Question

A bullet of mass $10\text{ g}$ moving horizontally with a velocity of $400\text{ m/s}$ strikes a wooden block of mass $3.99\text{ kg}$ suspended by a long string and gets embedded in it. The vertical height to which the block rises is ($g = 10\text{ m/s}^2$):

• A. 0.2 m
• B. 0.1 m
• C. 0.05 m
• D. 0.4 m Correct Answer: (C) 0.05 m

Hint:

For any inelastic ballistic pendulum problem, you can combine momentum and energy conservation into one neat master shortcut formula for height: $$\mathbf{h = \frac{1}{2g} \left(\frac{m_1 \cdot v_1}{m_1 + m_2}\right)^2}$$ Plugging in the standard variables saves valuable time during exams and prevents multi-step rounding errors!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem describes a classical ballistic pendulum system, which involves a two-phase mechanical process: 1. Collision Phase: The bullet strikes and embeds itself inside the suspended block. Because this occurs over an incredibly brief moments of impact, external forces like tension are negligible horizontally. This represents a textbook perfectly inelastic collision, where total linear momentum is strictly conserved, but kinetic energy is lost to heat and deformation. 2. Swing Phase: Following the impact, the combined mass moves together as a single body. As it swings upward along an arc, the tension force in the supporting string acts perpendicular to the path of motion, doing no work. Therefore, mechanical energy is conserved during the upward swing, converting all initial kinetic energy into gravitational potential energy at its highest point.

Step 2: Key Formula or Approach:
- Phase 1 (Conservation of Linear Momentum): $$ m_1 v_1 + m_2 v_2 = (m_1 + m_2)V $$ Where: - $m_1$ = mass of the bullet, $v_1$ = initial velocity of the bullet. - $m_2$ = mass of the stationary block, $v_2 = 0$. - $V$ = common velocity of the combined mass immediately after impact. - Phase 2 (Conservation of Mechanical Energy): $$ \frac{1}{2}(m_1 + m_2)V^2 = (m_1 + m_2)gh \implies h = \frac{V^2}{2g} $$ Where $h$ is the vertical height attained by the system.

Step 3: Detailed Explanation:
Let's convert our given parameters into standard SI units (kg, m, s): - Mass of the bullet ($m_1$) = $10\text{ g} = \frac{10}{1000}\text{ kg} = 0.01\text{ kg}$ - Initial velocity of the bullet ($v_1$) = $400\text{ m/s}$ - Mass of the wooden block ($m_2$) = $3.99\text{ kg}$ - Initial velocity of the block ($v_2$) = $0\text{ m/s}$ - Acceleration due to gravity ($g$) = $10\text{ m/s}^2$ 1. Calculate the combined velocity ($V$) immediately after collision: $$ (0.01\text{ kg} \times 400\text{ m/s}) + (3.99\text{ kg} \times 0) = (0.01\text{ kg} + 3.99\text{ kg}) \times V $$ $$ 4 = 4.00 \times V $$ $$ V = \frac{4}{4} = 1\text{ m/s} $$ 2. Calculate the maximum vertical height ($h$) using energy conservation: $$ h = \frac{V^2}{2g} $$ $$ h = \frac{(1)^2}{2 \times 10} = \frac{1}{20}\text{ m} $$ $$ h = 0.05\text{ m} $$

Step 4: Final Answer:
Based on the text parameters, the vertical height is $0.05\text{ m}$. matching option (C).