Note 34 Multiple Processes
Most of the processes that you observed in the real world require more than one step to analyze.
Now that you have the primary tools, Newton’s second law, the work-energy theorem, and the
impulse-momentum theorem, we will look situations where more than one of these tool is
required.
Ballistic Pendulum
The ballistic pendulum is a machine that is used to measure a projectile’s launch speed in the
horizontal direction. It looks like this.
Process 1. The Launch
The process starts with the launching of the bearing. A bearing of mass, m, is launched. At the
muzzle of the launcher, the speed of the bearing is called vlaunch. This is the value we are
interested in measuring. We know nothing about the spring so we can’t just calculate the launch
speed.
v=0
vlaunch
launcher
pendulum
Process 2. The Catch
Next, the bearing is caught by the bob of the pendulum. This results in the combined object of the
bearing plus the pendulum bob traveling at a reduced speed vpendulum.
vlaunch
vpendulum
launcher
During the collision between bearing and the pendulum, there is a multitude of forces on the
bearing from all of the different surfaces of the pendulum. We don’t know the forces on the
bearing. All that we can say is that energy is lost since we can hear the energy coming out of the
system as sound and you can feel it in the vibration of the apparatus too.
Therefore, this is an inelastic collision whose energy loss is unknown. This means we need to use
the momentum approach and collect both the bearing and the bob into a system so that we don’t
need to know the internal forces.
There are external forces on the system. They are gravity and tension. It happens that they act in
opposite directions during this process so that the external net force is zero. Thus, momentum of
the system is conserved. The initial and the final momentum are the same.
pi = p f
⇒ mvlaunch = (M + m)v pendulum
page 1
Process 3. The Swing
The last process is the pendulum swinging up to the highest position along a circular arc.
v=0
vpendulum
launcher
h
The acceleration is not constant since the motion is circular. The momentum of the system
changes as the pendulum stops. This leaves the energy approach.
For the energy, two forces act on the system, gravity and tension. Gravity does conservative work
and tension does zero work as it is always perpendicular to the direction of motion. Thus, the
mechanical energy is conserved.
ΔK + ΔU = 0
⇒ K i +U i = K f +U f
1
2
(M + m)v pendulum
+ 0 = 0 + (M + m)gh
2
Analysis
Combining the two equations from processes 2 and 3, we can arrive at this by substituting out the
intermediate pendulum speed, vpendulum.
vlaunch =
M +m
2gh
m
If we are measuring the angle rotated by the pendulum rather than the height traveled by the
pendulum, then we have to convert the height to the angle. The conversion goes through the
length of the pendulum, L.
h = L − L cos θ = L(1 − cos θ)
Together, we have this for the launch speed in terms of the angle traveled.
vlaunch =
M +m
2gL(1 − cos θ) m
page 2
Example
A block of mass M is pushed toward a second mass of 0.50 kg on a frictionless surface. It is given
a speed of 1 m/s. This block collides with a second, stationary block elastically. After the collision,
the second block slides onto a rough surface with a coefficient of kinetic friction with the block of
0.30. It stops after sliding a distance of 0.60 meter.
M
0.4 kg
1 m/s
0 m/s
0.60 m
µk = 0.30
What is the mass M?
This is a two-process situation. The first is an elastic collision. The second is just an object
slowed by friction. In the first process, since we don’t know the force between the masses, we
have to resort to using the momentum approach for a system. In the second process, any method
will work. Since we are given the displacement of the second block, the more efficient step to take
is to use the energy approach. The Newton’s second law approach will work too.
The Collision
This collision is specifically a mass M colliding with another mass m that starts out not moving.
M
0.4 kg
1 m/s
0 m/s
v1,f
v2,f
There is a shortcut formula for this. We need only the speed of the second mass for the next
process.
⎛ 2m1 ⎞⎟
⎛ 2M ⎞⎟
2M
⎟⎟vi,1 = ⎜⎜
v f ,2 = ⎜⎜
⎟⎟1 =
⎜⎝ m + m ⎟⎠
⎜⎝ M + 0.4 ⎟⎠
M + 0.4
1
2
The Slide
With this speed, the second block slides to a stop.
v2,f
0.4 kg
0.4 kg
0 m/s
0.60 m
µk = 0.30
Applying the work-energy theorem,
Wnc = ΔK + ΔU
⇒ Wfriction = −K i
⇒
1 2
− µkmgΔx = − mv 2,f
2
page 3
2
2µk gΔx = v
2
2,f
⇒
⎛ 2M ⎞⎟
2µk gΔx = ⎜⎜
⎟
⎜⎝ M + 0.4 ⎟⎟⎠
2M
= 1.8783 ⇒
M + 0.4
⇒
2M
=
M + 0.4
2M = 1.8783M + 0.75132 ⇒
2µk gΔx
0.1217M = 0.75132
M = 6.17 kg
page 4