A small ball of mass 100 g is attached to a light and in extensible string of length 50 cm. The string is tied to a support O and the mass m released from point A which is at a horizontal distance of 30 cm from the support. Calculate the speed of the ball at its lowest point of the trajectry

To solve this problem, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy in a closed system—i.e., a system that does not have any external forces, such as friction or air resistance, acting on it—remains constant.

Here are the steps to solve the problem:

1. First, we need to calculate the height from which the ball is released. We can use the Pythagorean theorem for this. The length of the string is the hypotenuse of a right triangle, and the horizontal distance from the support to the point of release is one of the sides. The height h can be calculated as follows:

   h = sqrt(50^2 - 30^2) = 40 cm = 0.4 m

2. Next, we calculate the potential energy (PE) at the point of release. The potential energy is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Substituting the given values, we get:

   PE = 0.1 kg * 9.8 m/s^2 * 0.4 m = 0.392 J

3. When the ball reaches its lowest point, all the potential energy will have been converted into kinetic energy (KE). The kinetic energy is given by the formula KE = 1/2 mv^2, where m is the mass and v is the velocity. We can set the kinetic energy equal to the potential energy and solve for v:

   0.392 J = 1/2 * 0.1 kg * v^2

   Solving for v gives v = sqrt((2 * 0.392 J) / 0.1 kg) = 8.83 m/s

So, the speed of the ball at its lowest point of the trajectory is approximately 8.83 m/s.