Question

A massless string of length $2l$ = 6 m are connected to two balls that are lying on frictionless surface. A constant horizontal force $\overrightarrow{F}$=3N is applied to the midpoint of the string, perpendicular to it (see figure). The kinetic energy (in Joule) that is lost when the balls collide, assuming they stick together is:

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

Answer: (C)

9

Answer 2

The problem describes a massless string of length $2l = 6$ m connecting two balls on a frictionless surface. A constant horizontal force $\vec{F} = 3$ N is applied to the midpoint of the string, perpendicular to it. We need to find the kinetic energy lost when the balls collide, assuming they stick together.

1. Work Done by the Applied Force:

Let the initial configuration of the string be a straight line along the x-axis, with the midpoint at the origin $(0,0)$ and the balls at $(-l,0)$ and $(l,0)$. The total length of the string is $2l = 6$ m, so $l = 3$ m.

The force $\vec{F}$ is applied at the midpoint, perpendicular to the string. Let's assume the force is applied along the positive y-axis.

As the force pulls the midpoint, the string forms a V-shape. The balls move inwards along the x-axis.

The balls collide when they meet at the center, i.e., at $(0,0)$.

At this moment, the string is folded, and the midpoint has moved a distance $y$ from its initial position. Since the string length is conserved and each segment is of length $l$, the midpoint must be at $(0,l)$ when the balls meet at $(0,0)$.

This means the displacement of the midpoint in the direction of the force is $y = l = 3$ m.

The work done by the constant force $\vec{F}$ is given by: $W = F \times (\text{displacement of midpoint})$ $W = F \times l$

Given $F = 3$ N and $l = 3$ m: $W = 3 \text{ N} \times 3 \text{ m} = 9 \text{ J}$

According to the work-energy theorem, this work done by the external force is converted into the kinetic energy of the system just before the collision. The string is massless, so only the balls possess kinetic energy.

Therefore, the kinetic energy of the system just before collision is: $K_{before} = W = 9 \text{ J}$

2. Collision Analysis:

The two balls collide and stick together. This is a perfectly inelastic collision.

Let the masses of the two balls be $m_1$ and $m_2$.

Due to the symmetry of the setup (force applied at the midpoint of the string), the tension in both halves of the string will be the same. The horizontal component of the tension is the force accelerating the balls towards each other.

Let $v_1$ and $v_2$ be the speeds of the balls just before collision.

The horizontal forces on the balls are due to the tension in the string. If $T$ is the tension and $\theta$ is the angle the string makes with the initial (x) axis, then $T \cos\theta$ is the horizontal force on each ball.

$m_1 \frac{dv_1}{dt} = T \cos\theta$ and $m_2 \frac{dv_2}{dt} = T \cos\theta$.

Thus, $m_1 \frac{dv_1}{dt} = m_2 \frac{dv_2}{dt}$.

Integrating from rest, we get $m_1 v_1 = m_2 v_2$. This means the magnitudes of their momenta are equal just before collision.

Let the velocity of ball 1 be $\vec{v}_1 = (v_1, 0)$ and the velocity of ball 2 be $\vec{v}_2 = (-v_2, 0)$ (as they move towards each other).

The total momentum of the system (two balls) in the x-direction just before collision is: $P_{x,before} = m_1 v_1 + m_2 (-v_2) = m_1 v_1 - m_2 v_2$

Since $m_1 v_1 = m_2 v_2$, we have: $P_{x,before} = 0$

After the collision, the balls stick together, forming a combined mass $M_{total} = m_1 + m_2$. Let their common final velocity be $\vec{V}_f = (V_f, 0)$.

The total momentum of the system in the x-direction after collision is: $P_{x,after} = (m_1 + m_2) V_f$

Since there are no external horizontal forces acting on the two-ball system during the collision (the applied force $\vec{F}$ is perpendicular to the motion of the balls), the total momentum in the x-direction is conserved: $P_{x,before} = P_{x,after}$ $0 = (m_1 + m_2) V_f$

Since $m_1 + m_2 \neq 0$, it implies $V_f = 0$.

Therefore, after the collision, the combined mass of the balls comes to rest.

The kinetic energy of the system after collision is: $K_{after} = \frac{1}{2} (m_1 + m_2) V_f^2 = \frac{1}{2} (m_1 + m_2) (0)^2 = 0 \text{ J}$

3. Kinetic Energy Lost:

The kinetic energy lost during the collision is the difference between the kinetic energy before and after the collision: $\Delta K_{lost} = K_{before} - K_{after}$ $\Delta K_{lost} = 9 \text{ J} - 0 \text{ J} = 9 \text{ J}$

The kinetic energy lost is 9 J.