CBSE Class 9 Answered
if we increase kinetic energy by 20%, then find percentage increase in linear momentum
Asked by kaziryan.05 | 19 May, 2021, 07:43: AM
If v is speed of particle and m is mass of particle , then Kinetic energy K is given as
K = (1/2) m v2
Momentum p = ( m v )
If we eliminate v and write kinetic energy in terms of momentum , we get
K = p2 / ( 2 m )
If we take logarithm on both side and differentiate , we get
ln(K) = ln(1/2m) + 2 ln(p)
dK/K = 2 ( dp/p )
Hence if change in kinetic energy is 20% , dK/K = 0.2 .
Hence change in momentum ( dp/p ) = 0.1 or change in momentum 10%
Answered by Thiyagarajan K | 19 May, 2021, 09:11: AM
Application Videos
Concept Videos
CBSE 9 - Physics
Asked by topperlearningforcontent | 31 Jan, 2022, 10:14: AM
CBSE 9 - Physics
Asked by tajbindarkaurs | 14 Nov, 2021, 10:19: AM
CBSE 9 - Physics
Asked by kaziryan.05 | 19 May, 2021, 07:43: AM
CBSE 9 - Physics
Asked by parthmalikschool | 24 Nov, 2020, 12:45: AM
CBSE 9 - Physics
Asked by swatipuspapatel | 31 May, 2020, 12:23: PM
CBSE 9 - Physics
Asked by matoriark | 29 Mar, 2020, 02:21: PM
CBSE 9 - Physics
Asked by aprameyachaubeycr7 | 23 Mar, 2020, 02:01: AM
CBSE 9 - Physics
Asked by ujjwalsharma7667 | 28 Dec, 2019, 08:35: PM
CBSE 9 - Physics
Asked by hy0831422 | 21 Dec, 2019, 11:37: AM
CBSE 9 - Physics
Asked by shubhadanaik91 | 14 Nov, 2019, 09:02: PM