4. Two masses of 3 kg and 5 kg collide head-on, the former moving to the right with the speed 5 m/s and the latter moving to the left with the speed 3 m/s.After collision they stick together. What amount of mechanical energy is lost?
5. A cylinder with a uniform density is rolling over a plane. What fraction of its kinetic energy is rotational kinetic energy?
4. Their momenta cancel and the stuck-together mass won't move, because
5x3 - 3x5 = 0.
All mechanical energy is lost.
5. Compare (1/2) I w^2 (the rotational KE) with the sum of (1/2) Iw^2 and (1/2) M V^2.
Note that I = (1/2) M R^2 (for solid cylinders) and
V = R w (for rolling objects). You should end up with a fraction independent of M, R and w.
To answer both questions, we need to use the principles of conservation of momentum and conservation of energy.
4. First, we need to find the velocity of the masses after the collision. We can do this by using the conservation of momentum. The initial momentum is the sum of the momentum of both masses before the collision, and the final momentum is the sum of the momentum after the collision (since they stick together).
Initial momentum = (mass1 * velocity1) + (mass2 * velocity2)
Final momentum = (mass1 + mass2) * velocity_final
From the given information:
mass1 = 3 kg, velocity1 = 5 m/s (to the right)
mass2 = 5 kg, velocity2 = -3 m/s (to the left)
Using the conservation of momentum:
(3 kg * 5 m/s) + (5 kg * -3 m/s) = (3 kg + 5 kg) * velocity_final
15 kg m/s - 15 kg m/s = 8 kg * velocity_final
0 = 8 kg * velocity_final
Therefore, the velocity of the masses after the collision is 0 m/s (they come to a stop).
Now, to find the amount of mechanical energy lost, we need to calculate the initial and final mechanical energies. The mechanical energy is the sum of the kinetic energy of both masses.
Initial mechanical energy = (1/2) * mass1 * velocity1^2 + (1/2) * mass2 * velocity2^2
Final mechanical energy = (1/2) * (mass1 + mass2) * velocity_final^2
Using the given values:
Initial mechanical energy = (1/2) * 3 kg * (5 m/s)^2 + (1/2) * 5 kg * (-3 m/s)^2
Final mechanical energy = (1/2) * (3 kg + 5 kg) * (0 m/s)^2
Calculating the initial mechanical energy:
Initial mechanical energy = (1/2) * 3 kg * (25 m^2/s^2) + (1/2) * 5 kg * (9 m^2/s^2)
Initial mechanical energy = 37.5 J + 22.5 J
Initial mechanical energy = 60 J
Calculating the final mechanical energy:
Final mechanical energy = (1/2) * 8 kg * (0 m/s)^2
Final mechanical energy = 0 J
The amount of mechanical energy lost during the collision can be calculated as the difference between the initial and final mechanical energies:
Amount of mechanical energy lost = Initial mechanical energy - Final mechanical energy
Amount of mechanical energy lost = 60 J - 0 J
Amount of mechanical energy lost = 60 J
Therefore, 60 J of mechanical energy is lost during the collision.
5. To find the fraction of the kinetic energy that is rotational kinetic energy for a rolling cylinder with uniform density, we can consider the total kinetic energy and separate it into translational and rotational components.
The total kinetic energy of the rolling cylinder is given by:
Total kinetic energy = (1/2) * mass * velocity^2
To separate the kinetic energy into translational and rotational components, we can use the relationship between linear and angular velocity for a rolling object:
v = ω * r
where v is the linear velocity, ω is the angular velocity, and r is the radius of the cylinder.
The translational kinetic energy of the rolling cylinder is given by:
Translational kinetic energy = (1/2) * mass * v^2
The rotational kinetic energy is given by:
Rotational kinetic energy = (1/2) * moment of inertia * ω^2
For a solid cylinder rolling without slipping, the moment of inertia about its axis of rotation is given by:
Moment of inertia = (1/2) * mass * r^2
Substituting the relationship between linear and angular velocity (v = ω * r) and the moment of inertia, we can express the rotational kinetic energy in terms of the linear velocity:
Rotational kinetic energy = (1/4) * mass * v^2
Therefore, the fraction of the kinetic energy that is rotational kinetic energy is:
Fraction of rotational kinetic energy = Rotational kinetic energy / Total kinetic energy
Fraction of rotational kinetic energy = (1/4) * mass * v^2 / ((1/2) * mass * v^2)
Fraction of rotational kinetic energy = (1/4) / (1/2)
Fraction of rotational kinetic energy = 1/2
Therefore, half of the kinetic energy of a rolling cylinder with a uniform density is rotational kinetic energy.