objectives
1. Identify different types of collisions.
2. Determine how much kinetic energy is lost in
perfectly inelastic collisions.
3. Compare conservation of momentum and
conservation of kinetic energy in perfectly
inelastic and elastic collisions.
4. Find the final velocity of an object in perfectly
inelastic and elastic collisions.
Types of collisions
Inelastic collision
elastic collision
Inelastic collision
elastic collision
Elastic and inelastic collisions
• An inelastic collision is a collision in which the
kinetic energy of the system of objects is not
conserved. It is transformed into other nonmechanical forms of energy such as heat
energy and sound energy.
• In an elastic collision, the two objects bounce
off each other, and kinetic energy is
conserved.
A special case of inelastic collision –
perfectly inelastic collision
• The two objects stick together and have the same
speed after collision
m1v1 + m2v2 = m1v1’ + m2v2’
Since the two objects stick together after the collision,
v1’ = v2’ = v
m1v1 + m2v2 = (m1+ m2)v’
Example 1
• Cart A (50. kg) approaches cart B (100. kg initially at rest) with
an initial velocity of 30. m/s. After the collision, cart A locks
together with cart B. both travels with what velocity?
Before collision
A
after collision
B
A
mA = 50 kg, vA = 30. m/s
mB = 100. kg, vB = 0
B
vA’ = vB’ = v’ = ?
mAvA + mBvB
=
(50.kg)(30.m/s) + (100.kg)(0) =
mAvA’ + mBvB’
(50.kg +100.kg)v’
v’ = 10. m/s
mass increases by 3 times (50 kg to 150 kg), speed decrease by 3 times (30
m/s to 10 m/s)
Example 2
• A railroad diesel engine coasting at 5.0 km/h runs into a
stationary flatcar. The diesel’s mass is 8,000. kg and the flatcar’s
mass is 2,000. kg. Assuming the cars couple together, how fast are
they moving after the collision?
Before collision
diesel
after collision
Flat car
diesel
Flat car
v’ = 4.0 km/h
mass increases by 1.25 times (8000 kg to 10000 kg), speed decrease by 1.25
times (50 m/s to 40 m/s)
Example 3 (pay attention to directions)
• Cart A (50. kg) moving with an initial velocity of 30. m/s
approaches cart B (100. kg moving with initial velocity of 20.
m/s towards cart A. The two carts lock together and move as
one. Calculate the magnitude and the direction of the final
velocity.
Before collision
A
after collision
B
A
v’ = 3.3 m/s to the left
B
Example 4
• A block of mass M initially at rest on a frictionless horizontal
surface is struck by a bullet of mass m moving with horizontal
velocity v. What is the velocity of the bullet-block system after
the bullet embeds itself in the block?
Before collision
after collision
mv + M(0)
(m + M)v’
mv + 0
=
mv / (M+m) =
(m + M)v’
v’
Example 5
•
1.
2.
3.
4.
A woman with horizontal velocity v1 jumps
off a dock into a stationary boat. After
landing in the boat, the woman and the boat
move with velocity v2. Compared to velocity
v1, velocity v2 has
the same magnitude and the same direction
the same magnitude and opposite direction
smaller magnitude and the same direction
larger magnitude and the same direction
Example 6 – Sample Problem 6E
• A 1850 kg luxury sedan stopped at a traffic light is
struck from the rear by a compact car with a mass of
975 kg. the two cars become entangle as a result of
the collision. If the compact car was moving at a
velocity of 220 m/s to the north before the collision,
what is the velocity of the entangled mass after the
collision?
7.59 m/s North
Class work
• Page 224 6E #1-5
1.
2.
3.
4.
5.
3.8 m/s S
1.8 m/s
4.27 m/s N
4.2 m/s Right
a. 3.0 kg b. 5.32 m/s
• Pp 234-235 #31-36, 44*, 45, 48
31.
32.
33.
34.
35.
44.
45.
48.
1 m/s
3.00 m/s
4.2 m/s
a. 1.80 m/s
a. 0.81 m/s E;
6.00 kg
23 m/s
14.5 m/s N
b. 2.16 x 104 J
b. 1400 J
Example 7 - Kinetic energy is not
constant in inelastic collisions
• Two clay balls collide head-on in a perfectly inelastic collision.
The first ball has a mass of 5. 00 kg and an initial velocity of
4.00 m/s to the right. The mass of the second ball is 0.250 kg.,
and it has an initial velocity of 3.00 m/s to the left. What is the
final velocity of the composite ball of clay after the collision?
