Initial Velocity of Bullet Hitting Ball on Vertical Post

To find the initial velocity $V$ of the bullet, we analyze the motion in two stages: projectile motion after collision and conservation of linear momentum during the collision.

1. Time of flight for projectile motion:
Both the ball and the bullet leave the post horizontally from a height $h = 5\text{ m}$. The vertical motion under gravity (taking $g = 10\text{ m/s}^2$) gives the time of flight $t$:
$h = \frac{1}{2}gt^2$
$5 = \frac{1}{2} \times 10 \times t^2 \implies t^2 = 1 \implies t = 1\text{ s}$

2. Horizontal velocities after collision:
Let $v_{\text{ball}}$ and $v_{\text{bullet}}$ be the horizontal velocities of the ball and bullet immediately after the collision, respectively.

• For the ball:
  $x_{\text{ball}} = v_{\text{ball}} \times t \implies 20 = v_{\text{ball}} \times 1 \implies v_{\text{ball}} = 20\text{ m/s}$

• For the bullet:
  $x_{\text{bullet}} = v_{\text{bullet}} \times t \implies 100 = v_{\text{bullet}} \times 1 \implies v_{\text{bullet}} = 100\text{ m/s}$

3. Conservation of linear momentum during collision:
Since no net external horizontal force acts on the bullet-ball system during the short duration of the collision, the horizontal linear momentum is conserved:
$p_i = p_f$
$m_{\text{bullet}} V = M_{\text{ball}} v_{\text{ball}} + m_{\text{bullet}} v_{\text{bullet}}$

Given:

• Mass of the bullet, $m_{\text{bullet}} = 0.01\text{ kg}$
• Mass of the ball, $M_{\text{ball}} = 0.2\text{ kg}$

Substituting the values into the conservation equation:
$0.01 \times V = (0.2 \times 20) + (0.01 \times 100)$
$0.01 \times V = 4 + 1$
$0.01 \times V = 5$
$V = \frac{5}{0.01} = 500\text{ m/s}$

Thus, the initial velocity $V$ of the bullet is $500\text{ m/s}$.

Correct Option: D