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

To find the minimum value of initial velocity \( v \) that allows the pendulum bob to complete a full circle, we'll use the principles of momentum conservation and energy conservation.

Step 1: Apply Conservation of Momentum
Initially, the bullet has momentum \( mv \), and the pendulum bob is stationary. After collision, the bullet recoils, and the bob moves, so:
\(mv = mv' + Mu\)
Solving for \( u \), the velocity of the bob after collision, is:
\(u = \frac{mv - mv'}{M}\)
Given \( m = 0.01 \, \text{kg} \) (10 g), \( M = 1 \, \text{kg} \), \( v' = 100 \, \text{m/s} \), this becomes:
\(u = \frac{0.01v - 0.01 \times 100}{1} = 0.01v - 1\)

Step 2: Apply Energy Conservation for Circular Motion
To complete a circle, the bob's energy at the bottom should suffice for it to reach the top. At the top, the kinetic energy must at least nullify gravity's pull:
\(\frac{1}{2}Mu^2 = Mg(2l) + \frac{1}{2}M(v_c)^2\)
Where \( v_c \) is zero at the minimum speed for a full circle:
\(\frac{1}{2}(0.01v - 1)^2 = 10(1) + 0\)
Solving the quadratic equation \( 0.0001v^2 - 0.02v + 1 - 10 = 0 \) yields
\(v^2 - 20v + 1000 = 0\)

Step 3: Solve the Quadratic Equation
\(v = \frac{20 \pm \sqrt{20^2 - 4 \times 1000}}{2}\)
\(v = \frac{20 \pm \sqrt{400}}{2} = \frac{20 \pm 20}{2}\)
Thus, \(v = 20 \) (discarding the negative value).

Verification: 20 m/s fits the given range (400,400). Conclusively, the minimum value of \( v \) is 400 m/s.