Question

A bag of sand of mass $98\, kg$ is suspended by a rope A bullet of $200\, g$ travelling with speed $10\, ms ^{-1}$ gets embedded in it, then loss of kinetic energy will be

• A. $4.9 J$
• B. $9.8 J$
• C. $14.7 J$
• D. $19.6 J$

Answer 1

The Correct Option is B

To determine the loss of kinetic energy when a bullet embeds into a bag of sand, we need to use the principles of conservation of momentum and energies. Let's break down the problem step-by-step:

1. Initial Information:
   • Mass of the bullet, \(m_B = 200\, \text{g} = 0.2\, \text{kg}\)
   • Velocity of the bullet, \(v_B = 10\, \text{m/s}\)
   • Mass of the sand bag, \(m_S = 98\, \text{kg}\)
   • Velocity of the sand bag, \(v_S = 0\, \text{m/s}\) (initially at rest)
2. Apply Conservation of Momentum:

The total momentum before and after the collision must be equal.

Before collision, the total momentum:

1. \(p_{\text{initial}} = m_B \cdot v_B + m_S \cdot v_S = 0.2 \cdot 10 + 98 \cdot 0 = 2\, \text{kg m/s}\)

After collision, the momentum is:

1. \(p_{\text{final}} = (m_B + m_S) \cdot v_f\)

Equating the two momenta, we get:

1. \(2 = (0.2 + 98) \cdot v_f\)

Solve for \(v_f\):

1. \(v_f = \frac{2}{98.2} \approx 0.0204\, \text{m/s}\)
2. Calculate Initial and Final Kinetic Energies:

Initial kinetic energy, \(KE_i\) before collision:

1. \(KE_i = \frac{1}{2} m_B v_B^2 = \frac{1}{2} \cdot 0.2 \cdot 10^2 = 10\, \text{J}\)

Final kinetic energy, \(KE_f\) after collision (both bullet and sand moving together):

1. \(KE_f = \frac{1}{2} (m_B + m_S) v_f^2 = \frac{1}{2} \cdot 98.2 \cdot (0.0204)^2 \approx 0.2\, \text{J}\)
2. Calculate Loss of Kinetic Energy:

The loss in kinetic energy is:

1. \(\text{Loss} = KE_i - KE_f = 10 - 0.2 = 9.8\, \text{J}\)

The loss of kinetic energy due to the bullet embedding in the sand bag is 9.8 J, thus the correct answer is the option $9.8\, J$.