Question

A block of mass $m = 1$ kg is attached to a free end of a spring whose other end is fixed with a wall performing simple harmonic motion as shown in Figure.

The position of the block from $O$ at any instant is $x = 2 + \frac{1}{\sqrt{2}} \sin(2t)$ where $x$ in meter and $t$ is in second.

A shell of same mass is released from smooth the circular path at a height $h = 80$ cm. The shell collides elastically with the block performing SHM and finally reaches up to height $5$ cm along circular path. Neglecting friction, find where the collision takes place.

Answer 1

The collision takes place at approximately $2.505 \text{ m}$ from point $O$.

Answer 2

The problem involves two parts: the motion of the shell and the simple harmonic motion (SHM) of the block, and an elastic collision between them.

1. Calculate shell velocities: Use conservation of energy to find the shell's speed before ($u_s$) and after ($v_s'$) collision. Account for direction using signs. $u_s = -\sqrt{2gh_1}$ and $v_s' = +\sqrt{2gh_2}$.
2. Apply elastic collision rules: For an elastic collision between two objects of equal mass, they exchange velocities. So, the block's velocity just before collision ($u_b$) is equal to the shell's velocity just after collision ($v_s'$).
3. Analyze SHM: Differentiate the block's position equation $x = 2 + \frac{1}{\sqrt{2}} \sin(2t)$ to get its velocity $v = \sqrt{2} \cos(2t)$.
4. Find collision time phase: Set the block's velocity $v$ equal to $u_b$ at the collision time $t_c$ to find $\cos(2t_c)$.
5. Determine collision position: Use the value of $\cos(2t_c)$ to find $\sin(2t_c)$ (using $\sin^2\theta + \cos^2\theta = 1$). Since the block starts at equilibrium and moves right, and the shell approaches from the right, the first collision occurs when the block is moving right and to the right of equilibrium. This implies $\sin(2t_c)$ is positive. Substitute this $\sin(2t_c)$ value into the position equation to find the collision position $x_c$.