Question

When the momentum of body increases by 10% its K.E. increase by 

(1) 21% 

(2) 40% 

(3) 44% 

(4) None

Answer 1

Correct option is (1) 21% 

Given data,

Final momentum = initial momentum + 10% of initial momentum.

We know Kinetic energy \((K E)_1=\frac{p^2}{2 m}\) ...... (1)

Where, p = momentum, m = mass of the body,

If we increase p by \(10 \%,\)

Then final \(p=p+\frac{p}{10}=\frac{11 p}{10},\) Putting this value of p in (1), we get

\((K E)_2=\frac{(11 p / 10)^2}{2 m}=\frac{121}{100} \cdot \frac{p^2}{2 m}=\frac{121}{100} \cdot(K E)_1\)

Change in kinetic energy, 

\(\Delta K E=(K E)_2-(K E)_1\)

\(=\frac{121}{100}(K E)_1-(K E)_1=(K E)_1\left(\frac{121}{100}-1\right)=\frac{21}{100}(K E)_1\)

So \(\frac{\Delta K E}{(K E)_1} \times 100=21\)

Hence kinetic energy will change by \(21 \%,\) If momentum will increase by \(10 \%.\)