Answer:
Given, \[m=100g=\frac{100}{1000}=0.1\,kg\] \[K.E.=20Jp=?\] We know, \[K.E.=\frac{{{p}^{2}}}{2m}\] \[\Rightarrow P=\sqrt{K.E.\times 2m}\] \[\therefore P=\sqrt{20\times 2\times 0.1}=\sqrt{4}=2kg-m{{s}^{-1}}\]
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