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Answer 1 · 2014-03-07T12:53:15

The momentum increases by 100 %.

Let initial kinetic energy be #K# and momentum be #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 "%"#