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

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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

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4 mnu

mnu

$\frac{1}{2} mnu$

2 mnu

answer is B.

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Detailed Solution

$\vec{F} = \frac{d \vec{p}}{dt} =$ Rate of change of momentum
As balls collide elastically hence, rate of change of momentum of ball $n [mu - (mu)] = 2 mnu$
$i . e . F = 2 mnu$

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