Momentum & Impulse Objectives
Focus Questions for the Unit:
• How do objects affect each other when they interact?
You Should KNOW:
1.) When one object exerts a force on another, both objects experience changes in velocity.
2.) When two objects exert force on each other, it often occurs as a result of the two colliding with or
recoiling from each other.
3.) When objects interact, the length of time they interact determines the magnitude of their force on
each other.
4.) An impulse applied to an object will produce a change in that objects velocity in proportion to the
object’s mass, in other words, a change in momentum.
5.) When objects collide or recoil, there is a conservation of momentum.
You Should BE ABLE TO:
6.) Develop skills and protocols in selecting and using proper formulas that require calculating
momentum, impulse, force, final velocity, and time
7.) Explain the ways interacting objects behave, such as when a golf club and a golf ball interact.
8.) Drawing and/or interpret collision diagrams.
You Should UNDERSTAND:
9.) When objects exert a force on each other, they obey the law of conservation of momentum.
Definitions – Linear Momentum & Impulse:

Linear momentum (always given by the symbol p) is defined as the product of mass and velocity.
Momentum is something that an object has at a point in time.
o Momentum = mass x velocity
o 𝒑 = 𝑚𝑣
 The SI units for momentum is kg m/s, or N s can also be used.
 Since velocity is a vector, momentum is also a vector. Therefore momentum has direction and can
be positive or negative (+/-).
Example:
A 70-kg man is riding on a 10-kg bicycle with a velocity of 7.0 m/s North.
What is the combined momentum of the man and the bicycle?
𝒑 = 𝑚𝑣
𝑚
𝒑 = (70𝑘𝑔 + 10𝑘𝑔) (7.0 ) 𝑁𝑜𝑟𝑡ℎ
𝑠
𝑚
𝒑 = (80𝑘𝑔) (7.0 ) 𝑁𝑜𝑟𝑡ℎ
𝒑 = 560
𝑘𝑔 𝑚
𝑠
𝑠
𝑁𝑜𝑟𝑡ℎ
 Impulse (given by the symbol J, but sometimes given by I) is the change in an object’s momentum
(given by Δp). An impulse, or change in momentum, is caused by a Force acting on a mass over
some period of time. Impulse is equal to the product of force and time.
o Impulse = change in momentum = force x time

o 𝐽 = 𝐹𝑛𝑒𝑡 𝑡 = ∆𝑝 = 𝑝𝑓 − 𝑝𝑖
The SI units for impulse is the same as the units for momentum; kg m/s or N s. Impulse is a vector
and can be a change in an object’s momentum in the positive or negative direction.
Example:
A 0.15-kg baseball is thrown with a speed of 26 m/s. A batter hits the baseball
with a bat, causing the ball to come off the bat in the opposite direction with a
speed of 30 m/s. The contact time between the bat and ball was 0.7
milliseconds.
a) What was the force on the baseball from the bat?
𝐽 = 𝐹𝑛𝑒𝑡 𝑡 = ∆𝑝 = 𝑝𝑓 − 𝑝𝑖
𝐹𝑛𝑒𝑡 𝑡 = 𝑝𝑓 − 𝑝𝑖
Note: velocity can be
positive or negative
𝑚
𝑚
𝐹𝑛𝑒𝑡 (0.0007𝑠) = (0.15𝑘𝑔) (−30 ) − (0.15𝑘𝑔)(26 )
𝑠
𝑠
𝐹𝑛𝑒𝑡 =
𝑘𝑔𝑚
𝑘𝑔𝑚
−3.9
𝑠
𝑠
−4.5
0.0007𝑠
= −12000𝑁
b) What was the impulse on the baseball?
𝐽 = 𝑝𝑓 − 𝑝𝑖
𝑚
𝑚
𝐽 = (0.15𝑘𝑔) (−30 𝑠 ) − (0.15𝑘𝑔) (26 𝑠 )
𝐽 = −8.4
𝑘𝑔𝑚
𝑠
Conservation of Momentum:

The law of conservation of momentum states that the total linear momentum of a system of
interacting particles or objects remains constant provided there is no net external force.

To see why, start by imagining two isolated particles A and B that collide with one another.
o The force from A onto B, FAB will cause B’s momentum to change by some amount over a
period of time, Δt..
o The equal but opposite force from B onto A, FBA will cause A’s momentum to change by an
equal but opposite amount over the same period of time, Δt.
o This means that the total momentum (momentum of A plus the momentum of B) will remain
the same. Total momentum is conserved.
𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = 𝑝𝑎𝑓𝑡𝑒𝑟
𝑚𝐴 𝑣𝐴𝑖 + 𝑚𝐵 𝑣𝐵𝑖 = 𝑚𝐴 𝑣𝐴𝑓 + 𝑚𝐵 𝑣𝐵𝑓
Elastic Collisions:
 A collision in which no mechanical energy is lost is called an elastic collision.
 In reality, collisions between everyday objects always lose some energy – the only real example of a
perfectly elastic collision is between molecules. For an elastic collision, relative velocity of approach
always equals the relative velocity of separation.
