Two particles X and Y joined by a rigit rod of negligible mass, move along a smooth horizontal plane. Particle X has mass 7.7kg and is pushed by an external force Fx of magnitude 73.01N. Particle Y has mass 5.68kg and is pulled along the plane by an external force Fy of magnitude 29.68N. It is found that the two particles accelerate along the plane a fixed distance apart.

Calculate the common acceleration of the particles.

Thank you!

To calculate the common acceleration of the particles, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

First, let's calculate the net force acting on particle X, which is given by the external force Fx minus the tension in the rod.

The tension in the rod is equal to the force pulling particle Y, Fy. Since the rod is rigid and of negligible mass, the tension is the same for both particles.

Therefore, the net force on particle X is Fx - Fy.

Using Newton's second law, we can calculate the acceleration of particle X:

Acceleration_x = (Fx - Fy) / mass_x

Where:
Fx = 73.01N (external force on particle X)
Fy = 29.68N (external force pulling particle Y)
mass_x = 7.7kg (mass of particle X)

Plugging in the values:

Acceleration_x = (73.01N - 29.68N) / 7.7kg

Acceleration_x = 43.33N / 7.7kg

Acceleration_x ≈ 5.63 m/s²

Since both particles are connected by a rigid rod and accelerate together, the common acceleration of the particles is the same as the acceleration of particle X:

Common acceleration = 5.63 m/s².

Therefore, the common acceleration of the particles is approximately 5.63 m/s².