If ke of a body increases by 0.1 the percentage increase in its momentum will be?

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Answer 1 · 2018-03-29T06:14:36

The increase in momentum is $= 0.05$ $=0.05$

The kinetic energy is $E = \frac{1}{2} m v^{2}$ $E=1/2mv^2$

The momentum is is $p = m v$ $p=mv$

$v = \sqrt{\frac{2 E}{m}}$ $v=sqrt((2E)/m)$

Therefore,

$p = m v = m \sqrt{\frac{2 E}{m}} = \sqrt{2 m E}$ $p=mv=msqrt((2E)/m)=sqrt(2mE)$

Taking logs

$ln p = \frac{1}{2} (ln 2 + ln m + ln E)$ $lnp=1/2(ln2+lnm+lnE)$

Differentiating

$(Deltap)/p=1/2((Deltam)/m)+1/2((DeltaE)/E)$

But,

$(DeltaE)/E=0.1$

Therefore,

$(Deltap)/p=1/2*0.1=0.05$