What % increase in momentum if increase of k.e is 23%
A. 46%
B. 33%
C. 50%
D. 11%
E. 46%
(1/2) m v2^2 = 1.23 (1/2) m v1^2
so
v2 = sqrt(1.23) v1
v2 = 1.11 v1
well, that is an 11% increase
Thx sir
Answer is correct but I cannot understand formula which you have written.
ke=1/2 m v^2
for ke increase of 23 precent, new KE is 1.23 of original
1.23*1/2 mvi^2=1/2 m v'^2
v'=vi*sqrt 1.23
To find the percentage increase in momentum when there is a 23% increase in kinetic energy (k.e.), we need to understand the relationship between momentum and kinetic energy.
Momentum (p) is defined as the product of an object's mass (m) and velocity (v):
p = m * v
Kinetic energy (k.e.) is defined as one-half the product of an object's mass and the square of its velocity:
k.e. = (1/2) * m * v^2
Given that the k.e. increases by 23%, we can express this as:
New k.e. = (1 + 0.23) * old k.e.
Since k.e. = (1/2) * m * v^2, we can write:
New k.e. = (1 + 0.23) * (1/2) * m * v^2
Simplifying, we get:
New k.e. = 1.115 * (1/2) * m * v^2
Since momentum is proportional to the square root of kinetic energy, we can write:
New p = sqrt(1.115) * old p
The percentage increase in momentum is then given by:
% increase in momentum = [(new p - old p) / old p] * 100
Plugging in the values, we have:
% increase in momentum = [(sqrt(1.115) * old p - old p) / old p] * 100
Calculating this expression, we find:
% increase in momentum ≈ 11%
Therefore, the correct answer is option D, 11%.