 In elastic collisions, the particles collide and bounce off of each other.
𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = 𝑝𝑎𝑓𝑡𝑒𝑟
𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 = 𝒎𝟏 𝒗𝟏𝒇 + 𝒎𝟐 𝒗𝟐𝒇
Note: Momentum of
the system remains
the same.
Kinetic Energy of the
system ALSO
remains the same
and is conserved for
elastic collisions.
𝐾𝐸𝑏𝑒𝑓𝑜𝑟𝑒 = 𝐾𝐸𝑎𝑓𝑡𝑒𝑟
1
1
1
1
2
2
2
2
𝑚1 𝑣1𝑖
+ 𝑚2 𝑣2𝑖
= 𝑚1 𝑣1𝑓
+ 𝑚2 𝑣2𝑓
2
2
2
2
Inelastic Collisions:
 A collision in which the particles or objects stick together called an elastic collision.
 In reality, collisions between everyday objects always lose some energy – the only real example of a
perfectly elastic collision is between molecules. For an elastic collision, relative velocity of approach
always equals the relative velocity of separation.
 In elastic collisions, the particles collide and bounce off of each other.
𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = 𝑝𝑎𝑓𝑡𝑒𝑟
𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 = (𝒎𝟏 + 𝒎𝟐 )𝒗𝒇
Note: Momentum
is conserved, BUT
Kinetic Energy is not
conserved.
𝐾𝐸𝑏𝑒𝑓𝑜𝑟𝑒 = 𝐾𝐸𝑎𝑓𝑡𝑒𝑟 + 𝐸𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑
Some Kinetic Energy
is lost/transferred to
sound, heat, and
deformation of the
object.
1
1
1
2
2
𝐸𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = { 𝑚1 𝑣1𝑖
+ 𝑚2 𝑣2𝑖
} − { (𝑚1 + 𝑚2 )𝑣𝑓2 }
2
2
2
𝐸𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = 𝐾𝐸𝑏𝑒𝑓𝑜𝑟𝑒 − 𝐾𝐸𝑎𝑓𝑡𝑒𝑟
Practice Questions – Momentum & Impulse
1.) In the diagram, a 60.-kilogram rollerskater exerts a 10.-newton force
on a 30.-kilogram rollerskater for 0.20 second. What is the
magnitude of the impulse applied to the 30.-kilogram rollerskater?
(1) 50. N•s
(2) 6.0 N•s
(3) 2.0 N•s
(4) 12 N•s
2.) 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
(1) the same magnitude and the same direction
(2) the same magnitude and opposite direction
(3) smaller magnitude and the same direction
(4) larger magnitude and the same direction
3.) At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.-kilogram
cannon. What is the recoil speed of the cannon?
(1) 75 m/s
(2) 3.0 m/s
(3) 15 m/s
(4) 5.0 m/s
4.) A 2.0-kilogram laboratory cart is sliding across horizontal frictionless surface at a constant velocity
of 4.0 meters per second east. What will be cart’s velocity after a 6.0-newton westward force
acts on it for 2.0 seconds?
(1) 2.0 m/s east
(2) 10. m/s east
(3) 2.0 m/s west
(4) 10. m/s west
5.) A force of 6.0 N changes the momentum of a moving object by 3.0 kilogram•meters per second.
How long did the force act on the mass?
(1) 1.0 s
(2) 0.25 s
(3) 2.0 s
(4) 0.50 s
6.) A 3.0-kilogram steel block is at rest on a frictionless horizontal surface. A 1.0-kilogram lump of
clay is propelled horizontally at 6.0 meters per
second toward the block as shown in the
diagram. Upon collision, the clay and steel
block stick together and move to the right with a
speed of
(1) 1.5 m/s
(2) 2.0 m/s
(3) 3.0 m/s
(4) 6.0 m/s
7.) In the diagram below, 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?
8.) Which situation will produce the greatest change of momentum for a 1.0-kilogram cart?
(1) accelerating it from rest to 3.0 m/s
(2) accelerating it from 2.0 m/s to 4.0 m/s
(3) applying a net force of 5.0 N for 2.0 s
(4) applying a net force of 10.0 N for 0.5 s
9.) In the diagram, scaled vectors represent the momentum of each of two masses, A and B, sliding
toward each other on a frictionless, horizontal surface.
Which scaled vector best represents the momentum of the system after the masses collide?
10.) A 1000-kilogram car traveling due east at 15 meters per second is hit from behind and receives
a forward impulse of 6000 newton-seconds. Determine the magnitude of the car’s change in
momentum due to this impulse. Show all work including equation, substitution, and answer
with units.
B> Problems
11) A compact car, mass = 725 kg, is moving at 100. km/h.
a) What is the momentum of the car?
b) At what velocity must a larger car, mass = 2175 kg, be traveling in order to have the same
momentum?
12) A force of 6.0 N acts on a 3.0 kg object for 10.0 s.
a) What is the object's change in momentum?
b) What is the new velocity of the object if it started at rest?
