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 |
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 |
0 = 4.60v1' + 3.86v2' (1) |
Second find the horizontal momentum, ph : |
m1v1 = m1v1'cos50° + m2v2'cos40° |
6.0 |
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. |