Question

Two sleeves A and B of masses $m_1$ and $m_2$ connected with the help of a thread of length $5l$ can slide on two parallel frictionless rods fixed at a distance $3l$ from each other. The sleeve A and B are projected with velocities $v_1$ and $v_2$ ($v_1>v_2$) towards the right. Neglect the size of the sleeve and diameter of the rods and calculate the impulses of the tensile force developed in the thread and the force of normal reactions of the rod when the thread becomes taut. Consider the thread to be perfectly inelastic and almost inextensible.

Answer 1

Impulse of tensile force: $J_T = \frac{5}{4} \frac{m_1 m_2 (v_1 - v_2)}{m_1 + m_2}$

Impulse of normal reaction: $J_{Normal} = \frac{3}{4} \frac{m_1 m_2 (v_1 - v_2)}{m_1 + m_2}$

Answer 2

The problem involves calculating the impulses of tensile force and normal reactions when a thread connecting two sleeves sliding on parallel rods becomes taut. Here's a breakdown of the solution:

1. Geometry:

   • When the thread becomes taut, it forms the hypotenuse of a right-angled triangle.
   • The vertical side is the distance between the rods, $h = 3l$.
   • The length of the thread, $L = 5l$.
   • Using the Pythagorean theorem, the horizontal distance between the sleeves is found to be $x = 4l$.
   • The angle $\theta$ the thread makes with the horizontal is calculated as $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.

2. Impulse due to tension:

   • Since $v_1 > v_2$, sleeve A is faster than B. The thread pulls A backward and B forward.
   • $J_T$ is the magnitude of the impulsive tension in the thread.
   • Impulses on A and B are expressed in terms of unit vectors along the thread.

3. Impulse-Momentum Theorem:

   • $V_1$ and $V_2$ are the velocities of A and B immediately after the thread becomes taut.
   • Applying the impulse-momentum theorem to each sleeve, considering both the tensile impulse and the normal reaction impulse.
   • Equations are separated into x and y components.

4. Constraint from inextensibility:

   • The component of the relative velocity of the two sleeves along the thread must be zero immediately after the impulse.
   • This leads to the conclusion that $V_1 = V_2$.

5. Conservation of momentum in x-direction:

   • The total momentum of the system (A+B) in the x-direction is conserved.
   • The final common velocity $V_f$ is found to be $V_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$.

6. Calculate the impulse of tensile force ($J_T$):

   • Using the equations from the impulse-momentum theorem and the expression for $V_f$, the impulse of the tensile force is calculated: $J_T = \frac{5}{4} \frac{m_1 m_2 (v_1 - v_2)}{m_1 + m_2}$.

7. Calculate the impulses of normal reactions:

   • Using the y-component equations from the impulse-momentum theorem, the magnitudes of the impulses of the normal reactions are found to be equal: $|J_{NA}| = |J_{NB}| = \frac{3}{4} \frac{m_1 m_2 (v_1 - v_2)}{m_1 + m_2}$.