Question

Two balls, A and B, with masses ma and mb are connected by a taut, massless string, and are moving along a horizontal frictionless plane. The distance between the centers of the two balls is L. At a certain instant, the velocity of ball B has magnitude v and is directed perpendicular to the string and parallel to the horizontal plane, and the velocity of ball A is zero.

Answer 1

Answer:

$T = (\frac{m_a m_b}{m_a + m_b})\frac{v^2}{L}$

Explanation:

As we know that the motion of the balls with respect to the position of center of mass is pure rotational motion

so here we can find the position of center of mass with respect to ball B as

$r_{cm} = \frac{m_a}{m_a + m_b} L$

also we know that the velocity of the center of mass is given as

$v_{cm} = \frac{m_b v}{m_a + m_b}$

now relative velocity of ball B with respect to center of mass is given as

$v_{b/cm} = v_b - v_{cm}$

$v_{b/cm} = v - \frac{m_b v}{m_a + m_b}$

$v_{b/cm} = \frac{m_a v}{m_a + m_b}$

now tension in the string is given as

$T = \frac{m_b v_{b/cm}^2}{r_{c}}$

$T = \frac{m_b (\frac{m_a v}{m_a + m_b})^2}{\frac{m_a}{m_a + m_b} L}$

$T = (\frac{m_a m_b}{m_a + m_b})\frac{v^2}{L}$

Answer 2

The magnitude of tension acting on the string due to two balls at each end is  ${\dfrac{m_{A} \times m_{B}}{m_{A}+m_{B}}} \times \dfrac{v^2}{L}$.

Given data:

The mass of ball A is, $m_{A}$.

The mass of ball B is, $m_{B}$.

Magnitude of velocity of ball B is, v.

Distance between the centers of ball is, L.

The motion of ball with respect to its center of mass is rotation. Then tension in the string due to each ball is,

$T=\dfrac{m_{B} \times v^{2}_{c}}{r_{c}}$.........................................................(1)

Here, $v_{c}$  is the relative velocity of ball B with respect to center of mass and its value is,'

$v_{c}=\dfrac{m_{A} \times v}{m_{A}+m_{B}}$

And, $r_{c}$ is the position of center of mass with respect to ball B. And its value is,

$r_{c}=\dfrac{m_{A} \times L}{m_{A}+m_{B}}$

Substituting the values in equation (1) as,

$T=\dfrac{m_{B} \times (\dfrac{m_{A} \times v}{m_{A}+m_{B}})^{2}}{\dfrac{m_{A} \times L}{m_{A}+m_{B}}}\\T={\dfrac{m_{A} \times m_{B}}{m_{A}+m_{B}}} \times \dfrac{v^2}{L}$

Thus, we can conclude that the magnitude of  tension in the string is ${\dfrac{m_{A} \times m_{B}}{m_{A}+m_{B}}} \times \dfrac{v^2}{L}$ .

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