Question

The K.E of a body is 1000j if its momentum decreases by 50% Then 1/K.E of body becomes
A/500j
B/750j
C/250j
D/1250j​

Answer 1

KE of body = 1000 Joule. And then momentum decreases by 50%.

So, we will say :

$KE = \dfrac{ {P}^{2} }{2m} = 1000 \: joule$

Now, new KE will be :

$KE_{2}= \dfrac{ {(50\% \times P)}^{2} }{2m}$

$\implies KE_{2}= \dfrac{ { (P/2)}^{2} }{2m}$

$\implies KE_{2}= \dfrac{1}{4} \times \dfrac{ { P}^{2} }{2m}$

$\implies KE_{2}= \dfrac{1}{4} \times KE$

$\implies KE_{2}= \dfrac{1}{4} \times 1000$

$\implies KE_{2}= 250 \: joule$

So, new KE is 250 Joule. (OPTION C) ✔️