91Ó°ÊÓ

Three particles A, B, C of masses \(m, 2 m, 3 m\) respectively lie at rest in that order in a straight line on a smooth horizontal table. The distance between consecutive particles is \(a .\) A slack light inelastic string of length \(2 a\) connects \(\mathrm{A}\) and \(\mathrm{B}\). An exactly similar slack string connects \(\mathrm{B}\) and \(\mathrm{C}\). If \(\mathrm{A}\) is projected in the direction CBA with speed \(V\), find the time which elapses before \(C\) begins to move. Find also the speed with which C begins to move. Show that the ratio of the impulsive tensions in \(\mathrm{BC}\) and \(\mathrm{AB}\) when \(\mathrm{C}\) is jerked into motion is \(3: 1 .\) Find the total loss of kinetic energy when C has started to move. (J.M.B.)

Step by step solution

01

Calculate the time for string AB to become taut

First, determine the time it takes for particle A to travel distance 'a' to make the string AB taut. Since A starts with an initial speed V, we use the equation of motion: \[ a = Vt \]Solving for time, we get: \[ t = \frac{a}{V} \]

02

Describe the motion after string AB becomes taut

After time \( \frac{a}{V} \), the string AB becomes taut pulling particle B. Since the string is inelastic, A and B will move together as a single system with a combined mass of \( m + 2m = 3m \). To find their common velocity (\(V_B\)), use the law of conservation of momentum: \[ mV = 3mV_B \]Solving for \(V_B\):\[ V_B = \frac{V}{3} \]

03

Calculate the time for string BC to become taut

When B starts moving, the distance between B and C is 'a'. Since B and A move with velocity \(\frac{V}{3}\) together, and C is still at rest, it will take time \(t_2\) for the string BC to become taut. Using the equation of motion: \[ a = \frac{V}{3} t_2 \]Solving for \(t_2\):\[ t_2 = \frac{3a}{V} \]

04

Total time elapsed before C begins to move

The total time elapsed before C starts to move is the sum of the times calculated in steps 1 and 3: \[ t_{total} = \frac{a}{V} + \frac{3a}{V} \]Simplifying, we get:\[ t_{total} = \frac{4a}{V} \]

05

Find the speed with which C begins to move

When string BC becomes taut, particles A, B, and C will move together with a common velocity. Use conservation of momentum:\[ (m + 2m) \frac{V}{3} = (m + 2m + 3m)V_C \]This simplifies to: \[ 3m \frac{V}{3} = 6mV_C \]Solving for \(V_C\):\[ V_C = \frac{V}{6} \]

06

Determine the ratio of impulsive tensions in BC and AB

The impulsive tension in string AB, when B starts moving is calculated as: \[ T_{AB} = m \left( v - 0 \right) \]For string BC, when C is jerked into motion: \[ T_{BC} = 3m \left( \frac{V}{6} - 0 \right) \]Simplifying both, we get: \[ T_{AB} = m \frac{V}{3} \ T_{BC} = 3m \frac{V}{6} = \frac{3mV}{6} \]The ratio \( \frac{T_{BC}}{T_{AB}} \) is: \[ \frac{\frac{3mV}{6}}{\frac{mV}{3}} = 3:1 \]

07

Calculate the total loss of kinetic energy

The initial kinetic energy is:\[ KE_{initial} = \frac{1}{2} mV^2 \]The final kinetic energy, when all three particles move together with velocity \( \frac{V}{6} \) is:\[ KE_{final} = \frac{1}{2} (m + 2m + 3m) \left( \frac{V}{6} \right)^2 \]Simplifying, we get: \[ KE_{final} = \frac{1}{2} (6m) \frac{V^2}{36} = \frac{1}{12} mV^2 \]Total loss of kinetic energy is:\[ KE_{loss} = KE_{initial} - KE_{final} = \frac{1}{2} mV^2 - \frac{1}{12} mV^2 = \frac{1}{2} mV^2 \left( 1 - \frac{1}{6} \right) = \frac{5}{12} mV^2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations

To tackle problems involving motion, we heavily rely on kinematic equations. These equations relate the variables of motion, such as displacement, initial velocity, final velocity, acceleration, and time.
In this problem, particle A travels distance 'a' with an initial velocity V to make the string taut. Using the equation: \[ a = Vt \]we find the time spent, \[ t = \frac{a}{V} \].
This relationship helps us understand how long it takes moving objects to perform specific actions.

conservation of momentum

The principle of conservation of momentum states that the total momentum in a closed system remains constant, provided there are no external forces acting on it. This principle is crucial in collision problems. In the given exercise, when particle A collides with particle B and they stick together: \[ mV = (m + 2m)V_B \] resulting in \[ V_B = \frac{V}{3} \].
By this principle, we solve for the velocities post-collision, ensuring that initial and final momentum are the same.

kinetic energy loss

In inelastic collisions, some kinetic energy is always converted into other forms of energy, often leading to a loss in the system's kinetic energy. Initially, particle A has kinetic energy: \[ KE_{initial} = \frac{1}{2} mV^2 \]. After the collision, and when particles A, B, and C move together, we calculate the final kinetic energy as: \[ KE_{final} = \frac{1}{12} mV^2 \] The loss of kinetic energy is thus: \[ KE_{loss} = KE_{initial} - KE_{final} = \frac{5}{12} mV^2 \]
This shows how the energy initially present decreases after collisions.

impulse and tension

Impulse is a change in momentum resulting from a force applied over a period. In this problem, impulsive tension arises as the inelastic string between particles suddenly becomes taut.

• The tension in string AB when A pulls B is: \[ T_{AB} = m \frac{V}{3} \].


• The tension in string BC when B pulls C is: \[ T_{BC} = 3m \frac{V}{6} = \frac{3mV}{6} \].

The ratio of impulsive tensions is \[ \frac{T_{BC}}{T_{AB}} = 3:1 \].
Understanding impulse helps in identifying how forces interact in physical systems during brief periods of force application.