Question

Three identical balls A,B and C each of mass m = 3kg are connected by strings AB and BC as shown in the figure. The whole system is placed on a smooth horizontal surface.

Now the ball B is given an initial velocity $v_0 = \sqrt{3}$ m/s, perpendicular to the string and along the horizontal surface. Find the tension (in newton) in the string just before the balls A and C collide.

Answer 1

2.5

Answer 2

The initial configuration of the system is three balls A, B, and C of mass $m=3$ kg, connected by strings AB and BC of length $L=1$ m each, arranged in a line on a smooth horizontal surface. Ball B is given an initial velocity $v_0 = \sqrt{3}$ m/s perpendicular to the string. Let the initial line of balls be the x-axis, with B at the origin (0,0), A at (-1,0), and C at (1,0). The initial velocity of B is along the y-axis, $\vec{v}_B(0) = (0, v_0)$. The initial velocities of A and C are zero.

Due to symmetry, the motion will be symmetric with respect to the y-axis. Ball B will move along the y-axis, so its x-coordinate remains 0. Balls A and C will move symmetrically with respect to the y-axis, so their x-coordinates are opposite, and their y-coordinates are equal. Let the coordinates at time t be $A(-x, y)$, $B(0, y_B)$, and $C(x, y)$. The lengths of the strings are constant: $|AB|^2 = (0 - (-x))^2 + (y_B - y)^2 = x^2 + (y_B - y)^2 = L^2 = 1$ $|BC|^2 = (x - 0)^2 + (y - y_B)^2 = x^2 + (y - y_B)^2 = L^2 = 1$

Balls A and C collide when their x-coordinates are equal, i.e., $-x = x$, which implies $x=0$.

Conservation of momentum in the y-direction: Initial y-momentum = $m v_0$. At time t, y-momentum = $m v_{Ay} + m v_{By} + m v_{Cy}$. Since $y_A = y_C = y$, $v_{Ay} = v_{Cy} = \dot{y}$. Let $v_{By} = \dot{y}_B$. $m \dot{y} + m \dot{y}_B + m \dot{y} = m v_0$, so $2 \dot{y} + \dot{y}_B = v_0$.

Conservation of energy: Initial kinetic energy = $\frac{1}{2} m v_0^2$. Initial potential energy = 0. At time t, kinetic energy = $\frac{1}{2} m (v_{Ax}^2 + v_{Ay}^2) + \frac{1}{2} m (v_{Bx}^2 + v_{By}^2) + \frac{1}{2} m (v_{Cx}^2 + v_{Cy}^2)$. $v_{Ax} = -\dot{x}$, $v_{Ay} = \dot{y}$, $v_{Bx} = 0$, $v_{By} = \dot{y}_B$, $v_{Cx} = \dot{x}$, $v_{Cy} = \dot{y}$. $\frac{1}{2} m ((-\dot{x})^2 + \dot{y}^2) + \frac{1}{2} m (0^2 + \dot{y}_B^2) + \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) = \frac{1}{2} m v_0^2$. $m \dot{x}^2 + m \dot{y}^2 + \frac{1}{2} m \dot{y}_B^2 + \frac{1}{2} m \dot{y}^2 = \frac{1}{2} m v_0^2$. $2 \dot{x}^2 + 3 \dot{y}^2 + \dot{y}_B^2 = v_0^2$.

From $x^2 + (y_B - y)^2 = 1$, differentiating with respect to time: $2x \dot{x} + 2(y_B - y)(\dot{y}_B - \dot{y}) = 0$, so $x \dot{x} + (y_B - y)(\dot{y}_B - \dot{y}) = 0$.

Just before collision, $x \to 0$. The equation becomes $(y_B - y)(\dot{y}_B - \dot{y}) = 0$. Since $x^2 + (y_B - y)^2 = 1$, as $x \to 0$, $(y_B - y)^2 \to 1$, so $|y_B - y| \to 1$. Thus, $y_B - y \neq 0$. Therefore, we must have $\dot{y}_B - \dot{y} = 0$, which means $\dot{y}_B = \dot{y}$. So, just before collision, the y-velocities of A, B, and C are equal: $v_{Ay} = v_{By} = v_{Cy} = \dot{y}$. From the conservation of y-momentum: $2 \dot{y} + \dot{y} = v_0$, so $3 \dot{y} = v_0$, $\dot{y} = v_0/3$. Just before collision, $v_{Ay} = v_{By} = v_{Cy} = v_0/3$.

