Question

a bullet of mass 10g strikes a wooden block with a velocity of 300m/s. after penetrating 20cm into it , its velocity drops to 200m/s . then average resistance offered by the block

Answer 1

V^2 = Vo^2 + 2a*d = 200^2.
300^2 + 2a*0.2 = 40,000,
0.4a = 40,000-90,000 = -50,000, a = -125,000 m/s^2.

F = M*a = 0.01 * (-125000) = -11250 N. Resisting the bullet.

Answer 2

First of all, we have to find the deceleration of the bullet.

$V^{2} = U^{2} + 2aS$ ------ (1)

Where, V = final velocity of the bullet = 200 m/s

            U = initial velocity of the bullet = 300 m/s

            a = acceleration

            S = distance traveled by bullet = 20 cm = 0.2 m

By using equation (1),

    $200^{2} = 300^{2} + 2* a* 0.2$

     $40000 = 90000 + 0.4a$

         $0.4a = -50000$

              $a = -50000/0.4$

              $a = -12500 = -1.25*10^{4} m/s^{2}$

Now we can use F=ma for bullet,

Where, F = resistance force

            m = mass of the bullet = 10g =0.1 kg

            a = acceleration of the bullet = 12500

    $F = 0.1 * 12500\\ F = 1250 N$

Answer : average resistance force = 1250 N