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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 5: MOMENTUM
CLASS NOTES
MOMENTUM
Momentum is the product of mass and velocity. It provides a way to describe and predict collisions and
other events:
p = mv
Momentum = mass[kg] x velocity[m/s]
CHANGE IN MOMENTUM
The change in momentum is the difference between final momentum and initial momentum:
p = pf - pi
FORCE-TIME GRAPHS
Another way to calculate the change in momentum p is to look at a force-time (Ft) graph. The change
in momentum is equal to the area underneath the curve.
FORCE-TIME GRAPHS
CONSERVATION OF MOMENTUM
If two particles of masses m1 and m2 form an isolated system, then the total momentum of the system
remains constant after a collision:
pi = pf
CONSERVATION OF MOMENTUM
In the case of a single object, the conservation of momentum can be written as:
m1v1 = m1’v1’
[(mass) x (initial velocity) =
(mass) x (final velocity)]
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 1. A railroad car of 1000 kg is rolling with a speed of 2.0 m/s when 2000 kg of coal is
dropped from a height of 1.0 m into the car. What is the final speed of the cart after the coal is in the
cart?
1A.
(1) m1v1 = m1’v1’
(2) v1’ = (m1v1) / (m1’)
(3) v1’ = [(1000 kg)(2 m/s)] / (3000 kg)
(4) v1’ = 0.67 m/s
Example 2 (“tricky” -- partial solution):
Two skaters, both of mass 50 kg, are at rest on skates on a frictionless ice pond. One skater throws a 0.2
kg Frisbee at 5 m/s to her friend, who catches it and throws it back at 5 m/s. When the first skater has
caught the returned Frisbee, what is the velocity of each of the two skaters?
2A.
(1) Define the x-axis: Motion to the right is positive, motion to the left is negative.
(2) First Skater’s Throw : Frisbee moves 5 m/s to the right.
(3) m1( - v1) = m2v2
(m1 = mass of skater, m2 = mass of Frisbee)
(4) - v1 = m2v2 / m1
(5) - v1 = (0.2 kg)(5 m/s) / 50 kg
(6) v1 = - 0.02 m/s
(7) v1 = 0.02 m/s to the left
MOMENTUM AND COLLISIONS
Momentum is transferred by collisions. After adjusting for friction, for any type of collision, the total
momentum before the collision equals the total momentum just after the collision.
ELASTIC COLLISIONS
In an elastic collision, objects bounce off each other. In a perfectly elastic collision, objects bounce off
each other without friction. Both momentum and kinetic energy are conserved:
m1v1 + m2v2 = m1v1’ + m2v2’
INELASTIC COLLISIONS
Because of friction, the observed final momentum does not equal the initial momentum of the system.
In a perfectly inelastic collision, the two colliding objects stick together following the collision:
m1v1 + m2v2 = (m1 + m2)V
Example 3. During a snowball fight, two balls, with masses of 0.4 kg and 0.6 kg respectively, are
thrown in such a manner that they meet head-on and combine to form a single mass. The magnitude of
initial velocity for each is 15 m/s. The final kinetic energy of the system just after the collision is what
percentage of kinetic energy just before the collision?
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3A.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) V = (m1v1 + m2v2 ) / (m1 + m2)
(3) V = [(0.4kg)(15m/s) + (0.6kg)(-15m/s)] / (0.4kg + 0.6kg)
(4) V = - 3 m/s
(5) KI = ½ m1(v1)2 + ½ m2(v2)2
(6) KI = ½ (0.4 kg)(15 m/s)2 + ½ (0.6 kg)(-15 m/s)2
(7) KI = 112.5 J
(8) KI = ½ m1(v1)2 + ½ m2(v2)2
(9) KF = ½ (0.4 kg + 0.6 kg)(- 3 m/s)2
(10) KF = 4.5 J
(11) % K = (KF / KI) x 100
(12) % K = (4.5 J / 112.5 J) x 100
(13) % K = 4%
Example 4. A 0.02 kg bullet is fired into, and becomes embedded in, a 1.7 kg wooden ballistic
pendulum. If the pendulum rises a vertical distance of 0.05 m, what is the initial speed of the bullet?