From the energy conservation equation: $2 \dot{x}^2 + 3 \dot{y}^2 + \dot{y}_B^2 = v_0^2$. Substitute $\dot{y} = \dot{y}_B = v_0/3$: $2 \dot{x}^2 + 3 (v_0/3)^2 + (v_0/3)^2 = v_0^2$. $2 \dot{x}^2 + 3 (v_0^2/9) + v_0^2/9 = v_0^2$. $2 \dot{x}^2 + v_0^2/3 + v_0^2/9 = v_0^2$. $2 \dot{x}^2 = v_0^2 - v_0^2/3 - v_0^2/9 = v_0^2 (1 - 1/3 - 1/9) = v_0^2 (\frac{9 - 3 - 1}{9}) = v_0^2 (\frac{5}{9})$. $\dot{x}^2 = \frac{5 v_0^2}{18}$. Just before collision, A and C are moving towards the y-axis, so $\dot{x} = -\sqrt{\frac{5}{18}} v_0$. $v_{Ax} = -\dot{x} = \sqrt{\frac{5}{18}} v_0$ and $v_{Cx} = \dot{x} = -\sqrt{\frac{5}{18}} v_0$.

At the moment just before collision, A and C are at $(0, y)$, and B is at $(0, y \pm 1)$. The strings are along the y-axis. The velocity of A is $\vec{v}_A = (\sqrt{\frac{5}{18}} v_0, v_0/3)$. The velocity of B is $\vec{v}_B = (0, v_0/3)$. The velocity of C is $\vec{v}_C = (-\sqrt{\frac{5}{18}} v_0, v_0/3)$.

Consider the motion of ball A relative to ball B. The relative velocity is $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B = (\sqrt{\frac{5}{18}} v_0, 0)$. Just before collision, A is at $(0, y)$ and B is at $(0, y \pm 1)$. The vector from B to A is $(0, y) - (0, y \pm 1) = (0, \mp 1)$. The radial velocity of A with respect to B is the component of $\vec{v}_{A/B}$ along the vector from B to A. If B is at $(0, y+1)$, the vector from B to A is $(0, -1)$. Radial velocity = $\vec{v}_{A/B} \cdot (0, -1) = (\sqrt{\frac{5}{18}} v_0, 0) \cdot (0, -1) = 0$. If B is at $(0, y-1)$, the vector from B to A is $(0, 1)$. Radial velocity = $\vec{v}_{A/B} \cdot (0, 1) = (\sqrt{\frac{5}{18}} v_0, 0) \cdot (0, 1) = 0$. In both cases, the radial velocity is 0. This means the length of the string is not changing at this instant, which is consistent with the constraint.

Consider the motion of A relative to B. The string AB is of length $L=1$. The motion of A relative to B is such that A is moving in a circle of radius 1 centered at B, or the distance between A and B is always 1. At the moment just before collision, A is at $(0,y)$ and B is at $(0,y \pm 1)$. The string AB is along the y-axis. The tension T in the string AB acts along the string. For ball A, the net force in the direction towards B (along the string) provides the centripetal acceleration. Let's assume B is at $(0, y_B)$ and A is at $(x_A, y_A)$. The string AB is along the vector from B to A. The velocity of A relative to B is $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. The component of $\vec{v}_{A/B}$ perpendicular to the string is the tangential velocity. The component of $\vec{v}_{A/B}$ along the string is the radial velocity. Just before collision, A is at $(0,y)$ and B is at $(0, y_B)$. The vector from B to A is $(0, y-y_B)$. The length is $|y-y_B| = 1$. Let's assume $y_B = y+1$. Then the vector from B to A is $(0, -1)$. The string is along the negative y-axis relative to B. $\vec{v}_A = (\sqrt{\frac{5}{18}} v_0, v_0/3)$, $\vec{v}_B = (0, v_0/3)$. $\vec{v}_{A/B} = (\sqrt{\frac{5}{18}} v_0, 0)$. The radial velocity is $\vec{v}_{A/B} \cdot (0, -1) = 0$. The tangential velocity magnitude is the magnitude of the component of $\vec{v}_{A/B}$ perpendicular to the vector from B to A. The direction perpendicular to $(0, -1)$ is $(\pm 1, 0)$. The tangential velocity is $\vec{v}_{A/B} - (\vec{v}_{A/B} \cdot \hat{r}) \hat{r}$, where $\hat{r}$ is the unit vector from B to A, which is $(0, -1)$. Tangential velocity vector = $(\sqrt{\frac{5}{18}} v_0, 0) - ((\sqrt{\frac{5}{18}} v_0, 0) \cdot (0, -1)) (0, -1) = (\sqrt{\frac{5}{18}} v_0, 0) - 0 = (\sqrt{\frac{5}{18}} v_0, 0)$. The magnitude of the tangential velocity is $v_{tan} = \sqrt{(\sqrt{\frac{5}{18}} v_0)^2 + 0^2} = \sqrt{\frac{5}{18}} v_0$. The centripetal acceleration of A towards B is $a_{rad} = \frac{v_{tan}^2}{L} = \frac{(\sqrt{\frac{5}{18}} v_0)^2}{1} = \frac{5 v_0^2}{18}$. The net force on A along the string is the tension T. This force provides the centripetal acceleration. So, $T = m a_{rad} = m \frac{5 v_0^2}{18}$.

