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Step-by-step Solution
Step 1: Identify the Physical Principles
The collision is perfectly elastic, so both linear momentum and angular momentum (about the center of the rod) are conserved. Also, the coefficient of restitution for a perfectly elastic collision is 1.
Step 2: Write down the Conservation of Linear Momentum
Before collision, the particle of mass m has velocity u , and the rod of mass M is at rest on a frictionless surface. After collision, the particle comes to rest, and the rod moves with linear velocity v and angular velocity \omega .
Conservation of linear momentum along the initial direction of motion of the particle gives:
m u = M v.
This implies
v = \frac{m u}{M}.
Step 3: Write down the Conservation of Angular Momentum
Take moments about the center of the rod (point O). Initially, the angular momentum is due to the particle only, since the rod is at rest. After collision, the rod has an angular velocity \omega .
The particle strikes the end of the rod, which is at a distance L/2 from the center. Then:
\text{Initial angular momentum} = m u \times \left( \frac{L}{2} \right).
After collision, the rod of mass M spins about its center with angular velocity \omega . The moment of inertia of a uniform rod about its center is:
I = \frac{M L^2}{12}.
Thus, the angular momentum of the rod after collision is:
\left( \frac{M L^2}{12} \right) \omega.
Hence,
m u \left( \frac{L}{2} \right) = \left( \frac{M L^2}{12} \right) \omega.
From this,
\omega = \frac{6 m u}{M L}.
(We will later substitute the linear velocity v for consistency, but for now this expression is useful.)
Step 4: Use the Coefficient of Restitution
A perfectly elastic collision means the relative speed of separation equals the relative speed of approach. Here, the particle comes to rest after collision, so the relative speed of separation between the end of the rod and the particle is the rod’s linear speed plus the tangential speed at the end of the rod due to rotation.
Let e be the coefficient of restitution. For a perfectly elastic collision, e = 1 . Therefore,
e = \frac{\text{relative speed after collision}}{\text{relative speed before collision}} = 1.
Before collision, the relative speed between the rod’s end and the particle is u . After collision, that relative speed is v + \frac{\omega L}{2}. Hence,
1 = \frac{v + \frac{\omega L}{2}}{u}.
This implies
v + \frac{\omega L}{2} = u.
Step 5: Substitute Known Relationships to Simplify
First, from linear momentum conservation (Step 2):
v = \frac{m u}{M}.
From angular momentum conservation (Step 3), we found
\omega = \frac{6 m u}{M L}
(if we directly used v instead of u , we would combine them accordingly, but let's proceed with the expression in terms of u ).
Step 6: Combine the Expressions (Coefficient of Restitution)
Using
v + \frac{\omega L}{2} = u,
substitute v and \omega :
\frac{m u}{M} + \frac{1}{2} \left( \frac{6 m u}{M L} \right) L = u.
This simplifies to:
\frac{m u}{M} + \frac{6 m u}{2 M} = u,
\frac{m u}{M} + \frac{3 m u}{M} = u,
\frac{4 m u}{M} = u.
Step 7: Solve for the Mass Ratio
Canceling u on both sides (assuming u \neq 0 ), we get:
\frac{4 m}{M} = 1 \quad \Rightarrow \quad \frac{m}{M} = \frac{1}{4}.
By the problem statement, \frac{m}{M} = \frac{1}{x}. Hence,
x = 4.
Step 8: Final Answer
The value of x is 4.
