Experiment 9
Conservation of Linear Momentum - Collisions
In this experiment you will be introduced to the definition of linear momentum.
You will learn the difference between an elastic and an inelastic collision. You will
explore how to determine the amount of momentum before and after an elastic and an
inelastic collision. In order to study these effects you will use a motion sensor. For the
collision you will use the frictionless track and carts equipped with the necessary
attachments to undergo elastic and inelastic collisions.
Objectives:
1. Measure the velocities of the carts before and after the collisions.
2. Calculate the momentum and the kinetic energies of the cart before and after the
collisions.
3. In the case of inelastic collisions, you will verify the law of conservation of
momentum, viz. compare the momentum before and after the collision for each
cart based on the measured masses and velocities.
4. In the case of elastic collisions, you will verify the law of conservation of
momentum, viz. compare the momentum before and after the collision for each
cart based on the measured masses and velocities. You will also study the effect
of such a collision on the kinetic energy of each cart.
Hypothesis:
How does the momentum before an elastic and an inelastic collision compare to the
momentum after the collision?
If two objects with equal but opposite momenta collide head-on inelastically, what is
their shared velocity after the collision?
Theory:
If a mass m moves along a straight line with a velocity v, the linear momentum of the
mass is defined as p = mv. Momentum is a vector quantity and thus has a magnitude and
a direction.
Inelastic collisions:
Figure 1 shows two carts with mass m1 and m2 and with initial velocities v1i and v2i
respectively headed for a collision (v1i > v2i) in one dimension (i.e the motions before and
after the collision are along a single axis). The two carts form the system under
consideration that is isolated from other effects. We write the law of conservation of
linear momentum for this two-body system as the following:
1
After
Before
v1i
v2i
m1
m2
v1f
v2f
m1
m2
Figure 1
Total momentum before the collision = Total momentum after the collision
m1 v 1i + m2 v 2i = m1 v 1 f + m2 v 2 f
(1)
where v1f and v2f are the final velocities of the carts after the collision. If the initial
velocities of the carts is known and one final velocity is known then using the equation
above we can calculate the other final velocity.
After
Before
v1i
m1
v2i = 0
m2
V
m1
m2
Figure 2
In the case of a completely inelastic collision the two carts stick together after the
collision as seen in Figure 2. In this case we can let the second cart (m2) stay at rest (v1i =
0) while the first cart moves with an initial velocity v1i. After the collision the two masses
stick together and move with a common velocity V. Conservation of momentum requires
that:
m1v1i = ( m1 + m2 )V
(2)
m1v1i
m1 + m2
In the equation above, if we know the masses and the initial velocity, we can calculate the
velocity of the masses after the collision. As expected, this velocity will be less than the
V =
2
m1
must be less than unity. In the inelastic
m1 + m 2
collision studied above, the linear momentum is conserved.
initial velocity of the first mass since
Elastic collisions:
After
Before
v1i
v2i
m1
m2
v1f
v2f
m1
m2
Figure 3
Figure 3 shows two carts with mass m1 and m2 and with initial velocities v1i and v2i
respectively headed for a collision in one dimension (i.e the motions before and after the
collision are along a single axis). The two carts form the system under consideration that
is isolated from other effects. We write the law of conservation of linear momentum for
this two-body system as the following:
m1 v 1i + m2 v 2i = m1 v 1 f + m2 v 2 f
(3)
We write the law of conservation of energy for this two-body system as the following:
1
1
1
1
2
2
2
2
m1v1i + m2 v2i = m1v1 f + m2 v2 f
2
2
2
2
(4)
where v1f and v2f are the final velocities of the carts after the collision. Equations (3) and
(4) can be solved simultaneously to give the following two equations for the velocities
after the collision in terms of the velocities before the collision.
v1 f =
m1 − m2
2m2
v1i +
v2i
m1 + m2
m1 + m2
(5)
v2 f =
m − m1
2m1
v1i + 2
v 2i
m1 + m2
m1 + m2
(6)
3
Exercises:
1. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere
1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then
released from rest. After swinging down, it undergoes an elastic collision with
sphere 2 with mass m2 = 75 g. What is the velocity of sphere 1 just before the
collision (v1i)?
h1
m1
m2
h2
From conservation of energy we have:
1
2
m1v1i = m 1 gh1
2
v1i = 2 gh 1 = 2 * 9 .8 * 0 .080 = 1 .252 m / s
2. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere
1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then
released from rest. After swinging down, it undergoes an elastic collision with
sphere 2 with mass m2 = 75 g. What is the velocity of sphere 1 just after the
collision (v1f)?
