An a-particle of mass m suffers onc-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is (a) 1.5 m (b) 4 m (c) 3.5 m (d) 2 m

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An a-particle of mass m suffers onc-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is (a) 1.5 m (b) 4 m (c) 3.5 m (d) 2 m

Pratik Kumar Jena 16th Dec, 2019

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NAVEEN MEENA Student Expert 16th Dec, 2019

Dear candidate, since it's elastic collision so by law of conservation of momentum -

mv0=mv2−mv1

12m(V1) ^2=0.36×12m(V0) ^2

v1=0.6v0

12m(V2)^ 2×0.64×12m(V0) ^2

V2=sqrt(mM) ×0.8V0

mV0=sqrt(mM)×0.8V0−m×0.6V0

⇒1.6m=0.8sqrt(mM)

4m^2=mM

Therefore the value of mass is 4

All the best

Hope this helps

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