1. Kinematic Analysis at Collision

Let the central mass be $M$ and the side masses be $m$. Initially, the system is linear. $M$ is projected with velocity $u$ perpendicular to the line of threads.

When the end discs are “about to collide”, they meet on the line of motion of $M$. At this instant, the threads are parallel to the direction of motion (vertical in our diagram).

Constraint: Since the threads are inextensible, the relative velocity between $M$ and $m$ along the thread must be zero. Since the threads are vertical, $M$ and $m$ must have the same vertical velocity component $V$. The masses $m$ will also have a horizontal collision velocity $v_x$.

2. Conservation Laws

Conservation of Linear Momentum (Vertical):

$$ Mu = (M + 2m)V \implies V = \frac{M}{M+2m}u $$

Conservation of Energy:

$$ \frac{1}{2}Mu^2 = \frac{1}{2}(M+2m)V^2 + 2\left(\frac{1}{2}mv_x^2\right) $$

Substituting $V$:

$$ Mu^2 = (M+2m)\left(\frac{Mu}{M+2m}\right)^2 + 2mv_x^2 $$ $$ Mu^2 = \frac{M^2 u^2}{M+2m} + 2mv_x^2 $$ $$ 2mv_x^2 = Mu^2 \left(1 – \frac{M}{M+2m}\right) = Mu^2 \left(\frac{2m}{M+2m}\right) $$ $$ v_x^2 = \frac{Mu^2}{M+2m} $$

3. Force Analysis (Tension)

We need the tension $T$ at this instant. Consider the forces in the vertical direction (y-axis). Let $a_M$ be acceleration of M and $a_{my}$ be vertical acceleration of m.

• Equation for $M$: $M a_M = -2T$
• Equation for $m$ (y-axis): $m a_{my} = T$

Kinematic Constraint on Acceleration:

Since the length of thread is constant, m would perform circular motion about M

$$ a_M – a_{my} = -\frac{v_x^2}{l} $$

Substituting accelerations ($a_M = -2T/M$ and $a_{my} = T/m$):

$$ \left(-\frac{2T}{M}\right) – \left(\frac{T}{m}\right) = -\frac{v_x^2}{l} $$ $$ T\left(\frac{2}{M} + \frac{1}{m}\right) = \frac{v_x^2}{l} $$ $$ T\left(\frac{2m + M}{mM}\right) = \frac{v_x^2}{l} $$

4. Final Calculation

Substitute $v_x^2 = \frac{Mu^2}{M+2m}$:

$$ T \frac{M+2m}{mM} = \frac{1}{l} \left( \frac{Mu^2}{M+2m} \right) $$ $$ T = \frac{mM}{M+2m} \cdot \frac{1}{l} \cdot \frac{Mu^2}{M+2m} $$ $$ T = \frac{m M^2 u^2}{l (M+2m)^2} $$

Numerical Substitution: $M=2.0, m=1.0, l=0.5, u=2$.

$$ T = \frac{1 \cdot 2^2 \cdot 2^2}{0.5 \cdot (2 + 2)^2} = \frac{16}{0.5 \cdot 16} = \frac{16}{8} = 2 \text{ N} $$

The tensile force in the threads is 2 N.