Question

In a ballistics test, a 21 g bullet traveling horizontally at 1200 m/s goes through a 35-cm-thick 300 kg stationary target and emerges with a speed of 950 m/s . The target is free to slide on a smooth horizontal surface.

Answer 1

Complete Question

In a ballistics test, a 21 g bullet traveling horizontally at 1200 m/s goes through a 35-cm-thick 300 kg stationary target and emerges with a speed of 950 m/s . The target is free to slide on a smooth horizontal surface.

a

How long is the bullet in the target? ____ s

b

What average force does the bullet exert on the target? _____ N

c

What is the target's speed just after the bullet emerges? _____ m/s

Answer:

a

$t =3.256*10^{-4} \ s$

b

$F = 16124 \ N$  

c

$v_1 = 0.0175 \ m/s$

Explanation:

From the question we are told that

  The mass of the bullet is  $m = 21 \ g = 0.021 \ kg$

   The horizontal speed is $u =1200 \ m/s$

   The thickness is  $s = 35 \ cm = 0.35 \ m$

   The mass of the block is  $m_1 = 300 \ kg$

   The emerging speed is  $v = 950 \ m/s$    

Generally from kinematic equation

    $v^2 = u^2 + 2as$

So

      $a = \frac{v^2 - u^2}{2s}$

=>  $a = \frac{ 950^2 - 1200^2}{2 * 0.35}$

=>   $a = -767857.14 \ m/s^2$

Gnerally the time taken is mathematically represented as

      $t = \frac{ v- u}{a}$

=>   $t = \frac{ 950 - 1200}{-767857.14}$

=>   $t =3.256*10^{-4} \ s$

Generally from law of momentum conservation we

     $m *u + m_1 * u_1 = m * v + m_1 * v_1$

$u_1$  is the velocity of the block at rest which is  $u_1 = 0 \ m/s$

=> $0.021 *1200 + 300 * 0 = 0.021 * 950 + 300 * v_1$

=> $v_1 = 0.0175 \ m/s$

Generally the force is mathematically represented as

    $F = \frac{m_1 * v_1 }{t}$

=> $F = \frac{300 * 0.0175 }{3.256 *10^{-4}}$

=> $F = 16124 \ N$