On a horizontal frictionless table, masses A and B (3 kg, 6 kg) slide to the right and left, respectively. They have speeds of 3 m/s and 1 m/s, respectively. The two masses collide, and bounce off each other. After the collision, they travel in opposite directions at speeds of 1.5 m/s and 1.250 m/s, respectively.
Calculate the kinetic energy of the A+B system:
before collision:
after collision
They tell you the masses of both and their speeds before and after the collision. All you have to do is calculate and add up the kinetic energies (1/2) M V^2 of each mass, before and after.
They have already provided velocities that satisfy the momentum conservation requirement, so you don't have to apply conservation laws to do this problem.
To calculate the kinetic energy of the A+B system before and after the collision, we need to use the formula for kinetic energy, which is given by:
Kinetic energy = 0.5 * mass * velocity^2
Before the collision:
1. Calculate the kinetic energy of mass A:
Kinetic energy of A = 0.5 * mass of A * velocity of A^2
Mass of A = 3 kg
Velocity of A = 3 m/s
Kinetic energy of A = 0.5 * 3 kg * (3 m/s)^2 = 13.5 J
2. Calculate the kinetic energy of mass B:
Kinetic energy of B = 0.5 * mass of B * velocity of B^2
Mass of B = 6 kg
Velocity of B = -1 m/s (since it is moving in the opposite direction)
Kinetic energy of B = 0.5 * 6 kg * (-1 m/s)^2 = 3 J
The total kinetic energy of the system before the collision is the sum of the kinetic energies of A and B:
Kinetic energy before collision = Kinetic energy of A + Kinetic energy of B
Kinetic energy before collision = 13.5 J + 3 J = 16.5 J
After the collision:
1. Calculate the kinetic energy of mass A:
Kinetic energy of A = 0.5 * mass of A * velocity of A^2
Mass of A = 3 kg
Velocity of A = -1.5 m/s (since it is moving in the opposite direction)
Kinetic energy of A = 0.5 * 3 kg * (-1.5 m/s)^2 = 3.375 J
2. Calculate the kinetic energy of mass B:
Kinetic energy of B = 0.5 * mass of B * velocity of B^2
Mass of B = 6 kg
Velocity of B = 1.25 m/s
Kinetic energy of B = 0.5 * 6 kg * (1.25 m/s)^2 = 4.6875 J
The total kinetic energy of the system after the collision is the sum of the kinetic energies of A and B:
Kinetic energy after collision = Kinetic energy of A + Kinetic energy of B
Kinetic energy after collision = 3.375 J + 4.6875 J = 8.0625 J
Therefore, the kinetic energy of the A+B system before the collision is 16.5 J, and after the collision, it is 8.0625 J.