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Step-by-Step Solution
Step 1: Identify the energies involved before collision
Initially, the simple pendulum (mass $m$) is released from a small angle $\theta_0$. At this angle, it possesses gravitational potential energy relative to its lowest position. By the time it reaches the lowest point (just before collision), all that potential energy converts into kinetic energy.
For small angular displacements, the change in height can be approximated, but more precisely,
$E_{\text{initial}} = m g \ell \bigl(1 - \cos(\theta_0)\bigr).$
Thus, the speed of the bob $v_0$ at the lowest point is found from
$ \frac{1}{2} m v_0^2 = m g \ell\bigl(1 - \cos(\theta_0)\bigr). $
Hence,
$ v_0 = \sqrt{2 g \ell \bigl(1 - \cos(\theta_0)\bigr)}. $
Step 2: Apply the formula for an elastic collision
During the collision, the bob (mass $m$) moving with speed $v_0$ elastically strikes a block (mass $M$) which is initially at rest on a smooth horizontal surface. For a one-dimensional elastic collision, the velocity of the bob after collision ($v'$) is given by:
$ v' = \frac{(m - M)}{(m + M)} \, v_0. $
Step 3: Relate the final velocity of the bob to its final angle $\theta_1$
After collision, the bob bounces back with speed $v'$. It then rises to a maximum angular displacement $\theta_1$. At this final angle, the bob’s kinetic energy converts into potential energy:
$ \frac{1}{2} m (v')^2 = m g \ell \bigl(1 - \cos(\theta_1)\bigr). $
Substitute $v'$ from Step 2 into this equation:
$ \frac{1}{2} m \Bigl(\frac{m - M}{m + M} \, v_0\Bigr)^2 = m g \ell \bigl(1 - \cos(\theta_1)\bigr). $
Step 4: Use the initial speed $v_0$ from Step 1
Recall $v_0^2 = 2 g \ell \bigl(1 - \cos(\theta_0)\bigr).$ Putting this into the above expression:
$ \frac{1}{2} m \Bigl(\frac{m - M}{m + M}\Bigr)^2 \bigl(2 g \ell \bigl(1 - \cos(\theta_0)\bigr)\bigr) = m g \ell \bigl(1 - \cos(\theta_1)\bigr). $
After simplifying, the $m g \ell$ factors can be managed, yielding a relationship in terms of $(m - M)/(m + M)$ and $(1 - \cos(\theta_0)) / (1 - \cos(\theta_1))$.
Step 5: Solve for $M$ using the ratio of angles
For small angles $\theta$, one can use the approximation $1 - \cos(\theta) \approx \frac{\theta^2}{2}$. Equivalently, a direct algebraic manipulation often yields the ratio in terms of $\theta_0$ and $\theta_1$. On solving, you obtain:
$ M = m \left(\frac{\theta_0 + \theta_1}{\theta_0 - \theta_1}\right). $
Hence, the mass of the block $M$ is expressed in terms of $m$, $\theta_0$, and $\theta_1$ as above.
Step 6: Final Answer
The correct expression for $M$ is:
$ M = m\left(\frac{\theta_0 + \theta_1}{\theta_0 - \theta_1}\right). $
