A 10 kg ball moving due west at 2m/sec collided with a 4kg ball that is moving due east at 3m/sec. you determine that immediately after the collision the 10 kg ball is moving due west at .57 m/sec.
a) this probably is a totally elastic collision
b) this definitely is NOT a totally inelastic collision
c) more info is needed
we are to show all of our work.
To determine whether the collision is totally elastic, totally inelastic, or whether more information is needed, we can use the principles of conservation of momentum and kinetic energy.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Momentum before collision = Momentum after collision
Momentum is defined as the product of mass and velocity:
Momentum = mass × velocity
For the 10 kg ball moving due west at 2 m/sec:
Momentum before collision = (10 kg) × (-2 m/sec) = -20 kg·m/sec
For the 4 kg ball moving due east at 3 m/sec:
Momentum before collision = (4 kg) × (3 m/sec) = 12 kg·m/sec
Total momentum before collision = -20 kg·m/sec + 12 kg·m/sec = -8 kg·m/sec
After the collision, the 10 kg ball is moving due west at 0.57 m/sec. The 4 kg ball's velocity is not given. Let's assume it is v m/sec in the opposite direction (due east).
For the 10 kg ball after the collision:
Momentum after collision = (10 kg) × (-0.57 m/sec) = -5.7 kg·m/sec
For the 4 kg ball after the collision:
Momentum after collision = (4 kg) × v = 4v kg·m/sec
Total momentum after collision = -5.7 kg·m/sec + 4v kg·m/sec = -8 kg·m/sec (conservation of momentum)
Solving for v:
-5.7 kg·m/sec + 4v kg·m/sec = -8 kg·m/sec
4v kg·m/sec = -8 kg·m/sec + 5.7 kg·m/sec
4v kg·m/sec = -2.3 kg·m/sec
v = (-2.3 kg·m/sec) / (4 kg·m/sec) = -0.575 m/sec
Therefore, the 4 kg ball's velocity after the collision is -0.575 m/sec, due east.
2. Conservation of kinetic energy:
In a totally elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Total kinetic energy before collision = Total kinetic energy after collision
The kinetic energy of an object is given by the formula:
Kinetic energy = (1/2) × mass × (velocity)^2
For the 10 kg ball moving due west at 2 m/sec:
Kinetic energy before collision = (1/2) × 10 kg × (2 m/sec)^2 = 20 J
For the 4 kg ball moving due east at 3 m/sec:
Kinetic energy before collision = (1/2) × 4 kg × (3 m/sec)^2 = 18 J
Total kinetic energy before collision = 20 J + 18 J = 38 J
After the collision:
For the 10 kg ball moving due west at 0.57 m/sec:
Kinetic energy after collision = (1/2) × 10 kg × (0.57 m/sec)^2 ≈ 1.63 J
For the 4 kg ball moving due east at -0.575 m/sec:
Kinetic energy after collision = (1/2) × 4 kg × (-0.575 m/sec)^2 ≈ 0.83 J
Total kinetic energy after collision = 1.63 J + 0.83 J ≈ 2.46 J
Since 38 J ≠ 2.46 J, the total kinetic energy is not conserved. Therefore, we can conclude that this is NOT a totally elastic collision (option b is correct).
Therefore, the answer is:
b) this definitely is NOT a totally elastic collision