if the increase in the kinetic energy of a body is 22% then the increase in the momentum is nearly what %?

Question

NITAI PATRA · Answer

E=1/2(mv 2 ) or,E=1/2(m 2 v 2 )/m or,E=P 2 /m as P=mv or,dE=PdP/m so, dE/E=2(dP/P) If dE/E is 22%. then dP/P is 11% as

Debayan Ghosh · Answer

P=mv E=1/2mv² P=√2mE, if m is constant So, P2/P1 = √E2/E1=√1.22E/E ›P2=1.1P1 ›P2 = P1+0.1 P1 =P1+ 10% of P1

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if the increase in the kinetic energy of a body is 22% then the increase in the momentum is nearly what %?

if the increase in the kinetic energy of a body is 22% then the increase in the momentum is nearly what %?

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if the increase in the kinetic energy of a body is 22% then the increase in the momentum is nearly what %?

if the increase in the kinetic energy of a body is 22% then the increase in the momentum is nearly what %?

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