The linear momentum of a body is increased by 10%.what is the percentage increase in its KE?

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Solution

$D e a r S t u d e n t, W e k n o w, K E = \frac{1}{2} m v^{2} a n d P = m v T h e r e f o r e, K E = \frac{1}{2 m} {(m v)}^{2} = \frac{P^{2}}{2 m} C o n s i d e r i n g t h e m a s s t o b e c o n s \tan t, i f t h e m o m e n t u m i s i n c r e a s e d b y 10 %, l e t t h e n e w m o m e n t u m b e, P' = P + 0.1 P = 1.1 P T h e n e w k i n e t i c e n e r g y i s, K E' = \frac{{(1.1 P)}^{2}}{2 m} \Rightarrow K E' = 1.21 \frac{P^{2}}{2 m} \Rightarrow K E' = 1.21 K E \Rightarrow \frac{K E'}{K E} = 1.21 \Rightarrow \frac{K E'}{K E} - 1 = 1.21 - 1 \Rightarrow \frac{K E' - K E}{K E} \times 100 = 0.21 \times 100 = 21 % S o, k i n e t i c e n e r g y i n c r e a s e s b y 21 % . R e g a r d s$

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