Solution Analysis
1. System Analysis and Frame of Reference
Consider the system consisting of the block of mass $M$ and the ball of mass $m$. The surfaces are frictionless.
Since there are no external horizontal forces acting on the system, the linear momentum of the system in the horizontal direction is conserved.
The ball is released from the horizontal position. As it swings down, it gains speed. The tension in the string is maximum when the ball is at the lowest point of its trajectory relative to the point of suspension.
2. Conservation of Linear Momentum
Let $\vec{v}$ be the velocity of the ball and $\vec{V}$ be the velocity of the block with respect to the ground at the instant the ball reaches the lowest point. Let $\vec{v}_{rel}$ be the velocity of the ball relative to the block. Since the ball swings left relative to the block, $\vec{v}_{rel}$ is directed to the left. To conserve momentum, the block must recoil to the right. Using the concept of Reduced Mass ($\mu$), we can analyze the energy in the center of mass frame or simply relate the relative velocity to the kinetic energy. The reduced mass of the system is given by: $$ \mu = \frac{mM}{m + M} $$
Let $\vec{v}$ be the velocity of the ball and $\vec{V}$ be the velocity of the block with respect to the ground at the instant the ball reaches the lowest point. Let $\vec{v}_{rel}$ be the velocity of the ball relative to the block. Since the ball swings left relative to the block, $\vec{v}_{rel}$ is directed to the left. To conserve momentum, the block must recoil to the right. Using the concept of Reduced Mass ($\mu$), we can analyze the energy in the center of mass frame or simply relate the relative velocity to the kinetic energy. The reduced mass of the system is given by: $$ \mu = \frac{mM}{m + M} $$
3. Conservation of Mechanical Energy
The system is conservative. The loss in gravitational potential energy of the ball as it falls through a vertical height $l$ is converted into the kinetic energy of the system. The total kinetic energy of a two-body system with zero total momentum can be expressed as: $$ K = \frac{1}{2}\mu v_{rel}^2 $$ Equating the loss in potential energy to the gain in kinetic energy: $$ mgl = \frac{1}{2} \mu v_{rel}^2 $$ $$ v_{rel}^2 = \frac{2mgl}{\mu} $$ Substituting $\mu = \frac{mM}{m+M}$: $$ v_{rel}^2 = \frac{2mgl (m+M)}{mM} = \frac{2gl(m+M)}{M} \quad \dots(1) $$
The system is conservative. The loss in gravitational potential energy of the ball as it falls through a vertical height $l$ is converted into the kinetic energy of the system. The total kinetic energy of a two-body system with zero total momentum can be expressed as: $$ K = \frac{1}{2}\mu v_{rel}^2 $$ Equating the loss in potential energy to the gain in kinetic energy: $$ mgl = \frac{1}{2} \mu v_{rel}^2 $$ $$ v_{rel}^2 = \frac{2mgl}{\mu} $$ Substituting $\mu = \frac{mM}{m+M}$: $$ v_{rel}^2 = \frac{2mgl (m+M)}{mM} = \frac{2gl(m+M)}{M} \quad \dots(1) $$
4. Dynamics at the Lowest Point
We apply Newton’s Second Law to the ball at the lowest point. The forces acting on the ball in the vertical direction are the Tension ($T$) acting upwards and gravity ($mg$) acting downwards. The vertical acceleration of the ball relative to the ground is the same as the vertical acceleration relative to the block (since the block moves purely horizontally). This acceleration is the centripetal acceleration required for the circular path relative to the suspension point. $$ T – mg = m \left( \frac{v_{rel}^2}{l} \right) $$ $$ T = mg + \frac{m}{l} v_{rel}^2 $$
We apply Newton’s Second Law to the ball at the lowest point. The forces acting on the ball in the vertical direction are the Tension ($T$) acting upwards and gravity ($mg$) acting downwards. The vertical acceleration of the ball relative to the ground is the same as the vertical acceleration relative to the block (since the block moves purely horizontally). This acceleration is the centripetal acceleration required for the circular path relative to the suspension point. $$ T – mg = m \left( \frac{v_{rel}^2}{l} \right) $$ $$ T = mg + \frac{m}{l} v_{rel}^2 $$
5. Final Calculation
Substitute the value of $v_{rel}^2$ from equation (1) into the tension equation: $$ T = mg + \frac{m}{l} \left[ \frac{2gl(m+M)}{M} \right] $$ $$ T = mg + 2mg \left( \frac{m+M}{M} \right) $$ $$ T = mg \left[ 1 + 2 \left( 1 + \frac{m}{M} \right) \right] $$ $$ T = mg \left[ 1 + 2 + \frac{2m}{M} \right] $$ $$ T = mg \left( 3 + \frac{2m}{M} \right) $$
Substitute the value of $v_{rel}^2$ from equation (1) into the tension equation: $$ T = mg + \frac{m}{l} \left[ \frac{2gl(m+M)}{M} \right] $$ $$ T = mg + 2mg \left( \frac{m+M}{M} \right) $$ $$ T = mg \left[ 1 + 2 \left( 1 + \frac{m}{M} \right) \right] $$ $$ T = mg \left[ 1 + 2 + \frac{2m}{M} \right] $$ $$ T = mg \left( 3 + \frac{2m}{M} \right) $$
Answer: The maximum tensile force in the string is
$$ T_{max} = mg \left( 3 + \frac{2m}{M} \right) $$