Question

A force of 50 dynes is acted on a body of mass 5g which is at rest for an interval of 3 seconds, the impulse is (in Ns):
(A) $1.5 \times {10^{ - 3}}$
(B) $1.5 \times {10^{ - 2}}$
(C) $2.5 \times {10^{ - 3}}$
(D) $1.5 \times {10^3}$

Answer 1

We know that the summary Newton's second law of motion is that the rate of change of momentum is called the force.
$F = \dfrac{{\Delta p}}{{\Delta t}}$
Where $\Delta p$ is the change in momentum and $\Delta t$ is the net time for which the force is acted on the object.
After rearranging we get the formula of impulse.
$F \times \Delta t = \Delta p$
This $\Delta p$ is the impulse.

Complete answer:
The force that is given in this question is not in standard form it is in dynes. And the options are in newton, so we need to convert the dynes into newton to reach the correct option.
Now the force value is 50 dynes
$F = 50dyne = 50 \times {10^{ - 5}} = 5 \times {10^{ - 4}}N$
(1 dyne is 0.00001 newton)
Now the time interval for which this impulse is acted upon is for 3 seconds.
$\Delta t = 3\sec$
Now impulse is
$\Delta p = F \times \Delta t$
$\Delta p = 5 \times {10^{ - 4}} \times 3 = 1.5 \times {10^{ - 3}}Ns$
So, the impulse is $1.5 \times {10^{ - 3}}Ns$ the option (D)

Thus, the correct option is (D).

Note:
Since this question is very simple and the force that is given to us in this question is constant throughout the time interval. But if the force varied with respect to time, we had to take an integral of force over the given amount of time in order to calculate the net impulse which is quite lengthy and difficult. So, in many questions the averaged force over time is given as in case of this one.