Question

A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = _________ m/s so that the pendulum describes a circle. (Assume the string to be inextensible and g = 10 m/s\(^2\)) 

 

Hint:

This is a classic two-part problem. Always work backward from the final condition (completing the circle) to find the required initial condition for that part (speed of the bob after collision). Then use that result in the first part of the problem (the collision) to find the ultimate unknown. Remember the critical speeds for vertical circular motion: \( \sqrt{gL} \) at the top, \( \sqrt{3gL} \) at the horizontal position, and \( \sqrt{5gL} \) at the bottom.

Answer 1

Correct Answer: 400

Step 1: Understanding the Question:
This problem involves two main physics concepts: a collision between a bullet and a pendulum bob, and the circular motion of the pendulum after the collision. We need to find the initial speed of the bullet required for the bob to just complete a vertical circle.
Step 2: Key Formula or Approach:
1. Condition for Completing a Vertical Circle: For a mass on a string to complete a vertical circle, its minimum speed at the lowest point must be \( u_{\text{min}} = \sqrt{5gL} \), where L is the length of the string.
2. Conservation of Linear Momentum: For the collision between the bullet and the bob, the total linear momentum just before and just after the collision is conserved. \( m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2 \).
Step 3: Detailed Explanation:
Part 1: Minimum speed for the bob
First, let's find the minimum speed \(u\) the bob must have just after the collision to complete the circle.
- Length of pendulum, \( L = 0.5 \) m.
- Acceleration due to gravity, \( g = 10 \) m/s\(^2\).
Using the formula for minimum speed at the bottom:
\[ u = \sqrt{5gL} = \sqrt{5 \times 10 \times 0.5} = \sqrt{25} = 5 \, \text{m/s} \] So, the bob must acquire a speed of 5 m/s immediately after being hit.
Part 2: Conservation of momentum
Now, we apply the conservation of linear momentum to the collision.
- Mass of bullet, \( m = 10 \, \text{g} = 0.01 \, \text{kg} \).
- Mass of bob, \( M = 1 \, \text{kg} \).
Let the initial direction of the bullet be positive.
- Initial velocity of bullet = \(v\).
- Initial velocity of bob = 0.
- Final velocity of bullet (recoils) = \( -100 \) m/s.
- Final velocity of bob = \( u = 5 \) m/s.
The momentum conservation equation is:
\( (\text{Momentum})_{\text{before}} = (\text{Momentum})_{\text{after}} \)
\[ mv + M(0) = m(-100) + Mu \] Substitute the known values:
\[ (0.01)v = (0.01)(-100) + (1)(5) \] \[ 0.01v = -1 + 5 \] \[ 0.01v = 4 \] \[ v = \frac{4}{0.01} = 400 \, \text{m/s} \] Step 4: Final Answer:
The minimum value of v is 400 m/s.