Question

K.E. of a body increases by $0.1\%$, the present increase in its momentum will be

• A. 0.0005
• B. 0.001
• C. 0.01
• D. 0.1

Answer 1

Answer: (A)

0.0005

Answer 2

Using K.E. = $\frac{p^2}{2m}$ , we get $2 \frac{\Delta p }{p} = \frac{\Delta E}{E}$ i.e. $\frac{\Delta p}{p} \times 100 = \frac{1}{2} \frac{\Delta E}{E} \times 100 = \frac{1}{2} \times$ 0.1 % i.e. change in K.E. = 0.05 %