A) 9.25 m/s.
B) 5 m/s.
C) 14.31 m/s.
D) 4.25 m/s.
Correct Answer: C
Solution :
\[F=\frac{dP}{dt}\Rightarrow dP=Fdt\] \[\therefore \]\[\int_{{}}^{{}}{dP=\int_{{}}^{{}}{F\,dt}\Rightarrow \Delta \Pi }=\int_{{}}^{{}}{F\,dt}\] So the change in moment\[\Delta P\]is equal to the area under the curve \[\therefore \]\[\Delta P=\frac{1}{2}\times 2\times 4+4\times 2+\frac{1}{2}(4+2.5)\]\[\times \,0.5+2.5\times 2\] \[=4+8+\frac{6.5}{4}+5=18.625\] \[\Rightarrow \]\[{{P}_{2}}-{{P}_{1}}=18.625\,\Rightarrow \,{{P}_{2}}={{P}_{1}}+18.625\] \[\therefore \]\[m{{v}_{2}}=m{{v}_{1}}+18.625\] \[\Rightarrow \]\[{{v}_{2}}={{v}_{1}}+\frac{18.625}{2}=5+\frac{18.625}{2}\] \[=14.3125\,m/s.\] Hence, the correction option is (c).
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