Question

· A stream of particles of mass m and separation d hits a perpendicular wall with a velocity vo and rebounds
along the original line of motion with a velocity v. The mass per unit length of the incident stream is
m
The force exerted by the stream on the wall is

Answer 1

Answer:

Correct option is

B

Nm+M

mvN

Mass of single particle is m

∴ Mass of N particles =Nm

N number of particles get embedded into the body of mass M

∴ Total mass of the combined system =M+Nm

Let the common velocity of the combined system be V

Applying conservation of momentum :

N(mv)+0=(M+mN)V

⟹V=

M+mN

mvN

Answer 2

Pressure exerted by positives on wall p=$\frac{F}{A}=2mnr^{2}cos^{2}$∅.

Explanation:

with the normal distance $(vdt)cos$∅

from wall will striate to the wall,

total volume of particles striking the wall $dv=(vdtcos$∅)A

$dv=Avcos$∅ dt

total. no. of particles striking the wall $N=ndr$

$=nAvcos$∅dt

force exerted by particles on wall

$F=\frac{dp}{dt} =\frac{Ndp}{dt}$

$F=\frac{(n Arcos0dt)(2 mr cos0)}{dt}$

$F=2mnAr^{2}cos^{2}$∅

$\frac{F}{A}=2mnr^{2}cos^{2}$∅.