What is the decrease in kinetic energy during the collision?
Class work
• Page 226 - 6F #1-3
1. a. 0.43 m/s W
2. a. 6.2 m/s S
3. a. 4.6 m/s S
b. 17 J
b. 3 J
b. 3.9 x 103 J
• Pp 234-235 #34-35, 49
34. a. 1.80 m/s
35. a. 0.81 m/s E;
49. a. 2.1 m/s E;
b. 2.16 x 104 J
b. 1400 J
b. 4.1 x 104 J
Elastic Collision
• In an elastic collision, two objects collide and return to their
original shapes with no change in total kinetic energy.
• Most collisions are neither elastic nor perfectly inelastic.
Elastic and perfectly inelastic collisions are limiting cases.
However, in this book, we assume that when objects bounces
off each other, the collision is elastic.
• Momentum and Kinetic energy remain constant elastic
collisions.
m 1 v1  m 2 v 2  m 1 v1 '  m 2 v 2 '
1
2
2
m 1 v1 
1
2
2
m 2v2 
1
2
m 1 v1 ' 
2
1
2
m 2v2 '
2
Example 8
• A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic
head-on collision with a 0.930 kg shooter marble moving to the left at
0.180 m/s. after the collision, the smaller marble moves to the left at
0.315 m/s. Assume that neither marble rotates before or after the
collision and that both marbles are moving on a frictionless surface.
What is the velocity of the 0.030 kg marble after the collision?
Confirm your answer by calculating the total kinetic energy before
and after the collision.
Another special case of collision - Explosions
• Total system momentum is conserved for collisions between
objects in an isolated system, there are no exceptions to this
law.
• This same principle of momentum conservation can be
applied to explosions.
Momentum before the
explosion is zero. so the
momentum after the
explosion is also zero.
m1v1 + m2v2 = m1v1’ + m2v2’
0 = m1v1’ + m2v2’
• Consider a homemade cannon.
p(before) = 0
•
•
•
•
p(after) = 0
p(cannon) +
p(ball) = 0
In the exploding cannon, total system momentum is
conserved.
The system consists of two objects - a cannon and a tennis
ball.
Before the explosion, the total momentum of the system is
zero since the cannon and the tennis ball located inside of it
are both at rest.
After the explosion, the total momentum of the system must
still be zero. If the ball acquires 50 units of forward
momentum, then the cannon acquires 50 units of backwards
momentum. The vector sum of the individual momentum of
the two objects is 0. Total system momentum is conserved.
Example 8
• A 2.0-kilogram toy cannon is at rest on a frictionless surface. A
remote triggering device causes a 0.005-kilogram projectile to
be fired from the cannon. Which equation describes this
system after the cannon is fired?
1. mass of cannon + mass of projectile = 0
2. speed of cannon + speed of projectile = 0
3. momentum of cannon + momentum of projectile = 0
4. velocity of cannon + velocity of projectile = 0
Example 9
• A 2-kilogram rifle initially at rest fires a 0.002kilogram bullet. As the bullet leaves the rifle
with a velocity of 500 meters per second,
what is the momentum of the rifle-bullet
system?
Example 10
• A 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon.
The cannon is at rest when it is ignited. Immediately after the
impulse of the explosion, a photogate timer measures the cannon to
recoil backwards a distance of 6.1 cm in 0.0218 seconds. Determine
the post-explosion speed of the cannon and of the tennis ball.
Given: Cannon:
m = 1.27 kg; d = 6.1 cm; t = 0.0218 s
Ball:
m = 56.2 g = 0.0562 kg
• The strategy for solving for the speed of the cannon is to recognize
that the cannon travels 6.1 cm at a constant speed in the 0.0218
seconds.
• vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded
• The strategy for solving for the post-explosion speed of the tennis
ball involves using momentum conservation principles. mball • vball =
- mcannon • vcannon
vball = 63.3 m/s
Example 11
• A 60. kg man standing on a stationary 40. kg boat throws a .20 kg
baseball with a velocity of 50. m/s. With what speed does the boat
move after the man throws the ball? Assume no friction between
the water and the boat.
v’(boat) = -0.1 m/s (in the opposite direction of the baseball)
Example 12
• In the diagram, a 100.-kilogram clown is fired
from a 500.-kilogram cannon. If the clown's
speed is 15 meters per second after the firing,
then what is the recoil speed (v) of the
cannon?
3.0 m/s