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
2/ The K. E of body decreases by
A/500j
B/750j
C/250j
D/1250j
Phy

Answer 1

Initial KE = 1000 Joules. And momentum decreases by 50%.

So, we know that :

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

Now, new momentum = 50% × P .

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

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

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

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

So, new KE of body is 250 J (Option C)

• KE of body decreases by 1000 - 250 = 750 Joule (Option B).

Hope It Helps.