1}{2} A_2 (T)^2 \]
Substituting \( A_2 \):
\[ S_Q = U_2 T - \frac{1}{2} \left(\frac{U_2}{T}\right) (T^2) \]
\[ S_Q = U_2 T - \frac{1}{2} U_2 T = \frac{1}{2} U_2 T \]

### Step 5: Set up the distance relationship
From the problem, we know:
\[ S_P = \frac{1}{3} S_Q \]
Substituting the expressions we found:
\[ U_1 T = \frac{1}{3} \left(\frac{1}{2} U_2 T\right) \]
Cancelling \( T \) from both sides:
\[ U_1 = \frac{1}{6} U_2 \]

### Step 6: Find the ratio of initial velocities
Thus, the ratio of their initial velocities is:
\[ \frac{U_1}{U_2} = \frac{1}{6} \]

### Final Answer
The ratio of the initial velocities \( U_1 : U_2 = 1 : 6 \).

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