A block of mass 0.50 kg is moving with a spee | vedclass

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Question

loss in $\mathrm{K} \mathrm{E}=\frac{\mathrm{m}_{1} \mathrm{m}_{2}}{2\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}\left(\mathrm{u}_{1}-\mathrm{u}_{2}\ri

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

loss in $\mathrm{K} \mathrm{E}=\frac{\mathrm{m}_{1} \mathrm{m}_{2}}{2\left(\mathrm{m}_{1}+\mathrm{m}_{2}ight)}\left(\mathrm{u}_{1}-\mathrm{u}_{2}ight)^{2}$

$=\frac{0.5 	imes 1}{2(0.5+1)}(2)^{2}=\frac{1}{1.5}=0.67 \mathrm{J}$

A block of mass $0.50\ kg$ is moving with a speedof $2.00$ $ms^{-1}$ on a smooth surface. It strikes another mass of $1.00\ kg$ and then they move together as a single body. The energy loss during the collision is ............. $\mathrm{J}$

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