If the kinetic energy of a body is increased by 4% the momentum:

(1)increases by 2%      (2)increases by 4%

(3)increases by 8%       (4)increases by 16%

Asked by ramanantra26 | 12th Sep, 2019, 11:51: PM

Expert Answer:

Let Kc be the change in kinetic energy, Ki be the initial kinetic energy, pc be the change in the momentum when kinetic energy is increased by 4% and p be the initial momentum. 
 
When the kinetic energy increases by 4%, 
Kc = Ki + 4% Ki
→Kc = Ki + 0.04 Ki
→Kc = 1.04 Ki 
 
Now, we know that 
K. E space equals space 1 half m v squared space
p equals m v space
therefore K. E space equals space fraction numerator p squared over denominator 2 m end fraction space
therefore fraction numerator p subscript c squared over denominator 2 m end fraction space equals 1.04 open parentheses fraction numerator p squared over denominator 2 m end fraction space close parentheses space
therefore p subscript c squared equals 1.04 space p squared space
therefore p subscript c equals space 1.019 space p space
Now, 
percent sign space c h a n g e space i n space t h e space m o m e n t u m space equals space fraction numerator P subscript c space minus space P over denominator P end fraction cross times 100 space equals space fraction numerator 1.019 space P minus space P over denominator P end fraction
space percent sign space c h a n g e space i n space t h e space m o m e n t u m equals space fraction numerator 0.019 P over denominator P end fraction cross times 100 space equals space 0.019 cross times 100 space equals space 1.9 percent sign
 
Thus, change in the momentum will be 1.9 %

Answered by Shiwani Sawant | 13th Sep, 2019, 11:04: AM