A particle A of mass 2kg lies on the edge of a table of height 1m. It is connected by a light inelastic string of length 0.7m to a second particle B of mass 3 kg which is lying on table 0.25m from the edge. If A is pushed gently so that it starts falling from table then, find the speed of B when it starts to move. Also find the impulsive tension in the string at that moment.

To find the speed of B when it starts to move, we need to consider the conservation of energy principle.

First, let's determine the potential energy of particle A when it is at the edge of the table. Since the height of the table is 1m and the mass of particle A is 2kg, the potential energy of A is given by:

Potential energy of A = mass of A * gravitational acceleration * height
= 2kg * 9.8m/s^2 * 1m
= 19.6 Joules

Next, let's calculate the extension of the string when particle A falls from the table. The string is initially 0.7m long, and when A falls, it will become shorter by the distance covered by A. The distance covered by particle A when it falls from height h is given by the Pythagorean theorem:

Distance covered by A = sqrt(h^2 + l^2)
= sqrt(1^2 + 0.7^2)
= sqrt(1.49)
= 1.22m (approx)

Thus, the string will extend by 0.7m - 1.22m = -0.52m (negative because it contracts).

Using Hooke's law, we can find the work done by the tension in the string, which is equal to the change in potential energy of A. The work done by the tension force is given by:

Work done by tension = (1/2) * k * (extension of the string)^2

where k is the spring constant. Since the string is light and inelastic, its spring constant can be considered very large, which means it will be very stiff. Therefore, the extension of the string will be very small compared to its natural length, and the tension can be considered constant.

Now, let's find the work done by the tension force:

Work done by tension = (1/2) * k * (-0.52m)^2
= 0.135k Joules (assuming k = 1)

Since the work done by the tension force is equal to the change in potential energy, we can express this as an equation:

0.135k = 19.6

Solving for k, we find:

k = 19.6 / 0.135
≈ 145.19 N/m (approx)

Since the tension in the string is equal to the spring force, which is given by Hooke's law, we can calculate the tension at that instant:

Tension in the string = k * extension of the string
= 145.19 N/m * -0.52m
= -75.51 N (approx)

Therefore, the impulsive tension in the string at the moment B starts to move is approximately 75.51 N (directed towards the table).

To find the speed of B when it starts to move, we can use the principle of conservation of linear momentum. Initially, the system is at rest, so the total momentum is zero. When B starts to move, the momentum of the system will still be zero because the particles are connected by a light inelastic string.

Let the speed of B when it starts to move be v. Since A has fallen 0.52m, B has moved 0.25m towards the edge, resulting in an extension of 0.52m + 0.25m = 0.77m. By the conservation of linear momentum principle, we have:

0 = mass of A * 0 + mass of B * v

Using the given masses (mass of A = 2kg and mass of B = 3kg), we get:

0 = 2kg * 0 + 3kg * v

Solving for v, we find:

v = 0 m/s

Therefore, the speed of B when it starts to move is 0 m/s.