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Answer 1 · 2014-03-07T12:53:15

The momentum increases by 100 %.

Let initial kinetic energy be $K$ $K$ and momentum be $p$ $p$ .

$K = ½mv^2$ (Equation 1)

If $K$ increases by 300 %, it increases by a factor of $(300 %)/(100%) = 3$ .

So the new kinetic energy is

$K' = K + 3K = 4K$

From Equation 1: $v = sqrt((2K)/m)$

The initial momentum $p = mv = m × sqrt((2K)/m) = sqrt(m^2 ×(2K)/m) = sqrt(2Km)$

The new momentum $p' = sqrt(2K'm)$

So $(p')/p = sqrt((2K'm)/(2Km))$

$(p')/p = sqrt((K')/K)$

$(p')/p= sqrt((4K)/K)$

$(p')/p = 2$

$p' = 2p$

% change = $(p' – p)/p × 100 % = (2p – p)/p × 100 % = 100 %$

% increase in momentum = 100%

Answer 2 · 2017-04-21T18:25:05

$T = p^2/(2m)$

$p = sqrt(2mT)$

$(p_2 - p_1)/p_1 = (sqrt(2mT_2) - sqrt(2mT_1))/sqrt(2mT_1)$

$= (sqrt(color(red)(T_2)) - sqrt(T_1))/sqrt(T_1)$

$= (sqrt(color(red)(4 T_1)) - sqrt(T_1))/sqrt(T_1) = 1$ or $100 "%"$