Question

If molecule of mass 5 x 10-25 kg is moving with a
velocity of 200 m/s inclined to horizontal at angle
O = 60°, then loss in momentum of molecule
during its collision with surface 's' will be (in kgm/s)
200m/s
a)
c)
10-22
10-24
b) 10-23
d) 10-25​

Answer 1

If molecule of mass 5 x 10¯²⁵ kg is moving with a velocity of 200 m/s inclined to horizontal at angle θ = 60°.

We have to find the loss in momentum of molecule during its collision with surface ‘s’.

Components of velocity along surface before and after collision are -200cos60° and 200cos60° respectively.

[here negative sign of component of velocity denotes that it is just directed opposite to the others. ]

Change in momentum = mass × change in velocity

= 5 × 10¯²⁵ (2 × 200 cos60°) kgm/s

= 5 × 10¯²⁵ × 400 × 1/2 kgm/s

= 1000 × 10¯²⁵ kgm/s

= 10¯²² kgm/s

Therefore loss in momentum of molecule during its collision with surface 's' will be 10¯²² kgm/s.

Hence option (a) 10¯²² is correct choice.