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NCERT Exemplar · Q18

Q.The position–time graph of a body of mass 2 kg2\ \text{kg} is a straight line rising from the origin: its position xx increases uniformly from 00 at t=0t=0 to 3 m3\ \text{m} at t=4 st=4\ \text{s}, and thereafter xx stays constant at 3 m3\ \text{m} for t>4 st>4\ \text{s}. What is the impulse on the body at t=0 st=0\ \text{s} and at t=4 st=4\ \text{s}?

Madhya Pradesh MpbseShort· 3mImportance★★★★★est
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The uniform slope tells us the body travels at a steady 0.75 m s−10.75\ \text{m s}^{-1} from t=0t=0 to t=4 st=4\ \text{s}, then rests. Impulse equals the sudden change in momentum, so it is +1.5 kg m s−1+1.5\ \text{kg m s}^{-1} at start and −1.5 kg m s−1-1.5\ \text{kg m s}^{-1} when it stops.

Concept: impulse = change in momentum

By the impulse–momentum theorem, J=Δp=m ΔvJ=\Delta p=m\,\Delta v. So we need the velocity just before and just after each instant.

Reading the velocity

For 0<t<4 s0<t<4\ \text{s} the graph is a straight line, so the velocity is constant and equals the slope:

v=ΔxΔt=3−04−0=0.75 m s−1.v=\frac{\Delta x}{\Delta t}=\frac{3-0}{4-0}=0.75\ \text{m s}^{-1}.

For t>4 st>4\ \text{s} the position is unchanging, so v=0v=0.

Impulse at t=0t=0

The body goes from rest (v=0v=0) to v=0.75 m s−1v=0.75\ \text{m s}^{-1}: …

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