A body of massm is moving in a straight line with momentum p. Starting at time t = 0, a force F = at acts in the same direction on the moving particle during time interval of T. So that its momentum changes from p to 2p. The value of T is

Options

A. `sqrt((2p)/(a)) `
B. `sqrt((p)/(a))`
C. `sqrt((2p)/(a))`
D. `(2p)/(a)`

Explanation

To solve the problem step by step, we will use the relationship between force, momentum, and time.

### Step 1: Understand the relationship between force and momentum
The force \( F \) acting on the body is given by the rate of change of momentum:
\[ F = \frac{dp}{dt} \]
where \( p \) is the momentum of the body.

### Step 2: Substitute the given force
According to the problem, the force is given as:
\[ F = at \]
where \( a \) is a constant and \( t \) is time.

### Step 3: Set up the equation
From the relationship established in Step 1, we can write:
\[ at = \frac{dp}{dt} \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ dp = at \, dt \]

### Step 5: Integrate both sides
Now, we will integrate both sides. The initial momentum is \( p \) at \( t = 0 \) and changes to \( 2p \) at \( t = T \):
\[ \int_{p}^{2p} dp = \int_{0}^{T} at \, dt \]

### Step 6: Solve the left side
The left side integrates to:
\[ 2p - p = p \]

### Step 7: Solve the right side
The right side integrates as follows:
\[ \int_{0}^{T} at \, dt = a \left[ \frac{t^2}{2} \right]_{0}^{T} = a \frac{T^2}{2} \]

### Step 8: Set the equations equal
Now we set the results of the integrals equal to each other:
\[ p = a \frac{T^2}{2} \]

### Step 9: Solve for \( T \)
Rearranging gives us:
\[ T^2 = \frac{2p}{a} \]
Taking the square root of both sides, we find:
\[ T = \sqrt{\frac{2p}{a}} \]

### Final Answer
Thus, the value of \( T \) is:
\[ T = \sqrt{\frac{2p}{a}} \]