Answers

2016-01-03T07:48:42+05:30
Let the k.e. in first case be x
x = 1/2*m*v^2=200j
in second case
let the kinetic energy be x'
v'=v
m'=2m
x'= 1/2*2m*v^2
x'= 2(1/2*m*v^2)
x'=2(x)
x'=2*200j
x'=400j
1 1 1
umm maybe wat u solvedis incorrect??
just check it out!!!...
yes Ashu31 wat u did is ryt am sorry!!!
please add as brainliest answer and hit the thnaks button
how do i add it as the  brainliest answer ?  cz im new to this!!
2016-01-03T09:13:00+05:30
Kinetic energy ( K_{e} ) of an object with mass m  and velocity v= \frac{1}{2} m v^{2} =200J(given)
when mass becomes double,then, m=2m
now  K_{e} = \frac{1}{2}*2mv^{2}
200J= \frac{1}{2}*2mv^{2}   \frac{200}{2J} = \frac{1}{2}*mv^{2}
100J= \frac{1}{2}*mv^{2}
∴ new  K_{e} =100J
0
k.e is not given for 2nd case as you have taken it = 200j
200j is for the 1st case when mass is m not 2m
they said u r doubling the mass naa...
and when we double the mass the k.e will not remain same 
k buddy wat u did is right !! no offence!! for sure !! ;-)i was mistaken!!!