Mr. Surrette Van Nuys High School

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:

Dp = pf - pi

FORCE-TIME GRAPHS

Another way to calculate the change in momentum Dp 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)]

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?

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) DUg = 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

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- (Spx) and y- (Spy) 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) Spx: 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) Spy: 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 (Dp) of the particle on which the force acts:

J = FDt = Dp

Momentum change = Force [N] x time [s]

EXAMPLE OF IMPULSE

As per the equation FDt = Dp, car air bags lengthen the stopping time of the passenger. This reduces the force applied during the stop and provides more safety.

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 = FDt

(2) F = J / Dt

(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 = FDt

(2) F = J / Dt

(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) Dp = FDt

(6) F = Dp / Dt

(7) Dp = pf – pi

(8) F = (pf – pi) / Dt

(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(Dv/Dt)

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PHYSICS