Given $m=3$ kg and $v_0 = \sqrt{3}$ m/s. $T = 3 \times \frac{5 (\sqrt{3})^2}{18} = 3 \times \frac{5 \times 3}{18} = 3 \times \frac{15}{18} = 3 \times \frac{5}{6} = \frac{5}{2} = 2.5$ N.

Let's verify this from the perspective of ball B. The forces on B are tension T from AB downwards and tension T from BC downwards. The net force on B is $2T$ downwards (assuming B is above A and C). The acceleration of B is $\vec{a}_B = (0, \ddot{y}_B)$. The equation of motion for B in the y-direction is $-2T = m \ddot{y}_B$.

Let's consider the equation of motion for A in the x-direction. The tension T has a component along the x-direction. Let $\theta$ be the angle between AB and the y-axis. Then the x-component of tension is $-T \sin \theta$. Equation of motion for A in x-direction: $-T \sin \theta = m a_{Ax} = m \ddot{x}_A = m (-\ddot{x})$. So $T \sin \theta = m \ddot{x}$. Equation of motion for A in y-direction: $T \cos \theta = m a_{Ay} = m \ddot{y}_A = m \ddot{y}$.

Just before collision, $x \to 0$, so $\sin \theta \to 0$ and $\cos \theta \to 1$. As $x \to 0$, $T \sin \theta = m \ddot{x}$. If T is finite, then $\ddot{x} \to 0$. $T \cos \theta = m \ddot{y}$. As $\theta \to 0$, $T = m \ddot{y}$.

From $x^2 + (y_B - y)^2 = 1$, differentiating twice: $2 \dot{x}^2 + 2x \ddot{x} + 2(\dot{y}_B - \dot{y})^2 + 2(y_B - y)(\ddot{y}_B - \ddot{y}) = 0$. Just before collision, $x=0$, $\dot{y}_B = \dot{y}$, $|y_B - y| = 1$. $2 \dot{x}^2 + 0 + 0 + 2(y_B - y)(\ddot{y}_B - \ddot{y}) = 0$. $2 \dot{x}^2 + 2(y_B - y)(\ddot{y}_B - \ddot{y}) = 0$. $\dot{x}^2 + (y_B - y)(\ddot{y}_B - \ddot{y}) = 0$. We found $\dot{x}^2 = \frac{5 v_0^2}{18}$. $\frac{5 v_0^2}{18} + (y_B - y)(\ddot{y}_B - \ddot{y}) = 0$.

From $2 \dot{y} + \dot{y}_B = v_0$, differentiating: $2 \ddot{y} + \ddot{y}_B = 0$. So $\ddot{y}_B = -2 \ddot{y}$. Substitute into the equation above: $\frac{5 v_0^2}{18} + (y_B - y)(-2 \ddot{y} - \ddot{y}) = 0$. $\frac{5 v_0^2}{18} + (y_B - y)(-3 \ddot{y}) = 0$. $\frac{5 v_0^2}{18} - 3 (y_B - y) \ddot{y} = 0$.

If B is above A and C, $y_B - y = 1$. Then $\frac{5 v_0^2}{18} - 3 (1) \ddot{y} = 0$, so $\ddot{y} = \frac{5 v_0^2}{54}$. If B is below A and C, $y_B - y = -1$. Then $\frac{5 v_0^2}{18} - 3 (-1) \ddot{y} = 0$, so $\frac{5 v_0^2}{18} + 3 \ddot{y} = 0$, $\ddot{y} = -\frac{5 v_0^2}{54}$.

From the equation of motion for A in y-direction, just before collision, $T = m \ddot{y}$. If $\ddot{y} = \frac{5 v_0^2}{54}$, then $T = m \frac{5 v_0^2}{54}$. If $\ddot{y} = -\frac{5 v_0^2}{54}$, then $T = -m \frac{5 v_0^2}{54}$. Tension must be positive, so this case is not possible unless the string is slack.

