Question

# A body of mass 1 kg crosses a point O with a velocity $60m{{s}^{-1}}$. A force of 10N directed towards O begins to act on it. It will again cross O in$\left( A \right)24s$ $\left( B \right)12s$ $\left( C \right)6s$ $\left( D \right)\text{Will never return to O}$

Hint: We will apply the formula of the force and the formula of displacement in order to solve the question further. This is because we are given the values of acceleration and mass of the given body. The formula of displacement is used to find the time of the object crossing at point O again.

Formula used:
$F=ma$, $s=ut+\dfrac{1}{2}a{{t}^{2}}$, where F is the force applied on the object, m is the mass and a is the acceleration, s is the displacement, u is the initial velocity, and t is time.

Velocity: A velocity of any object, either living or non-living thing, is the rate of change of the position of an object with respect to the time. It is a vector quantity so, if the direction or a magnitude of the object changes it directly affects its velocity. The SI unit of velocity is m/s.

Force: A force is a work done on one object by another. It is a kind of push or pull of any object by another object. If for example we suppose that a human is pushing a wall but it is resulting in no change in the wall’s position, then this means that the force of the wall is pushing the human with equal and opposite reaction. Another example of force is the stretching and squeezing of a sponge. The SI unit of force is defined as Newton. Force is also a vector quantity as it depends upon the direction and magnitude both.

Acceleration: By the term acceleration we mean, the change in the velocity of an object with respect to time. As the acceleration is also a vector quantity so if the direction or magnitude of any object is changing then, it will directly affect its acceleration. This is why it is a vector quantity. The Si unit of acceleration is written as $m/{{s}^{2}}$.

Now, we will apply the formula of Force as,
\begin{align} & F=ma \\ & \Rightarrow a=\dfrac{F}{m} \\ & \Rightarrow a=\dfrac{10}{1}m/s \\ & \Rightarrow a=10m/s \\ \end{align}
Since, the object is coming back to its initial point O so it makes displacement as 0.
Now, by the formula $s=ut+\dfrac{1}{2}a{{t}^{2}}$ we have,
\begin{align} & 0=60\times t+\dfrac{1}{2}\left( -10 \right){{t}^{2}} \\ & \Rightarrow 0=60t-5{{t}^{2}} \\ & \Rightarrow 0=5{{t}^{2}}-60t \\ & \Rightarrow 0=t\left( 5t-60 \right) \\ & \Rightarrow t=0\,\,or\,\,5t-60=0 \\ & \Rightarrow t=0\,\,or\,\,5t=60 \\ & \Rightarrow t=0\,\,or\,\,t=12 \\ \end{align}
As, t = 0 seconds is not possible so, it is rejected therefore, t = 12 seconds. Therefore, if any object having a mass of 1 kg is crossing a point again with a velocity of $60m{{s}^{-1}}$ and there is a force of 10 Newton which is directed in the direction towards the point O then, it will cross that point again in 12 seconds.

So, the correct answer is “Option B”.

Note:
The things to be on your finger tips for further questions like this are,
(1) Since, the given body again crosses the point O so, at this time the formula of displacement is used instead of the distance. As the SI unit of time is seconds this is why the time is considered in seconds.
(2) The units of acceleration, force and velocity should be cross checked first before starting the solution. As it can have an effect on the answer.
(3) SI units of force, acceleration, velocity and time: Newton, $m/{{s}^{2}}$, m/s and s.
(4) The time t = 0 is rejected because as soon as the object starts to move to the point it reaches back to the same point so, it will take some time which will never be equal to 0.