Question

A body A (m = 1 kg) is attached to the top of a vertically fixed spring of spring constant 1000 N m$^{-1}$ and then the mass is released at the initial position P with the spring being initially in its natural length $\ell_0$. When A comes back to original position P another body B of equal mass falls on A with a velocity v = 0.2$\sqrt{60}$ m s$^{-1}$ at t = t$_0$ and the collision is perfectly inelastic. Then,

• A. B will move in SHM for a duration of $\frac{2\pi}{15\sqrt{5}}$ s.
• B. B will detach from A after a duration of $\frac{\pi}{5\sqrt{5}}$ s after t$_0$.
• C. Maximum amplitude of oscillations of the combined mass is 4 x $10^{-2}$ m.
• D. The neutral point of SHM of the combined mass is 2 cm below P.

Answer 1

Answer: (C), (D)

c, d

Answer 2

Let's analyze the situation step by step:

1. Equilibrium Position of A:

   Initially, when only body A is attached, the spring extends until the spring force balances the weight of A. $kx_1 = mg$ $x_1 = \frac{mg}{k} = \frac{1 \times 10}{1000} = 0.01$ m = 1 cm

2. Velocity of A at P:

   When A returns to P, its velocity can be found using energy conservation. The potential energy lost by A is converted into kinetic energy and spring potential energy. However, since A starts at the natural length, the potential energy lost by A is converted into kinetic energy and spring potential energy when it reaches the lowest point, not P. Instead, we are given that A starts at its natural length $\ell_0$ and is released. When it returns to P, all the potential energy stored in the spring will convert to kinetic energy. The maximum extension is twice the equilibrium extension, so $x_{max} = 2x_1 = 2$ cm.

   The velocity at P can be found using conservation of energy from the lowest point:

   $\frac{1}{2}mv^2 = \frac{1}{2}kx_1^2$ $v = \sqrt{\frac{k}{m}}x_1 = \sqrt{\frac{1000}{1}} \times 0.01 = \sqrt{1000} \times 0.01 = \sqrt{10} \times 0.1 \approx 0.316$ m/s

3. Velocity of Combined Mass (A+B) Immediately After Collision:

   Using conservation of momentum for the inelastic collision at t = t$_0$: $mv + mv_B = (m+m)v'$ $v' = \frac{v + v_B}{2} = \frac{0.316 + 0.2\sqrt{60}}{2} = \frac{0.316 + 0.2 \times 7.746}{2} = \frac{0.316 + 1.549}{2} \approx \frac{1.865}{2} \approx 0.9325$ m/s

4. New Equilibrium Position of A+B:

   The new equilibrium position is where the spring force balances the combined weight of A and B. $kx_2 = (m+m)g$ $x_2 = \frac{2mg}{k} = \frac{2 \times 1 \times 10}{1000} = 0.02$ m = 2 cm (This confirms option (d))

5. Amplitude of SHM of A+B:

   The amplitude A can be determined using energy conservation. The kinetic energy of the combined mass at P plus the potential energy at P equals the potential energy at the maximum displacement from the new equilibrium. Let $x'$ be the displacement from the new equilibrium.

   $\frac{1}{2}(2m)v'^2 + \frac{1}{2}k(x_2 - x')^2 = \frac{1}{2}k(x')^2$

   However, a simpler way is to find the distance from the new equilibrium position. The new equilibrium position is 2 cm below P. At point P, the velocity is v'. Therefore, the amplitude of the new SHM is:

   $A = \sqrt{x^2 + (\frac{v'}{\omega})^2}$ where $x$ is the distance from the new equilibrium position. Here, $x = 2$ cm = 0.02 m, $v' = 0.9325$ m/s and $\omega = \sqrt{\frac{k}{2m}} = \sqrt{\frac{1000}{2}} = \sqrt{500} \approx 22.36$ rad/s.

   $A = \sqrt{(0.02)^2 + (\frac{0.9325}{22.36})^2} = \sqrt{0.0004 + (0.0417)^2} = \sqrt{0.0004 + 0.00174} = \sqrt{0.00214} \approx 0.046$ m = 4.6 cm

   Another approach: The initial displacement from the equilibrium position is 2 cm. $v' = 0.2\sqrt{60} / 2 = 0.1\sqrt{60}$ $\omega = \sqrt{\frac{k}{2m}} = \sqrt{\frac{1000}{2}} = \sqrt{500}$ $A = \sqrt{(0.02)^2 + \frac{(0.1\sqrt{60})^2}{500}} = \sqrt{0.0004 + \frac{0.01 \times 60}{500}} = \sqrt{0.0004 + \frac{0.6}{500}} = \sqrt{0.0004 + 0.0012} = \sqrt{0.0016} = 0.04$ m = 4 cm (This confirms option (c))

6. Time for B to Detach:

   B will detach when the spring force is zero and the acceleration is greater than g. This occurs at the topmost point of the oscillation. However, the problem states that the collision is perfectly inelastic. Thus, B will not detach from A.

7. Time for B to Move in SHM:

   This is not relevant as B does not move independently.

Therefore, options (c) and (d) are correct.