Question

Special cases

(a) If, e = 1 (perfectly elastic collision) (b) e = 0 (perfectly inelastic collision) $\phi = \theta$ and v' = v

Example 3: A bullet of mass m moving with a velocity $v_0$ strikes a stationary block of mass M suspenc by a string of length 'L'. The bullet gets embedded in the block. What is the maximum ar made by string after the impact.

Answer 1

$\theta = \arccos\left[1 - \frac{m^2 v_0^2}{2gL(m + M)^2}\right]$

Answer 2

The problem involves two distinct phases:

1. A perfectly inelastic collision: The bullet strikes the block and gets embedded. During this collision, linear momentum is conserved, but mechanical energy is not.
2. Pendulum motion: After the collision, the combined bullet-block system swings upwards like a simple pendulum. During this phase, mechanical energy is conserved.

Phase 1: Perfectly Inelastic Collision

Let V be the velocity of the combined bullet-block system immediately after the impact.

Before collision:

• Mass of bullet = m, velocity = v₀
• Mass of block = M, velocity = 0 (stationary)

After collision:

• Combined mass = (m + M), velocity = V

Applying the principle of conservation of linear momentum:

Initial momentum = Final momentum m * v₀ + M * 0 = (m + M) * V m * v₀ = (m + M) * V

Solving for V, the velocity of the combined system just after impact: $V = \frac{m v_0}{m + M}$

Phase 2: Pendulum Motion

The combined mass (m + M) starts with velocity V at the lowest point (its initial height can be considered h₁ = 0). It swings upwards to a maximum height h where its velocity momentarily becomes 0.

Applying the principle of conservation of mechanical energy between the lowest point (immediately after collision) and the highest point of the swing:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy $\frac{1}{2} (m + M) V^2 + (m + M) g (0) = \frac{1}{2} (m + M) (0)^2 + (m + M) g h$ $\frac{1}{2} (m + M) V^2 = (m + M) g h$ $h = \frac{V^2}{2g}$

Now, we need to relate this maximum height h to the maximum angle θ made by the string with the vertical.

From the geometry of a simple pendulum of length L swinging to a maximum angle θ, the vertical height h risen from the lowest point is given by: $h = L - L \cos\theta$ $h = L (1 - \cos\theta)$

Equating the two expressions for h: $L (1 - \cos\theta) = \frac{V^2}{2g}$ $1 - \cos\theta = \frac{V^2}{2gL}$ $\cos\theta = 1 - \frac{V^2}{2gL}$

Substitute the expression for V from Phase 1 into this equation: $V^2 = \left(\frac{m v_0}{m + M}\right)^2 = \frac{m^2 v_0^2}{(m + M)^2}$ So, $\cos\theta = 1 - \frac{m^2 v_0^2}{2gL(m + M)^2}$

Finally, the maximum angle θ is: $\theta = \arccos\left[1 - \frac{m^2 v_0^2}{2gL(m + M)^2}\right]$

The maximum angle made by the string after the impact is given by the above formula.

Explanation of the solution:

1. Conservation of Linear Momentum: Apply to the perfectly inelastic collision of the bullet and block to find their combined velocity V immediately after impact.
2. Conservation of Mechanical Energy: Apply to the subsequent pendulum motion. The kinetic energy of the combined mass at the lowest point is converted into potential energy at the maximum height h.
3. Geometry: Relate the maximum height h to the string length L and the maximum angle θ using pendulum geometry (h = L(1 - cosθ)).
4. Substitution: Substitute the expression for V into the energy conservation equation, then solve for θ.