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# Relation between Momentum and K.E

Relation between Momentum and K.E

Question:
If Momentum of a body is increased by 20%, what is percentage change in the K.E of the body?
a) 40%
b) 44%
c) 20%
d) 10%

44%

Solution:
Let the original momentum of the body be P, Then:
P=mv
Momentum is increased by 20%, so new momentum will be:
P' = P + 20%P
P' = P + 0.2P = 1.2P

K.E = 0.5 mv^2
=0.5 m^2v^2/m                          (multiply and divide by m)
K.E  = 0.5 P^2 / m                              ---------- (i)

New K.E = K.E'
= 0.5 (P')^2/m
= 0.5 (1.2P)^2 / m
= 0.5  (1.44) P^2 / m
= 1.44 K.E                              (using i)

Percentage Change in K.E = (New K.E - Old K.E) /Old K.E x 100%
= (1.44K.E - K.E)/K.E x 100%
= 44%

Second Method:

Momentum of a body is given by :
P = mv
P can be varied by varying either m or v. Since mass of a body is a constant quantity so we can change momentum of a body by changing its velocity.
Momentum will be increased by 20% if the velocity is changed by 20%, since its a linear relation between the two.
Let
mass = m = 10kg
velocity = v = 100m/s
Momentum with these values = P = 1000 kgm/s
K.E with these values = K.E = 50000 J

After 20% increase velocity becomes = v' = 120m/s
so new momentum = 1200 kgm/s
Change in momentum = (1200-1000)/1000  x 100%
= 0.2 x 100%
= 20%

New K.E = 72000 J
Change in K.E = (72000 - 50000)/50000  x 100%
= 0.44 x 100%
= 44%

Which agrees with the above result.

Formula to be used for finding the percentage change:

Percentage Change = New Value - Old value/Old value x 100%
In case of increment the change is positive.
In case of decrement the change is negative.