4A.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) m1v1 + 0 = (m1 + m2) V
(3) m1v1 = (m1 + m2) V
(4) v1 = (m1 + m2)(V) / m1
Determine value of V:
(5) Ug = K
(6) (m1 + m2)gh = ½ (m1 + m2)(V)2
(7) gh = ½ V2
(8) 2gh = V2
(9) V = (2gh)1/2
Substitute (2gh)1/2 for V:
(10) v1 = (m1 + m2)(2gh)1/2 / (m1) [From (4) above]
(11) v1 = (1.72 kg)[(2)(9.8 m/s2)(0.05 m)]1/2 / (0.02 kg)
(12) v1 = 85.1 m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
MOMENTUM IN TWO DIMENSIONS
Just like forces, momentum can occur in the x- and y- directions at the same time. In these cases, apply
the equation p = mv in both the x- (px) and y- (py) directions.
Example 5. A 1.00 kg duck decoy is flying overhead at 1.50 m/s when a hunter fires straight up. The
0.010 kg bullet is moving 100 m/s when it hits the decoy and stays lodged in the decoy’s body. What is
the momentum of the decoy and bullet immediately after the hit?
5A.
(1) px: m1v1 + m2v2 = (m1 + m2)Vx
(2) m1v1 + 0 = (m1 + m2)Vx
(3) m1v1 = (m1 + m2)Vx
(4) Vx = (m1v1) / (m1 + m2)
(5) Vx = [(1.00 kg)(1.50 m/s)] / (1.01 kg)
(6) Vx = 1.49 m/s
(7) py: m1v1 + m2v2 = (m1 + m2)Vy
(8) 0 + m2v2 = (m1 + m2)Vy
(9) m2v2 = (m1 + m2)Vy
(10) Vy = (m2v2) / (m1 + m2)
(11) Vy = [(0.01 kg)(100 m/s)] / (1.01 kg)
(12) Vy = 0.99 m/s
(13) V2 = Vx2 + Vy2
(14) V = (Vx2 + Vy2)1/2
(15) V = [(1.49 m/s)2 + (0.99 m/s)2]1/2
(16) V = 1.77 m/s
(17) p = (m1 + m2)V
(18) p = (1.01 kg)(1.77 m/s)
(19) p = 1.79 kg.m/s
IMPULSE AND MOMENTUM THEOREM
The time rate of change of the momentum of a particle is equal to the resultant force on the particle. The
impulse of a force J equals the change in momentum (p) of the particle on which the force acts:
JFt = p
Momentum change = Force [N] x time [s]
EXAMPLE OF IMPULSE
As per the equation Ft = p, car air bags lengthen the stopping time of the passenger. This reduces
the force applied during the stop and provides more safety.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Examples 6 - 7. During a collision, an impulse of 300 N.s is experienced by a car passenger.
6. Luckily, air bags are deployed, and the passenger comes to a complete stop in 3 seconds. Calculate
the average force experienced by the passenger.
6A.
(1) J = Ft
(2) F = J / t
(3) F = 300 N.s / 3 s
(4) F = 100 N (22.5 lbs)
7. Unfortunately, the air bags fail to deploy, and the seatbelts malfunction. The passenger collides
against the dash board in 0.1 seconds. Calculate the new average force experienced by the passenger.
7A.
(1) J = Ft
(2) F = J / t
(3) F = 300 N.s / 0.1 s
(4) F = 3000 N (674 lbs)
Example 8. A 0.6 kg tennis ball, initially moving at a speed of 12 m/s, is struck by a racket causing it to
rebound in the opposite direction at a speed of 18 m/s. A high speed movie film determines that the
racket and ball are in contact for 0.05 s. What is the average net force exerted on the ball by the racket?
8A.
(1) pi = mv
(2) pi = (0.6 kg)(12 m/s) = 7.2 kg.m/s
(3) pf = m(- v)
(4) pf = - (0.6 kg)(18 m/s) = - 10.8 kg.m/s
(5) p = Ft
(6) F = p / t
(7) p = pf – pi
(8) F = (pf – pi) / t
(9) F = (- 10.8 kg.m/s - 7.2 kg.m/s) / 0.05 s
(10) F = (- 18 kg.m/s) / 0.05 s
(11) F = - 360 N (opposite direction)
Example 9. (partial):
A fire hose directs a steady stream of 15 kg/sec of water with velocity 28 m/s against a flat plate. What
force is required to hold the plate in place?
9A.
(1) F = ma
(2) F = m(v/t)
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PHYSICS