Use Equation (5)
v1 f =
m1 − m2
0.030 − 0.075
v1i =
* 1.252 = −0.537m / s
m1 + m2
0.030 + 0.075
3. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere
1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then
released from rest. After swinging down, it undergoes an elastic collision with
sphere 2 with mass m2 = 75 g. What is the velocity of sphere 2 just after the
collision (v2f)?
4
Use Equation (6)
v2 f =
2m1
2 * 0.030
v1i =
* 1.252 = 0.715m / s
m1 + m2
0.030 + 0.075
4. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere
1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then
released from rest. After swinging down, it undergoes an elastic collision with
sphere 2 with mass m2 = 75 g. To what height h1’ does the sphere 1 swing to the
left after the collision?
Use the conservation of energy here to get:
1
2
m1v1 f
2
'
m1 gh1 =
'
h1 =
v1 f
2
=
2g
( −0.537) 2
= 0.0147m = 1.5cm
2 * 9.8
5. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere
1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then
released from rest. After swinging down, it undergoes an elastic collision with
sphere 2 with mass m2 = 75 g. To what height h2 does the sphere 2 swing to the
right after the collision?
Use the conservation of energy to get:
1
2
m2 v 2 f
2
m2 gh2 =
h2 =
v2 f
2
2g
=
(0.715) 2
= 0.0261m = 2.6cm
2 * 9.8
5
M
m
h
v
6. A ballistic pendulum is a device that was used to measure the speeds of bullets
before electronic timing devices were invented. The device consists of a large
block of wood of mass M = 5.4 kg, hanging from two long cords. A bullet of
mass m = 9.5 g is fired into the block, coming quickly to rest. The block+bullet
then swing upward, their center of mass rising a vertical distance h = 6.3 cm
before the pendulum comes momentarily to rest at the end of its arc. What was the
speed v of the bullet before it hit the block (v1i)?
Use the conservation of linear momentum:
m1v1i = ( m1 + m2 )V
v1i =
( m1 + m2 )V
m1
where m1 = m = 9.5g and m2 = M = 5.4 kg
The next step is to find V, the velocity of the block+bullet just after impact.
Since the mechanical energy (kinetic plus potential) is conserved after the
collision, we have:
1
( m + M )V 2 = ( m + M ) gh
2
V = 2 * g * h = 2 * 9.8 * 0.063 = 1.111m / s
Now we can find the speed of the bullet before it hit the block as follows
( m + m2 )V (0.0095 + 5.4)
v1i = 1
=
* 1.111 = 632.6m / s
m1
0.0095
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7. What is the linear momentum of an automobile with weight 1000 kg traveling at
60 km/hr?
p = mv = 1000 * 60 * 1000 / 3600 = 16666.67kg .m / s
8. Suppose that your mass is 80 kg. How fast would you have to run to have the
same momentum as a 1600 kg car moving at 1.2 km/hr?
m1v1 = m2 v 2
v1 =
5.5 m/s
1600 * 1.2
= 24km / hr
80
2.5 m/s
1.6 kg
V2f=?
2.4 kg
4.9 m/s
1.6 kg
BEFORE
2.4 kg
AFTER
9. Two blocks slide on a frictionless surface and collide. What is the velocity of the
1.6 kg block after the collision?
Since momentum is conserved we have:
m1v1i + m2 v2i = m1v1 f + m2 v 2 f
v1 f =
=
m1v1i + m2 v 2i − m2 v2 f
m1
1.6 * 5.5 + 2.4 * 2.5 − 2.4 * 4.9
= 1.9m / s
1.6
Now we calculate the kinetic energy of the system before the collision and after
the collision as follows:
7
Before :
After :
1
1
1
1
2
2
m1v1i + m2 v2i = * 1.6 * 5.52 + * 2.4 * 2.52 = 31.7 J
2
2
2
2
1
1
1
1
2
2
m1v1 f + m2 v 2 f = * 1.6 * 1.9 2 + * 2.4 * 4.9 2 = 31.7 J
2
2
2
2
10. Is the collision elastic?
Yes, the collision is not elastic as the kinetic energy before and after are the
same.
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