Answer:
We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] Applying the above formula for two cases, we get \[\Rightarrow \] ?????(1) \[K.E.=\frac{1}{2}\times m\times {{(2v)}^{2}}=2m{{V}^{2}}\] ?????(2) Dividing equation. (2) by (1), we get \[\frac{K.{{E}_{1}}}{K.{{E}_{2}}}=\frac{\frac{1}{2}m{{v}^{2}}}{2m{{v}^{2}}}=\frac{m{{v}^{2}}}{4m{{v}^{2}}}=\frac{1}{4}\] \[\therefore K.E{{.}_{1}}:K.E{{.}_{2}}=1:4\] Body-1 Body-2 \[{{m}_{1}}=m\] \[{{m}_{2}}=m\] \[{{v}_{1}}=v\] \[{{v}_{2}}=2v\]
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