Question

A body of mass 5 kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 sec. Find the increase in its kinetic energy.
A. 2.8 J
B. 3.2 J
C. 3.8 J
D. 4.4 J

Answer 1

This question can be answered by using newton’s second law. According to Newton's second law for a particular system, rate of change of momentum will be equal to the external force acting on the system. Here external force will be the applied force. We use a kinetic energy formula to find the initial and final kinetic energy to find the change.

Formula used:
${F_{ext}} = \dfrac{{d{p_{system}}}}{{dt}}$

Complete step by step answer:
when an object of some mass moves with some velocity then it has kinetic energy and momentum. Now due to some impact if the velocity of that object changes then we can tell that its kinetic energy has changed and momentum has also changed. Now the rate of change of that momentum is a force while change in that momentum is called an impulse.
Let us assume a body of mass(m) moving with the velocity(v) then it will be having momentum(p). Momentum is a vector which means it has the direction and magnitude. That momentum is the product of the mass and the velocity. Direction of momentum is nothing but the direction of velocity of the body. Its magnitude will be the product of its mass and its velocity.
We have the formula
${F_{ext}} = \dfrac{{d{p_{system}}}}{{dt}}$ where ‘p’ is the momentum of the mass here
External force is 0.2 newton and the time is 10 seconds and initial momentum is given as 10 and mass is 5 kg. So the change in momentum will be
${F_{ext}} = \dfrac{{d{p_{system}}}}{{dt}}$
\eqalign{ & \Rightarrow dp = Fdt \cr & \Rightarrow {p_f} - {p_i} = 0.2 \times 10 \cr & \Rightarrow {p_f} - 10 = 2 \cr & \Rightarrow {p_f} = 12 \cr & \Rightarrow m{v_f} = 12 \cr & \Rightarrow {v_f} = \dfrac{{12}}{5} \cr & \therefore {v_f} = 2.4m/s \cr}
Kinetic energy in terms of momentum will be $\dfrac{{{p^2}}}{{2m}}$
Difference in final and initial kinetic energy will be
\eqalign{ & \Rightarrow \Delta KE = \dfrac{{{p_f}^2}}{{2m}} - \dfrac{{{p_i}^2}}{{2m}} \cr & \Rightarrow \Delta KE = \dfrac{{{{12}^2}}}{{2 \times 5}} - \dfrac{{{{10}^2}}}{{2 \times 5}} \cr & \Rightarrow \Delta KE = 4.4joules \cr & \therefore \Delta KE = 4.4J \cr}
Hence option D will be the correct answer.

Note:
Change in momentum is called as impulse but if that change happens very fast i.e within micro second or milliseconds then rate of change of momentum will be very high which means that the force will be very high and that kind of forces will be called as impulsive forces. But here the force applied is small and applied for considerable time.