Let's reconsider the geometry. When B is given initial velocity upwards, it moves upwards, and A and C move upwards and inwards. When A and C collide at the y-axis, they are at the same y-coordinate. B is at a different y-coordinate. Since the strings have length 1, B must be 1 unit away from A and C along the y-axis. Since B started at the origin and moved upwards, its y-coordinate is greater than the initial y-coordinate of A and C (which is 0). As A and C move upwards, their y-coordinate increases. Consider the initial setup where A, B, C are on the x-axis. B is at (0,0), A at (-1,0), C at (1,0). B is given velocity $(0, v_0)$. So B moves to $(0, y_B)$ with $y_B > 0$. A moves to $(-x, y)$ and C moves to $(x, y)$, with $x$ decreasing and $y$ increasing. When A and C collide, $x=0$. So A and C are at $(0, y)$. Since the string length is 1, the distance between B and A (or B and C) is 1. So $\sqrt{(0-0)^2 + (y_B - y)^2} = 1$, $|y_B - y| = 1$. Since B moved upwards and A and C also moved upwards, and B started between A and C in the initial line, it is likely that B is above A and C when they collide. So $y_B > y$. Thus $y_B - y = 1$.

So, $\ddot{y} = \frac{5 v_0^2}{54}$. Then $T = m \ddot{y} = m \frac{5 v_0^2}{54} = 3 \times \frac{5 (\sqrt{3})^2}{54} = 3 \times \frac{5 \times 3}{54} = 3 \times \frac{15}{54} = 3 \times \frac{5}{18} = \frac{5}{6}$ N.

Let's check the centripetal acceleration approach again. The relative velocity of A with respect to B is $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. The radial direction is along the line AB. The tension provides the force in this direction. The component of $\vec{v}_{A/B}$ along AB is the radial velocity. The component perpendicular to AB is the tangential velocity. The centripetal acceleration of A towards B is $\frac{v_{tan}^2}{L}$. The force in the radial direction is T. So $T = m \frac{v_{tan}^2}{L}$. Just before collision, A is at $(0,y)$, B is at $(0,y_B)$, with $|y_B - y| = 1$. Let's assume $y_B > y$, so B is above A. The vector from B to A is $(0, y-y_B) = (0, -1)$. The unit vector from B to A is $\hat{r} = (0, -1)$. $\vec{v}_A = (\sqrt{\frac{5}{18}} v_0, v_0/3)$, $\vec{v}_B = (0, v_0/3)$. $\vec{v}_{A/B} = (\sqrt{\frac{5}{18}} v_0, 0)$. Radial velocity $v_{rad} = \vec{v}_{A/B} \cdot \hat{r} = (\sqrt{\frac{5}{18}} v_0, 0) \cdot (0, -1) = 0$. Tangential velocity vector $\vec{v}_{tan} = \vec{v}_{A/B} - v_{rad} \hat{r} = (\sqrt{\frac{5}{18}} v_0, 0) - 0 = (\sqrt{\frac{5}{18}} v_0, 0)$. Magnitude of tangential velocity $v_{tan} = \sqrt{\frac{5}{18}} v_0$. Centripetal acceleration $a_{rad} = \frac{v_{tan}^2}{L} = \frac{(\sqrt{\frac{5}{18}} v_0)^2}{1} = \frac{5 v_0^2}{18}$. Tension $T = m a_{rad} = m \frac{5 v_0^2}{18} = 3 \times \frac{5 (\sqrt{3})^2}{18} = 3 \times \frac{5 \times 3}{18} = 3 \times \frac{15}{18} = 3 \times \frac{5}{6} = \frac{5}{2} = 2.5$ N.

Let's check the case where B is below A and C, $y_B < y$, so $y_B - y = -1$. The vector from B to A is $(0, 1)$. The unit vector is $\hat{r} = (0, 1)$. Radial velocity $v_{rad} = \vec{v}_{A/B} \cdot \hat{r} = (\sqrt{\frac{5}{18}} v_0, 0) \cdot (0, 1) = 0$. Tangential velocity magnitude is the same, $v_{tan} = \sqrt{\frac{5}{18}} v_0$. Centripetal acceleration is the same, $a_{rad} = \frac{5 v_0^2}{18}$. Tension $T = m a_{rad} = 2.5$ N.

