A model rocket rests on a frictionless horizontal surface and is joined by a string of length `l` to a fixed point so that the rocket moves in a horizontal circular path of radius `l`. The string will break if its tensiion exceeds a value `T`. The rocket engine provides a thrust `F` of constant magnitude along the rocket's direction of motion. the rocket has a mass `m` that does not change with time. Answer the following questions based on the above passage.
Starting from rest at `t=0` at what later time `t_(1)` is the rocket travelling so fast that the string breaks. Ignnore any air resistance.

Options

A. `((2mlT)/F^(2))^(1//2`
B. `((mlT)/(F^(2)))^(1//2)`
C. `((mlT)/(2F^(2)))^(1//2)`
D. `((mlF)/(T^(2)))^(1//2)`

Explanation

To solve the problem step-by-step, we need to analyze the forces acting on the rocket and the conditions under which the string will break.

### Step 1: Understand the Forces Acting on the Rocket
The rocket is propelled by a thrust force \( F \) and is also subject to centripetal force due to its circular motion. The tension in the string provides this centripetal force.

### Step 2: Establish the Relationship Between Thrust and Acceleration
From Newton's second law, we can express the acceleration \( a \) of the rocket as:
\[ a = \frac{F}{m} \]
where \( F \) is the thrust and \( m \) is the mass of the rocket.

### Step 3: Relate Acceleration to Velocity
Since the rocket starts from rest, we can integrate the acceleration to find the velocity \( v \) as a function of time \( t \):
\[ \frac{dv}{dt} = \frac{F}{m} \]
Integrating both sides gives:
\[ v = \frac{F}{m} t \]

### Step 4: Determine the Conditions for String Breakage
The tension \( T \) in the string must not exceed a certain value \( T \) for the string to remain intact. The tension can be expressed as:
\[ T = \frac{mv^2}{l} \]
where \( l \) is the length of the string (radius of the circular path).

### Step 5: Set Up the Equation for Breaking Condition
At the moment the string breaks, we have:
\[ \frac{mv^2}{l} = T \]
Substituting \( v \) from our earlier equation:
\[ \frac{m \left( \frac{F}{m} t \right)^2}{l} = T \]
This simplifies to:
\[ \frac{F^2 t^2}{ml} = T \]

### Step 6: Solve for Time \( t_1 \)
Rearranging the equation gives:
\[ t^2 = \frac{Tl}{F^2} \]
Taking the square root of both sides, we find:
\[ t_1 = \sqrt{\frac{Tl}{F^2}} \]

### Final Answer
Thus, the time \( t_1 \) at which the rocket is traveling so fast that the string breaks is:
\[ t_1 = \sqrt{\frac{Tl}{F^2}} \]