A body of mass 'm' at rest is applied with a force F = (2 + 4t)N. The change in momentum of the interval one second to two second is

Options

A. ` 3kg - ms^(-1)`
B. `4 kg - ms^(-1)`
C. `6 kg ms^(-1)`
D. `8 kg ms^(-1)`

Answer

To solve the problem, we need to find the change in momentum of a body of mass 'm' when a force \( F = (2 + 4t) \, \text{N} \) is applied over the time interval from 1 second to 2 seconds. Here’s a step-by-step solution:

### Step 1: Understanding the relationship between force and momentum
The change in momentum (\( \Delta P \)) of an object is given by the integral of force over time:
\[ \Delta P = \int F \, dt \]

### Step 2: Substitute the given force into the equation
Given the force \( F = 2 + 4t \), we can substitute this into the momentum change equation:
\[ \Delta P = \int (2 + 4t) \, dt \]

### Step 3: Set the limits of integration
We need to find the change in momentum from \( t = 1 \, \text{s} \) to \( t = 2 \, \text{s} \). Therefore, we will integrate from 1 to 2:
\[ \Delta P = \int_{1}^{2} (2 + 4t) \, dt \]

### Step 4: Calculate the integral
Now we calculate the integral:
\[ \Delta P = \left[ 2t + 2t^2 \right]_{1}^{2} \]
This is because the integral of \( 4t \) is \( 2t^2 \) and the integral of \( 2 \) is \( 2t \).

### Step 5: Evaluate the integral at the limits
Now we evaluate the expression at the upper and lower limits:
\[ \Delta P = \left( 2(2) + 2(2^2) \right) - \left( 2(1) + 2(1^2) \right) \]
Calculating this gives:
\[ \Delta P = \left( 4 + 8 \right) - \left( 2 + 2 \right) \]
\[ \Delta P = 12 - 4 = 8 \, \text{kg m/s} \]

### Conclusion
Thus, the change in momentum of the body from 1 second to 2 seconds is:
\[ \Delta P = 8 \, \text{kg m/s} \]