A) 0.5 N-s
B) 0.1 N-s
C) 0.3 N-s
D) 1.2 N-s
Correct Answer: C
Solution :
If a constant force \[\overset{\to }{\mathop{F}}\,\] is applied a body for a short interval of time \[\Delta t\], then the impulse of this force will be \[\overset{\to }{\mathop{F}}\,\times \Delta \,t\]. Also from Newton?s law \[\overset{\to }{\mathop{F}}\,=m\overset{\to }{\mathop{a}}\,\] Where \[m\]is mass and a is acceleration. Given, \[m=150\,\,g=150\times {{10}^{-3}}\,kg,\]\[a=20\,\,m/{{s}^{2}}\] \[\therefore \] \[F=0.15\times 20=3\,N\] Impulse\[=F\Delta \,t=3\times 0.1=0.3\,N-s\] Note: Impulse is a vector quantity, its direction is the same as that of the force.You need to login to perform this action.
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