A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10–2 kg moving with a speed of 2 × 102 ms–1 . The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g = 10 ms-2 )
- 0.30 m
- 0.20 m
- 0.35 m
- 0.40 m
The Correct Option is B
Approach Solution - 1
Applying the conservation of momentum:
\(mu = (M + m)V\)
Substituting the given values:
\(10^{-2} \times 2 \times 10^2 = (1 + 10^{-2}) \times V\)
\(V = 2 \, \text{ms}^{-1}\)
The height to which the bob rises can be calculated using:
\(h = \frac{V^2}{2g}\)
\(h = \frac{2^2}{2 \times 10} = 0.2 \, \text{m}\)
Approach Solution -2
The problem describes a scenario where a bullet strikes and gets embedded in the bob of a simple pendulum. We need to find the maximum vertical height the combined system of the bob and bullet rises to after the collision.
Concept Used:
This problem involves two main physical principles applied in sequence:
1. Conservation of Linear Momentum: The collision between the bullet and the wooden bob is a perfectly inelastic collision (since the bullet gets embedded). In such a collision, linear momentum is conserved, but kinetic energy is not. The formula for conservation of momentum is:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) V \]
where \(m_1, m_2\) are the masses, \(u_1, u_2\) are their initial velocities, and \(V\) is the common final velocity of the combined mass.
2. Conservation of Mechanical Energy: After the collision, the combined mass swings upwards. During this swing, the gravitational force is a conservative force, and air resistance is assumed to be negligible. Therefore, the total mechanical energy (the sum of kinetic and potential energy) is conserved. The kinetic energy just after the collision is converted into gravitational potential energy at the highest point of the swing.
\[ \frac{1}{2} M_{\text{total}} V^2 = M_{\text{total}} g h \]
where \(M_{\text{total}}\) is the total mass, \(V\) is the velocity at the lowest point, \(g\) is the acceleration due to gravity, and \(h\) is the maximum height reached.
Step-by-Step Solution:
Step 1: List the given values.
Mass of the wooden bob, \(M = 1 \, \text{kg}\).
Mass of the bullet, \(m = 10^{-2} \, \text{kg} = 0.01 \, \text{kg}\).
Initial speed of the bullet, \(u = 2 \times 10^2 \, \text{ms}^{-1} = 200 \, \text{ms}^{-1}\).
The bob is initially at rest, so its initial speed is \(U = 0\).
Acceleration due to gravity, \(g = 10 \, \text{ms}^{-2}\).
Step 2: Apply the principle of conservation of linear momentum to the collision.
Let \(V\) be the common velocity of the bullet and bob combined, just after the collision. The initial momentum of the system is just the momentum of the bullet, and the final momentum is that of the combined mass.
\[ m u + M U = (m + M) V \]
Substituting the given values:
\[ (0.01 \, \text{kg}) \times (200 \, \text{ms}^{-1}) + (1 \, \text{kg}) \times (0) = (0.01 \, \text{kg} + 1 \, \text{kg}) V \] \[ 2 = (1.01) V \]
Solving for \(V\):
\[ V = \frac{2}{1.01} \, \text{ms}^{-1} \]
Step 3: Apply the principle of conservation of mechanical energy to the subsequent swing.
The kinetic energy of the combined mass \((m+M)\) at the lowest point is converted into potential energy at the maximum height \(h\). The total mass is \(M_{\text{total}} = m+M = 1.01 \, \text{kg}\).
\[ \frac{1}{2} (m+M) V^2 = (m+M) g h \]
The term \((m+M)\) cancels from both sides, simplifying the equation to:
\[ \frac{1}{2} V^2 = g h \]
We can rearrange this to solve for the height \(h\):
\[ h = \frac{V^2}{2g} \]
Step 4: Substitute the value of \(V\) from Step 2 into the equation for \(h\) and calculate the final result.
\[ h = \frac{\left(\frac{2}{1.01}\right)^2}{2 \times 10} = \frac{\frac{4}{(1.01)^2}}{20} \] \[ h = \frac{4}{20 \times (1.01)^2} = \frac{1}{5 \times 1.0201} \] \[ h = \frac{1}{5.1005} \, \text{m} \] \[ h \approx 0.196 \, \text{m} \]
Since the mass of the bullet is very small compared to the bob, approximating \(1.01 \approx 1\) gives \(V \approx 2 \, \text{ms}^{-1}\) and \(h = \frac{2^2}{2 \times 10} = \frac{4}{20} = 0.2 \, \text{m}\).
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