Question

A quill of 0.100 g is falling with a velocity of (-0.05 j) m/s. When blown from lower side, its velocity changes to (0.20i + 0.15j) m/s. The change in its momentum will be ........... kgm/s
(A) 2 × 10⁻² i + 2 × 10⁻² j
(B) 2 × 10⁻⁵i + 2 × 10⁻⁵ j
(C) 2 × 10⁻² i + 1 × 10⁻² j
(D) 2 × 10⁻² i - 2 × 10⁻² j

Answer 1

Answer:

2 * 10⁻² i +  2 * 10⁻² j

Explanation:

Earlier Momentum =  mV

m = 100 gm = 0.1 kg

v = (-0.05 j) m/s

momentum = 0.1 * (-0.05) j  = -0.005 j

Momentum after blow

0.1 * (0.2 i + .15j)

= 0.02 i + 0.015 j

Change in Momentum = 0.02 i + 0.015 j - (-0.005 j)

= 0.02 i  + 0.02j

= 2 * 10⁻² i +  2 * 10⁻² j

Answer 2

Answer:

(B) 2*10^-5i + 2*10^-5j

Explanation:

They had asked answer in kgm/s. Here we have tp read carefully and must use proper units.

mass of quill= 1*10^-4 kg

so, change in momentum = Pf-Pi

                                           =1*10^-4(0.20i+0.15j)  - (-0.05j)

                                           =2*10^-5i + 1.5*10^-5j + 0.5*10^-5j

                                           =(2*10^-5i + 2*10^-5j) kgm/s