Question

A block of wood of mass 3 M is suspended by a string of length $\frac{10}{3}$ m. A bullet of mass M hits it with a certain velocity and gets embedded in it. The block and the bullet swing to one side till the string makes 120° with the initial position. The velocity of the bullet is ($g = 10 \ ms^{-2}$)

• A. $\frac{40}{\sqrt{3}} \ ms^{-1}$
• B. $20 \ ms^{-1}$
• C. $30 \ ms^{-1}$
• D. $40 \ ms^{-1}$

Answer 1

Answer: (D)

The velocity of the bullet is $40 \ ms^{-1}$.

Answer 2

The problem involves two main stages: an inelastic collision followed by a pendulum swing.

Stage 1: Inelastic Collision

When the bullet hits the block and gets embedded, it's an inelastic collision. Linear momentum is conserved during this collision.

Let:

• Mass of bullet = m_b = M
• Initial velocity of bullet = v (this is what we need to find)
• Mass of block = m_w = 3M
• Initial velocity of block = 0
• Combined mass after collision = m_c = m_b + m_w = M + 3M = 4M
• Velocity of the combined system immediately after collision = V

By conservation of linear momentum:

$m_b v + m_w (0) = m_c V$

$Mv = (4M)V$

$V = \frac{v}{4}$ (Equation 1)

Stage 2: Pendulum Swing

After the collision, the combined block and bullet system swings upwards. During this swing, mechanical energy is conserved. The system starts with kinetic energy KE at the bottom and converts it into potential energy PE at the maximum height of the swing.

• Length of the string = $L = \frac{10}{3} \ m$
• Angle made with the initial vertical position = $θ = 120°$
• Acceleration due to gravity = $g = 10 \ ms^{-2}$

The vertical height h gained by the center of mass of the system can be calculated using the length of the string and the angle of swing:

$h = L(1 - \cos\theta)$

Substitute the given values:

$h = \frac{10}{3} (1 - \cos(120°))$

We know that $\cos(120°) = -\frac{1}{2}$.

$h = \frac{10}{3} (1 - (-\frac{1}{2}))$

$h = \frac{10}{3} (1 + \frac{1}{2})$

$h = \frac{10}{3} (\frac{3}{2})$

$h = 5 \ m$

Now, apply conservation of mechanical energy:

Initial Kinetic Energy = Final Potential Energy

$\frac{1}{2} m_c V^2 = m_c g h$

$\frac{1}{2} (4M) V^2 = (4M) g h$

$\frac{1}{2} V^2 = g h$

$V^2 = 2gh$

Substitute the value of h and g:

$V^2 = 2 \times 10 \times 5$

$V^2 = 100$

$V = \sqrt{100}$

$V = 10 \ ms^{-1}$

Combining the stages:

Now, substitute the value of V back into Equation 1:

$V = \frac{v}{4}$

$10 = \frac{v}{4}$

$v = 10 \times 4$

$v = 40 \ ms^{-1}$

The velocity of the bullet is $40 \ ms^{-1}$.