LINEAR momentum velocity of second PARTICLE after collision WOULD be \[\frac{V}{3}\] Coefficient of restitution is   \[e=\frac{relative\,\,velocity\,\,ofseparation}{relative\,\,velocity\,\,of\,approach}=\frac{V/3}{V}=\frac{1}{3}\]Loss in kinetic ENERGY             \[=\frac{1}{2}v{{V}^{2}}-\frac{1}{2}(3m){{\LEFT( \frac{V}{3} \right)}^{2}}=\frac{1}{3}m{{V}^{2}}\]

"> LINEAR momentum velocity of second PARTICLE after collision WOULD be \[\frac{V}{3}\] Coefficient of restitution is   \[e=\frac{relative\,\,velocity\,\,ofseparation}{relative\,\,velocity\,\,of\,approach}=\frac{V/3}{V}=\frac{1}{3}\]Loss in kinetic ENERGY             \[=\frac{1}{2}v{{V}^{2}}-\frac{1}{2}(3m){{\LEFT( \frac{V}{3} \right)}^{2}}=\frac{1}{3}m{{V}^{2}}\]

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A particle of mass m collides inelastically with another particle of mass 3m which is at rest. After the collision first particle comes to rest. Then choose the wrong answer

NEET Physics in NEET 10 months ago

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[d] From conservation of LINEAR momentum velocity of second PARTICLE after collision WOULD be \[\frac{V}{3}\] Coefficient of restitution is   \[e=\frac{relative\,\,velocity\,\,ofseparation}{relative\,\,velocity\,\,of\,approach}=\frac{V/3}{V}=\frac{1}{3}\]Loss in kinetic ENERGY             \[=\frac{1}{2}v{{V}^{2}}-\frac{1}{2}(3m){{\LEFT( \frac{V}{3} \right)}^{2}}=\frac{1}{3}m{{V}^{2}}\]

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