Problem :
Two balls with equal masses, m, and equal speed, v, engage in a head on
elastic collision. What is the final velocity of each ball, in terms of m and
v?
Though we could go through the formal application of the equations of linear
momentum, it is easier to think about this problem conceptually. Since the balls
of equal mass are moving at equal and opposite speeds, the total linear momentum
of the system is zero. For linear momentum to be conserved after the collision,
both balls must rebound with the same velocity. If one ball had more speed than
the other, there would be a net linear momentum and our conservation principle
would be invalid. Having established that both balls rebound with the same
speed, we must find what that speed is. Since the collision is elastic, kinetic
energy must be conserved. If the final velocity of each ball were more, or less,
than its initial velocity, kinetic energy would not be conserved. Thus we can
state that the final velocity of each ball is equal in magnitude and opposite in
direction to their respective initial velocities.
Problem :
Two balls, each with mass 2 kg, and velocities of 2 m/s and 3 m/s collide head
on. Their final velocities are 2 m/s and 1 m/s, respectively. Is this collision
elastic or inelastic?
To check for elasticity, we need to calculate the kinetic energy both before and
after the collision. Before the collision, the kinetic energy is
(2)(2)2 + (2)(3)2 = 13. After, the kinetic energy is
(2)(2)2 + (2)(1)2 = 5. Since the kinetic energies are not
equal, the collision is inelastic.
Problem :
Two balls of mass m1 and m2, with velocities v1 and v2 collide head
on. Is there any way for both balls to have zero velocity after the collision?
If so, find the conditions under which this can occur.
First of all, the collision must be inelastic, as the final kinetic energy must
be zero, clearly less than the initial kinetic energy. Secondly, we can state
that the collision is completely inelastic, as both objects with zero velocity
must remain at the site of the collision, i.e., they must stick together. The
final principle we must check is that momentum is conserved. Clearly the final
momentum of the system must be zero, as neither ball is moving. Thus the same
value must be true before the collision. For this to happen, both masses must
have equal and opposite momentum, or m1v1 = m2v2. Thus, in a completely
inelastic collision in which m1v1 = m2v2, both masses will be stationary
after the collision.
Problem :
A car of 500 kg, traveling at 30 m/s rear ends another car of 600 kg, traveling
at 20 m/s. in the same direction The collision is great enough that the two cars
stick together after they collide. How fast will both cars be going after the
collision?
This is an example of a completely inelastic collision. Since the two cars stick
together, they must move with a common velocity after the collision. Thus simply
using the conservation of momentum is enough to solve for our one unknown
variable, the velocity of the two cars after the collision. Relating the initial
and final moments:
| po | = | pf |
|
| m1v1 + m2v2 | = | Mvf |
|
| (500)(30) + (600)(20) | = | (1100)vf |
|
| vf | = | 24.5m/s |
|
Thus both cars will travel at 24.5 m/s, in the same direction as their initial
travel.
Problem :
One pool ball traveling with a velocity of 5 m/s hits another ball of the same
mass, which is stationary. The collision is head on and elastic. Find the final
velocities of both balls.
Here we use our two conservation laws to find both final velocities. Let's call
the pool ball that is initially moving ball 1, and the stationary one ball 2.
Relating the kinetic energies before and after the collision,
 mv1o2 + mv2o2 | = | mv1f2 + mv2f2 |
|
 m | = | mv1f2 + mv2f2 |
|
| | | Canceling the fractions and masses, |
|
| 25 | = | v1f2 + v2f2 |
|
We also know that momentum must be conserved. The initial momentum is provided
entirely by ball 1, and has a magnitude of
5m. The final momentum has
contributions from both balls. Relating the two,
5m = mv1f + mv2f
Implying that
m1f + m2f = 5
Notice the similarity of the two equations we have. Though our kinetic energy
equation includes the velocities squared, both equations include the sum of the
velocities being equal to a constant. The systematic approach to this problem is
to substitute for
m1f into our first equation using our second equation.
However we can use a shortcut. Let's see what happens when we square our second
equation:
| (m1f+m2f)2 | = | 25 |
|
| m1f2 + m2f2 +2m1fm2f | = | 25 |
|
But we know from our kinetic energy equation that
25 = v1f2 + v2f2
.
Substituting this in we find that
2m1fm2f = 0
Thus we know that one of the final velocities must be zero. If the final
velocity of ball 2 were zero, then the collision never would have taken place.
Thus we can infer that
v1f = 0 and, consequently,
v2f = 5. This problem
states a general principle of collisions: when two bodies of the same mass
collide head on in an elastic collision, they exchange velocities.
