Answer:
We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] Applying the above formula for both cases, we get \[K.E{{.}_{1}}=\frac{1}{2}\times m\times {{(2)}^{2}}=2m\] ????..(1) \[K.E{{.}_{2}}=\frac{1}{2}\times m\times {{(6)}^{2}}=18m\] ????..(2) Dividing equation (1) by (2), we get \[\frac{K.E{{.}_{1}}}{K.E{{.}_{2}}}=\frac{2m}{18m}=\frac{1}{9}\] Therefore, \[K.E{{.}_{1}}:K.E{{.}_{2}}=1:90\] Body-1 Body-2 \[{{m}_{1}}=m/s\] \[{{m}_{2}}=m\,kg\] \[{{v}_{1}}=2\,\,m/s\] \[{{v}_{2}}=6\,\,m/s\]
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