A 10.0 g marble slides to the left with a
velocity of magnitude 0.400 m/s on the
frictionless horizontal surface of an icy
Town sidewalk and makes a head-on
collision with a larger 30.0 g marble
sliding to the right with a velocity of
magnitude 0.200 m/s (see the figure). If
the collision is elastic, find the velocity of each marble (magnitude and direction)
after the collision. (Since the collision is head-on, all the motion is along a line).
To find the velocity of each marble after the collision, we can use the principles of conservation of momentum and kinetic energy in an elastic collision.
1. Conservation of momentum:
In an isolated system, the total momentum before the collision is equal to the total momentum after the collision.
Let's assign variables to the velocities of the marbles:
- Velocity of the 10.0 g marble before the collision: V1f
- Velocity of the 30.0 g marble before the collision: V2f
- Velocity of the 10.0 g marble after the collision: V1f'
- Velocity of the 30.0 g marble after the collision: V2f'
The conservation of momentum equation can be written as:
(m1 * V1f) + (m2 * V2f) = (m1 * V1f') + (m2 * V2f')
Plugging in the given values:
(10.0 g * 0.400 m/s) + (30.0 g * (-0.200 m/s)) = (10.0 g * V1f') + (30.0 g * V2f')
Note that we assign a negative sign to the velocity of the 30.0 g marble before the collision since it's moving in the opposite direction.
2. Conservation of kinetic energy:
In an elastic collision, both momentum and kinetic energy are conserved.
The kinetic energy equation can be written as:
(1/2 * m1 * V1f²) + (1/2 * m2 * V2f²) = (1/2 * m1 * V1f'²) + (1/2 * m2 * V2f'²)
Plugging in the given values, we have:
(1/2 * 10.0 g * (0.400 m/s)²) + (1/2 * 30.0 g * (-0.200 m/s)²) = (1/2 * 10.0 g * V1f'²) + (1/2 * 30.0 g * V2f'²)
Now we have two equations with two variables (V1f' and V2f') that we can solve simultaneously to find the velocities after the collision.