Question

A force exerts an impulse $I$ on a particle and changes its speed from $u$ to $2u$. If the applied force and the initial velocity are oppositely directed along the same line then find the work done by the force.
A) $\dfrac{3}{2}Iu$
B) $\dfrac{1}{2}Iu$
C) $Iu$
D) $2Iu$

Answer 1

Hint:Impulse refers to the change in momentum of a particle when a force is applied to it. Since the speed of the particle changes, its kinetic energy must have also changed. The work-energy theorem suggests that this change in kinetic energy will be equal to the work done by the force.

Formulas used:
-The change in momentum of a body is given by, $\Delta p = m{v_f} - m{v_i}$ where $m$ is the mass of the particle, ${v_f}$ is the final velocity of the body and ${v_i}$ is its initial velocity.
-The kinetic energy of an object is given by, $K = \dfrac{1}{2}m{v^2}$ where $m$ is the mass of the object and $v$ is the velocity of the object.

Complete step by step answer.
Step 1: List the key points of the problem at hand.
An applied force causes the speed of a particle to change from $u$ to $2u$. An impulse $I$ is imparted to the particle. This implies that the momentum of the particle has changed. The force is directed opposite to the initial velocity.
The initial velocity of the particle is ${v_i} = u$ and its final velocity is ${v_f} = 2u$
Step 2: Find the change in the momentum of the particle.
 The change in momentum of a body is given by, $\Delta p = m{v_f} - m{v_i}$ -------- (1)
where $m$ is the mass of the particle, ${v_f}$ is the final velocity of the body and ${v_i}$ is its initial velocity.
Substituting for ${v_i} = - u$ and ${v_f} = 2u$ in equation (1) we get, $\Delta p = 2mu - \left( { - mu} \right) = 3mu$
Thus the impulse imparted to the particle is $I = \Delta p = 3mu$ .
Step 3: Find the change in the kinetic energy of the particle.
The kinetic energy of an object is given by, $K = \dfrac{1}{2}m{v^2}$ where $m$ is the mass of the object and $v$ is the velocity of the object.
Since the initial velocity of the particle is ${v_i} = - u$, the initial kinetic energy of the particle will be ${K_i} = \dfrac{1}{2}m{\left( { - u} \right)^2} = \dfrac{1}{2}m{u^2}$
Similarly, as the final velocity of the particle is ${v_f} = 2u$, the final kinetic energy of the particle can be obtained as ${K_f} = \dfrac{1}{2}m{\left( {2u} \right)^2} = \dfrac{4}{2}m{u^2}$
The change in kinetic energy is expressed as $\Delta K = {K_f} - {K_i}$ ---------- (2)
Substituting for ${K_i} = \dfrac{1}{2}m{u^2}$ and ${K_f} = \dfrac{4}{2}m{u^2}$ in equation (2) we get, $\Delta K = \dfrac{4}{2}m{u^2} - \dfrac{1}{2}m{u^2} = \dfrac{{3m{u^2}}}{2}$
Now substitute the obtained relation for impulse $I = 3mu$ in the above expression to get, $\Delta K = \dfrac{1}{2}Iu$
Then by the work-energy theorem, the work done by the force will be $W = \Delta K = \dfrac{1}{2}Iu$

Thus the correct option is B.

Note: It is mentioned in the question that the direction of the force is opposite to the direction of the initial velocity. Here, we assume the direction of force to be along the positive x-direction and the initial velocity to be along the negative x-direction. Thus we substitute ${v_i} = - u$ in equation (1). The direction is arbitrary and can be taken in the reverse order. The solution however will not change.