13) An 1100. kg car moves at 22.0 m/s. Neglecting friction, what breaking force is needed to stop
the car in 20.0 s?
14) A net force of 2000. N acts on a rocket of mass 1000. kg. How long must the force be applied in
order to increase the
rocket's velocity from 10.0 m/s to 200.0 m/s?
15) A snow scooter has a mass of 250 kg. A constant force acts upon the scooter for 1.0 minute.
The scooter's velocity
changes from 6.0 m/s to 28 m/s.
a) What change in momentum occurs?
b) What is the magnitude of the force acting on the scooter?
16) A car weighing 15,680 N and moving at a speed of 20. m/s is slowed to a stop by a constant
force of 640 N.
a) What is the initial momentum of the car?
b) How long must the breaking force be applied to the car?
17) A constant force of 300. N acts on a 600.0 kg mass for 68 s. The initial velocity of the mass is
10.0 m/s. The direction of
the force and the velocity is the same. What is the final velocity of the mass?
18) A rocket of mass 2.0 x 104 kg starting from rest is accelerated by a net force of 1.5 x 105 N for
15.0 s. What is the final
velocity of the rocket?
19) A snowmobile has a mass of 250 kg. A constant force acts for 1.0 minute changing the velocity
from 28.0 m/s to 6.0 m/s.
What is the magnitude and direction of the force?
A> Conceptual Questions
1) Explain two ways a heavy truck and a person on a skateboard can have the same momentum.
2) In stopping an object, how does the time of impact effect the magnitude of the force used to stop
the object?
3) What is the relationship between an impulse on an object and the change in momentum of that
object?
4) For a constant force, if the time the force is applied to an object doubles, the impulse will
__________.
5) From question #4, by how much is the object’s momentum changed?
6) In a car crash, what is the advantage of an air bag in terms of impulse/change in momentum?
7) Why is the impulse greater on a falling object that bounces instead of an equally massive object
that doesn’t bounce?
8) Explain why a tossed egg can be caught with a sagging sheet but not a piece of plywood?
9) A bug and the windshield of a moving car collide. Indicate which of the following statements are
TRUE.
a. The impact force on the bug and the car are the same magnitude.
b. The impulse on the bug and the car are the same magnitude.
c. The change in momentum of the bug and the car are the same magnitude.
d. The change in velocity of the bug and the car are the same.
C> Conservation of Momentum:
1) A 95 kg fullback, running at 8.2 m/s, collides in midair with a 128 kg defensive tackle moving in
the opposite direction.
Both players end up with zero speed.
(a) What as the fullback’s momentum before the collision?
(b) What as the change in the fullback’s momentum?
(c) What was the change in the tackle’s momentum?
(d) What was the tackle’s original momentum?
(e) How fast was the tackle moving originally?
2) Marble A, mass 5.0 g, moves at a speed of 20.0 cm/s. It collides with a second marble, B, mass
10.0 g, moving at 10.0
cm/s in the same direction. After the collision, marble A continues with a speed of 8.0 cm/s in the
same direction.
(a) Calculate the marbles’ momenta before the collision.
(b) Calculate the momentum of marble A after the collision.
(c) Calculate the momentum of marble B after the collision.
(d) What is the speed of marble B after the collision?
3) A 0.105 kg puck moving at 48 m/s is caught by a stationary 75 kg goalie. With what speed does
the goalie with the puck
slide on the ice?
4) A 35.0 g bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block
and bullet fly off together
at 8.6 m/s. What was the original velocity of the bullet?
5) A 35.0 g bullet moving at 475 m/s strikes a 2.5 kg wooden block. The bullet passes through the
block leaving at 275 m/s.
If the block was originally at rest, with what velocity is the block moving after the bullet leaves?
6) A 0.50 kg ball traveling at 6.0 m/s collides head on with a 1.00 kg ball moving in the opposite
direction at 12.0 m/s. The
0.50 kg ball bounces back with a velocity of 14.0 m/s after the collision. What is the velocity
(magnitude & direction) of the
1.00 kg ball after the collision?
7) A green ball (m = 10.0 g) moving with a velocity of 20.0 cm/s catches up to and collides with a red
ball (m = 20.0 g) moving
along the same line with a velocity of 10.0 cm/s. After the collision, the green ball is still moving in its
original direction with
a velocity of only 8.0 cm/s. Determine the final velocity of the red ball.
8) A 700. kg car traveling at 20.0 m/s collides with a stationary 1400 kg truck. The two vehicles
interlock and travel together.
What is the final velocity of the car?
9) A 40.0 kg projectile leaves a 2000. kg launcher with a velocity of 800. m/s. What is the recoil
velocity of the launcher?
10) Upon launching, a model rocket expels 50.0 g of fuel from its exhaust at an average velocity of
600. m/s. If the rocket
moves upward at 7.5 m/s, what is the mass of the rocket?
11) Two campers dock a canoe. One camper (80.0 kg) steps onto the dock moving at 4.0 m/s. The
canoe and other camper
move backwards at 2.9 m/s. If the second camper’s mass is 75.0 kg, what is the mass of the canoe?