Question

Use 2nd law motion, derive the solution between force and acceleration of bullet of 10g strikes a sand bag at a speed of 10^3 ms^-1 and get embedded after travelling 5cm .Calculate the resistive force exerted by the sand on the bullet

Answer 1

Second law if motion states that rate of change if momentum is directly proportional to applied unbalanced force in the direction of force.

Here the sand bag will offer resistive force to the motion of the bullet and hence it stops after travelling 5 cm into it.

we have m = 10 g ; u = 10^3 m/s

v = 0 m/s , s = 5 cm = 0.05 m

As we know that from 2nd law

F = ( mv - mu )/t

or, F = m ( v - u )/t = m × a

Also, v^2 - u^2 = 2as

or, 0 - (10^3)^2 = 2×a×0.05

or, - 10^6 = 0.10a

so, a = -10^6/0.10 = -10^7 m/s^2

Hence, F = 10g × (-10^7 m/s^2)

= 0.01kg × (-10^7 m/s^2) = - 10^5 N

So, the sand bag will offer a force of 10^5 N in the opposite direction (-) of the motion of the bullet.

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\bf{\therefore 10^5 \:N}$

$\bf{\underline {Given:}}$

• A bullet of 10 g strikes a sand bag at a speed of$\:10^3 \:m\:s^{-1}$ .
• Gets embedded after travelling 5 cm.

$\bf{\underline {To\:Find:}}$

• The resistive force exerted by the sand on the bullet.

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

According to second law of motion, force is di proportional to rate of change of momentum.

So, $\:\:\:\:\:\:\:\:F=\dfrac{dp}{dt}$

We know that,

$p\:=\:mv$

So, $\:\:\:F=\dfrac{d(mv)}{dt} =m\dfrac{dv}{dt}$

$F\:=\:ma$ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\therefore {\dfrac{dv}{dt}} \:=\:a\right]$

Mass of bullet, $m\:=\:10g\:=\:\dfrac{10}{1000}\:kg\:=\:0.01\:kg$

Initial velocity of bullet, $u\:=\:10^3\:m\:s^{-1}$

Distance covered by bullet before coming to rest,

$s=5 cm = \dfrac{5}{100}m$

We know that,

$\:\:\:\:\:\:\:v^2 \:= \:u^2 \:- \:2as$

$\:\:\:\:\:\:\:0\:=\:(10^3)^2-2 \:a × \dfrac{5}{100}$

$\:\:\:\:\:\:\: 2 \:a × \dfrac{5}{100} \:= \:- (10^3)^2$

$\:\:\:\:\:\:\:a \:= \:- 10^7 \:m \:s^{-2}$

We know that,

Force, F = Mass × Acceleration

$\:\:\:\:\:\:\: = </strong><strong>0.01</strong><strong> </strong><strong>\:</strong><strong>×</strong><strong> </strong><strong>\:</strong><strong>10</strong><strong>^</strong><strong>7</strong><strong>\:</strong><strong> </strong><strong>kg</strong><strong> </strong><strong>\:</strong><strong>m</strong><strong>\:</strong><strong> </strong><strong>s</strong><strong>^</strong><strong>{</strong><strong>-</strong><strong>2</strong><strong>}</strong><strong>$

$\:\:\:\:\:\:\:= </strong><strong>\</strong><strong>bf</strong><strong> </strong><strong>10</strong><strong>^</strong><strong>5</strong><strong> </strong><strong>\:</strong><strong>N</strong><strong>$