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Question

Physics Question on laws of motion

In the figure given the position-time graph of a particle of mass 0.1 kg0.1\, kg is shown. The impulse at t=2 st=2\, s is :

A

0.2 kg m s−10.2\, kg\, m\, s ^{-1}

B

−0.2 kg m s−1-0.2\, kg\, m\, s ^{-1}

C

0.1 kg m s−10.1\, kg\, m\, s ^{-1}

D

−0.2 kg m s−1-0.2\, kg\, m\, s ^{-1}

Answer

−0.2 kg m s−1-0.2\, kg\, m\, s ^{-1}

Explanation

Solution

The impulse is given by I=I = change in momentum =Δ=mΔv=\Delta=m \Delta v =mΔxΔt=m \frac{\Delta x}{\Delta t} Given m=0.1 kg,ΔxΔt=4.02ms−1m=0.1\, kg , \frac{\Delta x}{\Delta t}=\frac{4.0}{2} ms ^{-1} ∴I=0.142=−0.2 kg−ms−1\therefore I=0.1 \frac{4}{2}=-0.2\, kg - ms ^{-1}