The direction of centripetal acceleration of A relative to B is towards B. If B is above A, the direction is upwards. If B is below A, the direction is downwards. Tension acts from A towards B. If B is above A, tension on A is upwards. If B is below A, tension on A is downwards. The force equation for A in the direction from A to B is T. The acceleration of A relative to B in this direction is the radial acceleration. Acceleration of A is $\vec{a}_A = (\ddot{x}_A, \ddot{y}_A) = (-\ddot{x}, \ddot{y})$. Acceleration of B is $\vec{a}_B = (0, \ddot{y}_B)$. Relative acceleration $\vec{a}_{A/B} = \vec{a}_A - \vec{a}_B = (-\ddot{x}, \ddot{y} - \ddot{y}_B)$. The radial acceleration is the component of $\vec{a}_{A/B}$ along the vector from B to A. If B is above A, vector from B to A is $(0, -1)$. Radial acceleration = $\vec{a}_{A/B} \cdot (0, -1) = (-\ddot{x}, \ddot{y} - \ddot{y}_B) \cdot (0, -1) = -(\ddot{y} - \ddot{y}_B) = \ddot{y}_B - \ddot{y}$. This radial acceleration is equal to $-\frac{v_{tan}^2}{L}$ because it is towards the center B. So, $\ddot{y}_B - \ddot{y} = -\frac{v_{tan}^2}{L} = -\frac{5 v_0^2}{18}$. From $2 \ddot{y} + \ddot{y}_B = 0$, $\ddot{y}_B = -2 \ddot{y}$. $-2 \ddot{y} - \ddot{y} = -\frac{5 v_0^2}{18}$, so $-3 \ddot{y} = -\frac{5 v_0^2}{18}$, $\ddot{y} = \frac{5 v_0^2}{54}$. $\ddot{y}_B = -2 \ddot{y} = -\frac{10 v_0^2}{54} = -\frac{5 v_0^2}{27}$. The net force on A in the y-direction is the tension T (since the string is vertical). So $T = m a_{Ay} = m \ddot{y} = 3 \times \frac{5 v_0^2}{54} = 3 \times \frac{5 \times 3}{54} = \frac{5}{6}$ N. This is not matching the centripetal force calculation.

Let's check the direction of tension. Tension on A is towards B. If B is above A, tension is upwards. If B is below A, tension is downwards. The centripetal acceleration of A relative to B is towards B. The force on A along the string is T. This force causes the radial acceleration. So, $T = m a_{rad} = m \frac{v_{tan}^2}{L}$. This is correct. The issue might be in the application of the equation of motion $T = m \ddot{y}$. This is only valid if the tension is the only force in the y-direction and the string is vertical.

Let's recheck the relative acceleration approach. The radial acceleration of A relative to B is $\vec{a}_{A/B} \cdot \hat{r}$. This is the acceleration along the line AB. The force on A along AB is T. So, $T = m (\vec{a}_{A/B} \cdot \hat{r})$. Let's assume B is above A, so $\hat{r} = (0, -1)$. $T = m ((-\ddot{x}, \ddot{y} - \ddot{y}_B) \cdot (0, -1)) = m (-(\ddot{y} - \ddot{y}_B)) = m (\ddot{y}_B - \ddot{y})$. We have $\ddot{y}_B - \ddot{y} = -\frac{v_{tan}^2}{L} = -\frac{5 v_0^2}{18}$. So, $T = m (-\frac{5 v_0^2}{18}) = -m \frac{5 v_0^2}{18}$. Tension cannot be negative.

Let's check the direction of radial acceleration. The radial acceleration of A relative to B is towards B. So, the component of $\vec{a}_{A/B}$ along the vector from A to B should be equal to $\frac{v_{tan}^2}{L}$. The vector from A to B is $(0, y_B - y)$. If $y_B > y$, this is $(0, 1)$. The unit vector is $(0, 1)$. Radial acceleration = $\vec{a}_{A/B} \cdot (0, 1) = (-\ddot{x}, \ddot{y} - \ddot{y}_B) \cdot (0, 1) = \ddot{y} - \ddot{y}_B$. So, $\ddot{y} - \ddot{y}_B = \frac{v_{tan}^2}{L} = \frac{5 v_0^2}{18}$. From $2 \ddot{y} + \ddot{y}_B = 0$, $\ddot{y}_B = -2 \ddot{y}$. $\ddot{y} - (-2 \ddot{y}) = \frac{5 v_0^2}{18}$, so $3 \ddot{y} = \frac{5 v_0^2}{18}$, $\ddot{y} = \frac{5 v_0^2}{54}$. $\ddot{y}_B = -2 \ddot{y} = -\frac{10 v_0^2}{54} = -\frac{5 v_0^2}{27}$. The force on A in the direction from A to B is T. So $T = m (\ddot{y} - \ddot{y}_B) = m \frac{5 v_0^2}{18} = 3 \times \frac{5 \times 3}{18} = 2.5$ N. This matches the centripetal force calculation. So, the tension is 2.5 N.