A particle of mass 1 kg is initially at rest. A force starts acting on it in one direction whose magnitude changes with time. The (F - t) graph is shown in figure. The velocity of the particle at the end of 10 s is
A
150 m/s
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B
120 m/s
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C
100 m/s
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D
112.5 m/s
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Solution
The correct option is D112.5 m/s Given: Velocity of particle at (t=0); vi=0 Let velocity of particle at (t=10 s) is vf. Using Impulse(J) = change in momentum ΔP
As we know area under (F -t) graph gives change in momentum.
∴ we have, m(vf−vi)= Area under (F - t) graph from t=0 to t=10 s
⇒m(vf−vi)= Area under (F - t) graph from (t=0 to t=5 s)+Area under (F - t) graph from (t=5 to t=10 s) ⇒mvf=12×5×15+5×15 ⇒1.vf=12×5×15+5×15 ∴vf=112.5 m/s