A dumb bell of 10 kg mass falls from a height of 0.8 m . What is themomentum transferred by the dumb bell while hitting the ground?

To solve this problem, we first need to find the velocity of the dumbbell just before it hits the ground. We can use the equation of motion:

v^2 = u^2 + 2gs

where: v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and s is the distance.

The dumbbell is dropped, so the initial velocity u is 0. The acceleration due to gravity g is approximately 9.8 m/s^2, and the distance s is 0.8 m.

Substituting these values into the equation gives:

v^2 = 0 + 29.80.8 v^2 = 15.68 v = sqrt(15.68) = 3.96 m/s

The momentum p of an object is given by the product of its mass m and its velocity v, so:

p = mv p = 10 kg * 3.96 m/s = 39.6 kg*m/s

Therefore, the momentum transferred by the dumbbell while hitting the ground is 39.6 kg*m/s.