Question

A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation \[ F = F_0 \left[ 1 - \left( \frac{t - T}{T} \right)^2 \right] \] Where \( F_0 \) and \( T \) are constants. The force acts only for the time interval \( 2T \). The velocity \( v \) of the particle after time \( 2T \) is:

• A. \( \frac{2F_0 T}{M} \)
• B. \( \frac{F_0 T}{2M} \)
• C. \( \frac{4F_0 T}{3M} \)
• D. \( \frac{F_0 T}{3M} \)

Hint:

The velocity can be found by integrating the acceleration, where acceleration is the force divided by the mass of the particle. Pay attention to the time-dependence of the force in such problems.

Answer 1

The Correct Option is C

Step 1: The force \( F(t) \) varies with time as: \[ F(t) = F_0 \left( 1 - \left( \frac{t - T}{T} \right)^2 \right) \] The work done by the force is given by the integral of force over displacement. Since the force is time-dependent, we first need to calculate the velocity using the relationship between force and acceleration, i.e., \( F = ma \). 
Step 2: Acceleration is the rate of change of velocity, \( a = \frac{dv}{dt} \). \[ F(t) = M \frac{dv}{dt} \quad \Rightarrow \quad \frac{dv}{dt} = \frac{F_0}{M} \left( 1 - \left( \frac{t - T}{T} \right)^2 \right) \] 
Step 3: Integrating both sides with respect to time from 0 to \( 2T \) (since force acts for time interval \( 2T \)): \[ v(2T) = \int_0^{2T} \frac{F_0}{M} \left( 1 - \left( \frac{t - T}{T} \right)^2 \right) dt \] Solving the integral yields: \[ v(2T) = \frac{4F_0 T}{3M} \] \bigskip