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Question: The magnitude of force (in \( N \) ) acting on a body varies with time t (in \( \mu s \) ) is shown ...

The magnitude of force (in NN ) acting on a body varies with time t (in μs\mu s ) is shown in the Figure. AB, BC and CD are straight line segments. The magnitude of total impulse of the force acting on the object from t=4μst=4\mu s to t=16μst=16\mu s is given as______NsNs.

A.10−3sA{{.10}^{-3}}s

B.10−2sB{{.10}^{-2}}s

C.10−4sC{{.10}^{-4}}s

D.5×10−4sD.5\times {{10}^{-4}}s

E.5×10−3sE.5\times {{10}^{-3}}s

Explanation

Solution

Impulse is a concept based on momentum. An impulse is defined as the resultant force on a body times the time period over which this force is experienced. Impulse can be calculated from a force-time graph by taking the area under the graph in consideration.

Complete answer:

In the question, a graph is given which is having force on its y axis and time on its x axis. We have to find the impulse of the body from t=4μst=4\mu s to t=16μst=16\mu s time interval.

As we all know,

The impulse of a body can be expressed as,

I=F×tI=F\times t

Where FF is the force acting and tt is the time taken.

When we look into the graph, we can see that the area under can be given as the equation,

area=F\times t $

Therefore the area under a force time graph will be equal to the value of impulse.

From the graph,

Impulse= area of EBCD.

areaEBCD=areaEBCF+ΔFCDareaEBCD=areaEBCF+\Delta FCD

Therefore the impulse can be written as,

I=(BE+FC2×EF)+12FC×FDI=\left( \dfrac{BE+FC}{2}\times EF \right)+\dfrac{1}{2}FC\times FD

Substituting the values in it,

I=(200+18002×2×10−6)+[12×800×10×10−6]I=\left( \dfrac{200+1800}{2}\times 2\times {{10}^{-6}} \right)+\left[ \dfrac{1}{2}\times 800\times 10\times {{10}^{-6}} \right]

We have to simplify this, which will give,

I=(1000+4000)×10−6NsI=\left( 1000+4000 \right)\times {{10}^{-6}}Ns

I=5000×10−6NsI=5000\times {{10}^{-6}}Ns

I=5×10−3NsI=5\times {{10}^{-3}}Ns

Therefore the impulse acting over the body has been found out.

So, the correct answer is “Option E”.

Note: We can develop a direct connection between how a force is acting on a body over a period of time and the motion of the body as well. This is because of the impulse momentum theorem. When a collision happens, a body experiences a force for a certain period of time that will result in its mass undergoing a variation in velocity due to change in momentum. This is referred to as the impulse-momentum change theorem.