Question

answer the following question

Caution⚠️: Don't Spam. If it happens ur account will be suspended​

Answer 1

F bullet

= d p/dt

= Pf- Pi/dt

= 0- mv/∆t

= - mv/∆t

if N number of bullets are fired then

F by bullet= mv N/∆T

here,n=N/∆t (it represents number of bullets fired per unit time)

F by bullet=mn v

As the plate is in mid air there are two forces acting on the plate

mg in downward direction and to mnv in upward direction

mnv=mg

I hope this help ( ╹▽╹ )

Answer 2

As per the question , metallic plate has mass M . Mass of bullets is m , velocity of bullets be v and rate of bullets be n bullets per second.

To find:

Force exerted by bullets on plate.

Concept:

We shall be using Conservation of Momentum concept for solving this type of questions .

Calculation:

Force exerted by the bullets on the wall will be equal to the rate of change of momentum of bullets.

Let number of bullets be N.

Let force be F

$F = \sf{net \: momentum \: change}$

$= > F = \sf{ \dfrac{momentum \: change}{time} }$

$= > F = \sf{ \dfrac{N(initial \: P - final \: P)}{time} }$

$= > F = \sf{ \dfrac{N(mv - 0)}{time} }$

$= > F = \sf{ \dfrac{N}{time} } \times (mv)$

$= > F = \sf{ n \: \times (mv)}$

$= > F = \sf{ nmv}$

So final answer :

Force on metallic plate is (nmv) units.