| 1) A ball with a mass of 5.0 kg moves to the right with a speed of 4.0
m/s and collides with a stationary ball whose mass |
| 5.0 kg. The collision is not head-on and after
the collision the first ball moves in a direction of 45° to the left
of its |
| original path with a velocity of 2.8 m/s. The
second ball moves in a direction of 45° below the horizontal. Determine |
| the momentum and velocity of each ball after
the collision. |
| |
| (a) m1 =
5.0 kg m2
= 5.0 kg |
|
v1 = 4.0 m/s v2
= 0 |
| θ1
= 45°
θ2 = 45° |
| v1'
= 2.8 m/s |
| |
| The conservation
of momentum still applies as indicated by equations (1) and (2). However,
the collision |
| takes place in
two dimensions (x and y) so vector addition must be used. It may
be easier to visualize the |
| the solution
by entering these values into the simulator and focus on the green vectors
for both the red and |
| blue ball. The
diagrams can be thought of as either velocity or momentum diagrams. When
comparing the |
| answers from the simulator
to those you work out, beware there are inherent
rounding errors that occur |
| due
to the difference in floating point processors. |
| |
| Δp
= 0 (1) |
| pi
= pf (2) |
| |
| Referencing the
simulator, the initial momentum of the system is defined by the red ball
as it is the only |
| ball moving. |
| |
| pi
= m1v1 =
5.0 kg • 4.0 m/s = 20. kg•m/s |
| |
| After the collision,
you must consider the horizontal and vertical components of momentum of
both balls. |
| |
| The vertical momentum
of the red ball is given by: |
|
|
| pv1'
= m1v1sinθ1 = 5.0 kg • 4.0
m/s • 0.707 = 14 kg•m/s |
| |
| Similarly, the
vertical momentum of the blue ball after the collision is -14 kg•m/s.
Keep in mind that |
| momentum is a
vector and if up is positive then down must be negative. This is good
news because if the |
| initial vertical
momentum of the system is zero, so must the final vertical momentum be
equal to zero. |
| |
| The horizontal
momentum of the red ball after the collision is given by: |
| |
| ph1'
= m1v1'cosθ1 = 5.0 kg • 4.0
m/s • 0.707 = 14 kg•m/s |
| |
| Because of symmetry
considerations, masses and angles, the momentum of the blue ball is also
14 kg•m/s. |
| |
| The velocity of
each ball is given by: |
| |
| pv1'
= m1v1' |
| |
v1'
= pv1' / m1 = 14 kg•m/s / 5.0 kg
= 2.8 m/s |
| |
| Because of symmetry
considerations, the velocity of the blue ball is also 2.8 m/s. Don't you
wish that |
| everything was
so symmetrical? |
| |
| 2) A ball with a mass of 6.0 kg moves to the right with a speed of 3.0
m/s and collides with a stationary ball whose mass |
| is 6.0 kg. After the collision, the first ball
moves in a direction of 50° above the horizontal with a velocity of
2.8 m/s |
| and the second ball moves in a direction of 40°
below the horizontal. Determine the momentum and velocity of each |
| ball after the collision. |
| |
| (a) m1 =
6.0 kg
m2 = 6.0 kg |
|
v1 = 3.0 m/s
v2 = 0 |
|
v1' = ? v2'
= ? |
|
θ1 = 50°
θ2 = 40° |
| |
| Δp
= 0 |
| pi
= pf |
| |
| First find the
vertical momentum, pv : |
| |
| m1v1
= m1v1'sin50° + m2v2'sin40° |
0 = 6.0 kg
• v1' • 0.766 + 6.0
kg • v2' • 0.643 |
| 0 = 4.60v1'
+ 3.86v2' (1) |
|
|
| Second find
the horizontal momentum, ph : |
| |
| m1v1
= m1v1'cos50° + m2v2'cos40° |
6.0 kg
• 3.0 m/s = 6.0 kg• v1'
• 0.643 + 6.0 kg • v2'
• 0.766 |
| 18 m/s = 3.86v1'
+ 4.60v2'
(2) |
| |
| Not bad, two equations
with two unknowns, so we will solve equation (1) above in terms of v2': |
| |
| v1'
= -0.839 • v2' |
| |
| Now, all you have
to do substitute v1' into equation (2) and solve for v2': |
| |
| 18 m/s = -3.24v2'
+ 4.60v2'
|
| m2v2'
= 18 kg•m/s / 1.36 = 13 kg•m/s |
| v2'
= 13 kg•m/s / 6.0 kg = 2.2 m/s |
| |
| Substitute
v2' into equation (2) and solve for v1': |
| |
| 18 m/s = 3.86v1'
+ 4.60 • 2.2 m/s |
| v1'
= 2.0 m/s |
| m1v1'
= 6.0 kg • 2.0 m/s = 12 kg•m/s |
| |
| Problem 2 was
solved somewhat different
than problem 1. This was to show that sometimes there is more |
| than one way to
solve a problem. An important consideration in this problem is that an
elastic collision |
| was not
assumed. Remember a couple of ideas: An isolated system is one in which
there is no net force |
| acting on it and
the conservation of momentum always applies for such a system.
The kinetic energy of an |
| isolated system
need not be conserved. |