n small balls each of mass m impinge elastically each second on a surface with velocity u. The force experienced by the surface will be

mnu

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 mnu

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 mnu

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

mnu

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 2 mnu
Given :
mass of each ball $= m$ kg
velocity= $u m / s$
In elastic collision :
Change in momentum $= m u - (- m u)$
$= m u + m u$
$= 2 m u$
For $N$ balls change in momentum:
$Δ p = N \times 2 m u$
Rate of change of momentum $= Δ p / t = N 2 m u / 1 s e c$
$= N 2 m u$
According to Netwon's Second Law
$F = Δ p / t$
$F = 2 N m u$
Hence,
option $B$ is correct .

Suggest Corrections

Similar questions

n balls each of mass m impinge elastically each second on a surface with velocity u. The average force experienced by the surface will be

n

m

u

$n$ small balls, each of mass $m$ , impinge elastically each second on a surface with velocity $u$ . The force experienced by the surface will be

Join BYJU'S Learning Program