If Kinetic energy of a moving object increases by 21%, then its momentum increases by
কোন গতিশীল বস্তুর গতিশক্তি 21% বৃদ্ধি পেলে তার ভরবেগের বৃদ্ধি হয়,
Answers
Answer:
10 %
Explanation:
KE = 1/2 mv^2
increased by 21 % = 1/2 mv^2 + 1/2 mv^2 * 21/100
= 1/2 mv^2 ( 1 + 0.21 )
= 1.21 ( 1/2 mv^2 )
= 1.21 kE
we know
KE = p^2 / 2m
so
Δp^2 / 2m = 1.21 p^2 / 2m
so
Δp ^2 = 1.21 p^2
Δp = √1.21 p^2 ⇒ 1.1 p
% change in momentum is Δp - p / p * 100
= 1.1 p - p / p * 100
= 0.1 * 100
= 10 %
Answer:
If the kinematic energy of a moving object increases by 21%, then its momentum increase by 10%
Explanation:
Kinetic energy , KE in relation to momentum is given by the following formula:
k = P²/2m
Momentum from the above formula:
p² = 2mk
Where, k = kinetic energy
P = momentum
m = mass
Let the initial Kinetic energy be k,
If the kinetic energy increase by 21%, then the new kinetic energy is:
21% + 100% = 121%
121/100 × k = 1.21k
Initial kinetic energy = k,
p² = 2mk
Final kinetic energy = 1.21k
Final momentum therefore will be:
p² = 2m1.21k
⇒Percentage change is given by:
(Final value - initial value)/ initial value × 100%
Change in momentum
Initial momentum², p²₁ = 2mk
Final momentum², p²₂= 2m1.21k = 2.42mk
= (2.42mk - 2mk) / 2mk × 100%
= 0.42mk/2mk × 100%
= 0.21 × 100%
= 21%
Therefore change in square of momentum is 21%
Initial momentum, p²₁ = 100% ⇒ p = √100 = p₁ = 10% = 0.1p
New momentum, p²₂= 100% +21% =121% √121% = 11% = 0.11p
0.11p - 0.1 p / 0.1 p × 100%
0.01p / 0.1p × 100%
0.1 × 100 % = 10%
Therefore the change in momentum if kinetic energy changes by 21% is 10%