Question

Assertion: $n$ small balls each of mass $m$ colliding elastically each second on surface with velocity
$u$ .The force experienced by the surface is $2mnu$ .
Reason: On elastic collision, the ball rebounds with the same velocity
A. If both assertion and reason are true and the reason is the correct explanation of the assertion.
B. If both assertion and reason are true but the reason is not the correct explanation of the
assertion.
C. If assertion is true statement but reason is false
D. If both assertion and reason are false statements

Answer 1

Hint: Here we have to see whether the statement is correct or not. Then we have to see whether the reason provided for the statement is correct or not. If both are correct then, we have to see whether the first statement is explained by the second statement or not.

Step by step answer:First let us see what elastic collision is-
An elastic collision is a collision, in which as a result of the collision, there is no net loss of kinetic energy in the system. In elastic collisions, momentum and kinetic energy are often conserved quantities.
Given,
Number of small balls $ = n$
Mass of the small balls $ = m$
Velocity of the small balls $ = u$
Total force experienced by the surface $ = 2mnu$

In elastic collisions, momentum and kinetic energy are often conserved quantities. Let us assume that two adjacent trolleys are travelling at equal speed towards each other. They dash, with no loss of momentum, bouncing off each other. This collision is fully elastic because it has lost no control.
The laws of momentum and energy govern collisions between objects. If a collision happens in an isolated system, the overall momentum of the entity system is preserved.
There is no loss of energy in elastic collisions. Therefore, after the surface collision, the kinetic energy would be equal to that before the collision. Therefore, there will be no change in heights, just direction changes.
Momentum shift per ball
$
{\Delta P}= m\left( {u - \left( { - u} \right)} \right) = 2mu$
In $t = 1\,\sec $ time $n$ balls hit.
Thus, net average force on surface
$
Force = \dfrac{{\Delta P}}
{{\Delta t}} = \dfrac{{1mu}}
{{\dfrac{1}
{n}}} = 2mnu$

From the above discussion we can conclude that option A is correct.

Note: Since there is no loss of energy here due to elastic collision, the statement and reason both are true. Otherwise if there had been inelastic collision, the statements would have not been true.