A)
B)
C)
D)
Correct Answer: C
Solution :
[c] From 0 to 2 s: at any time t, F=10 t \[\Rightarrow a=F/m=10\,t/m\] \[\Rightarrow \int\limits_{0}^{v}{dv}=\int_{0}^{t}{\frac{10t}{m}dt}\Rightarrow v=\frac{5{{t}^{2}}}{m}\] Momentum: \[p=mv=5{{t}^{2}}\] At \[t=2s,\]\[p=5{{(2)}^{2}}=20kg\] \[m{{s}^{-1}}\], v=20/m From 2 to 4 s: F=40-10 t \[\int_{20/m}^{v}{dv=\int_{2}^{t}{\frac{40-10t}{m}dt}}\] \[\Rightarrow v=\frac{1}{m}[40t-40-5{{r}^{2}}]\] \[p=mv=40t-40-5{{t}^{2}}\]
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