Two buckets each of mass 5kg are connected one each end of a light inextensible string going over a smooth pulley. A piece of putty of mass 2kg is dropped through a height of 20m into one bucket and become attached. Calculate

(i) The initial velocity of the system after impact
(ii) The acceleration with which the buckets move
(iii) The energy lost by the 2kg mass due to the impact.

for initial velocity, conservation of momentum

let clockwise rotation be +
initial=final
2*vi=(10+2)Vf
you want to solve for vf.
vi=sqrt(2*9.8*20)

acceleration: force/mass=2g/12

Yes

Answer

To answer these questions, we can utilize the principle of conservation of energy. We will first calculate the potential energy gained by the 2kg mass as it falls to a height of 20m and then use that information to find the initial velocity of the system after the impact.

(i) Calculate the potential energy gained by the 2kg mass:
The potential energy gained by an object at a height is given by the formula: potential energy = mass * gravity * height.

The mass of the object is 2kg, gravity is approximately 9.8 m/s², and the height is 20m.
Potential energy gained = 2kg * 9.8 m/s² * 20m = 392 Joules

Since the system is isolated, this potential energy is transferred to the system as kinetic energy. Therefore, the initial kinetic energy of the system after impact is also 392 Joules.

To find the initial velocity of the system, we can use the formula for kinetic energy:

Initial kinetic energy = (1/2) * (mass1 + mass2) * velocity^2

Here, mass1 and mass2 are both 5kg (the total mass of the system), and the initial kinetic energy is 392 Joules.

Rearranging the equation, we get:

velocity^2 = (2 * initial kinetic energy) / (mass1 + mass2)
= (2 * 392 J) / 10kg
= 78.4 m²/s²

Taking the square root of both sides, we find:

velocity ≈ √78.4 ≈ 8.85 m/s

Therefore, the initial velocity of the system after impact is approximately 8.85 m/s.

(ii) Calculate the acceleration with which the buckets move:
Since the system is connected by a light inextensible string that goes over a smooth pulley, the acceleration of the system is the same for both buckets. Let's denote the acceleration as "a".

Using the equation for the acceleration of an object connected by a string, we have:

mass2 * acceleration = mass2 * gravity - tension

Since the mass of the 2kg bucket is given as mass2 and there is no vertical acceleration (due to the smooth pulley and light inextensible string), the equation simplifies to:

2kg * a = 2kg * 9.8 m/s² - tension

We need to find the tension in the string. The tension is equal to the weight of the other bucket that is hanging. The hanging mass is also 5kg, so the tension can be calculated as:

tension = mass2 * gravity = 5kg * 9.8 m/s² = 49 N

Substituting the known values back into the equation, we have:

2kg * a = 2kg * 9.8 m/s² - 49 N

Simplifying further:

2kg * a = 19.6 N - 49 N
2kg * a = -29.4 N

Dividing both sides by 2kg, we find:

a = -29.4 N / 2kg
a = -14.7 m/s²

The negative sign indicates that the acceleration is in the opposite direction of the gravitational force, as the hanging mass is heavier. So the acceleration with which the buckets move is approximately 14.7 m/s² in the opposite direction to gravity.

(iii) Calculate the energy lost by the 2kg mass due to the impact:
The energy lost by the 2kg mass is equal to the potential energy it gained minus the kinetic energy it had afterward.

Potential energy gained = 392 Joules (calculated earlier)

We already found the initial velocity of the system (8.85 m/s). The kinetic energy of the 2kg mass can be calculated using the formula:

Kinetic energy = (1/2) * mass * velocity^2

The mass is 2kg, and the velocity is 8.85 m/s.

Kinetic energy = (1/2) * 2kg * (8.85 m/s)^2
= 78 Joules (approximately)

Therefore, the energy lost by the 2kg mass due to the impact is:

Energy lost = Potential energy gained - Kinetic energy
= 392 Joules - 78 Joules
= 314 Joules