Ch. 7 Impulse and Momentum

Phys 131 Recitation

For Recitation Practice

Chapter 7

FOC: 1, 2, 10 & 15.

Problems: 4, 11, 12, 19 & 64.

FOC 1Section 7.1 The Impulse–Momentum Theorem

1. Two identical cars are traveling at the same speed. One is heading due east and the

other due north, as the drawing shows. Which statement is true regarding the kinetic

energies and momenta of the cars? (a) They both have the same kinetic energies and the

same momenta. (b) They have the same kinetic energies, but different momenta. (c) They

have different kinetic energies, but the same momenta. (d) They have different kinetic

energies and different momenta.

1. (b) Kinetic energy, , is a scalar quantity and is the same for

both cars. Momentum, is a vector quantity that has a magnitude

and a direction. The two cars have different directions, so they

have different momenta. 2 1 2 mv , mv

FOC 22. Six runners have the mass (in multiples of m0), speed (in multiples of v0), and direction

of travel that are indicated in the table. Which two runners have identical momenta? (a) B

and C (b) A and C (c) C and D (d) A and E (e) D and F

Runner Mass Speed Direction of Travel

A 102

m v0 Due north

B m0 v0 Due east

C m0 2v0 Due south

D 2m0 v0 Due west

E m0 1

02v Due north

F 2m0 2v0 Due west

2. (d) Momentum is a vector quantity that has a magnitude and a direction. The magnitudes(m0v0) and directions (due north) are the same for both runners.

FOC 10Section 7.2 The Principle of Conservation of Linear Momentum

10. As the text discusses, the conservation of linear momentum is applicable only when

the system of objects is an isolated system. Which of the systems listed below are isolated

systems?

1. A ball is dropped from the top of a building. The system is the ball.

2. A ball is dropped from the top of a building. The system is the ball and the earth.

3. A billiard ball collides with a stationary billiard ball on a frictionless pool table. The

system is the moving ball.

4. A car slides to a halt in an emergency. The system is the car.

5. A space probe is moving in deep space where gravitational and other forces are

negligible. The system is the space probe.

(a) Only 2 and 5 are isolated systems. (b) Only 1 and 3 are isolated systems. (c) Only 3

and 5 are isolated systems. (d) Only 4 and 5 are isolated systems. (e) Only 5 is an isolated

system.

10. (a) The net external force acting on the ball/earth system is

zero. The gravitational forces that the ball and earth exert on

each other are internal forces, or forces that the objects within

the system exert on each other. The space probe is also an

isolated system, since there are no external forces acting on it.

FOC 15Section 7.4 Collisions in Two Dimensions

15. Object 1 is moving along the x axis with an initial momentum of +16 kg m/s, where

the + sign indicates that it is moving to the right. As the drawing shows, object 1 collides

with a second object that is initially at rest. The collision is not head-on, so the objects move

off in different directions after the collision. The net external force acting on the two-object

system is zero. After the collision, object 1 has a momentum whose y component is −5 kg

m/s. What is the y component of the momentum of object 2 after the collision? (a) 0 kg

m/s (b) +16 kg m/s (c) +5 kg m/s (d) −16 kg m/s (e) The y component of the

momentum of object 2 cannot be determined.

Pr. 44. In a performance test, each of two cars takes 9.0 s to accelerate from rest to 27 m/s. Car A has a mass

of 1400 kg, and car B has a mass of 1900 kg. Find the net average force that acts on each car during the test.

___________________________________________________________________

4. REASONING The impulse-momentum theorem, as expressed in Equation 7.4, states

that the impulse acting on each car is equal to the final momentum of the car minus its

initial momentum:

SF( )Dt

Impulse

= mvf

Finalmomentum

- mv0

Initialmomentum

or SF =mv

f- mv

0

Dt

where F is the net average force that acts on the car, and t is the time interval during

which the force acts.

SOLUTION We assume that the velocity of each car points in the +x direction. The

net average force acting on each car is:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1400 kg 27 m/s 1400 kg 0 m/sCar A 4200 N

9.0 s

1900 kg 27 m/s 1900 kg 0 m/sCar B 5700 N

9.0 s

m m

t

m m

t

− + − = = = +

− + − = = = +

f 0

f 0

v vF

v vF

Pr. 11*11. A student (m = 63 kg) falls freely from rest and strikes the ground. During the collision with the

ground, he comes to rest in a time of 0.040 s. The average force exerted on him by the ground is +18 000 N,

where the upward direction is taken to be the positive direction. From what height did the student fall?

Assume that the only force acting on him during the collision is that due to the ground.

SOLUTION We begin with the energy conservation principle 210 02

mv mgh+ =

21f f2

mv mgh+ (Equation 6.9b) applied to the student’s fall to the ground. Falling from

rest implies v0 = 0 m/s, and the student’s final velocity is the impact velocity: vf =

vimpact. Thus, we have

0 + mgh0

= 12

mvimpact2 + mgh

f or g h

0- h

f( )H

= 12

vimpact2 or H =

vimpact2

2g

(1) For the student’s collision with the ground, the impulse-momentum theorem gives

ground f 0F t mv mv = − (Equation 7.4). The collision brings the student to rest, so we

know that vf = 0 m/s, and Equation 7.4 becomes ground 0F t mv = − . Solving for the

impact speed v0, we obtain

ground

0 impact

F tv v

m

= = − (2)

Substituting Equation (2) into Equation (1) yields

( ) ( ) ( )

( )( )

2

ground 2 22groundimpact

2 22

18 000 N 0.040 s6.7 m

2 2 2 2 9.80 m/s 63 kg

F t

F tv mH

g g gm

− + = = = = =

Pr. 12*12. A golf ball strikes a hard, smooth floor at an angle of 30.0 and, as the drawing shows, rebounds at

the same angle. The mass of the ball is 0.047 kg, and its speed is 45 m/s just before and after striking the

floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the

vertical component of the ball’s momentum changes during impact with the floor, and ignore the weight of

the ball.)

SOLUTION The following figures show the initial and final velocities of the golf ball.

If we take up as the positive direction, then the vertical components of the initial and

final velocities are, respectively, 0 cos 30.0v= −

0yv and f

cos 30.0v= + f y

v . Then,

from Equation 7.4 the impulse is

f 0( ) ( cos 30.0 ) – (– cos 30.0 )t m m v v = − = + fy 0y

F v v

Since v v

0= =

f45 m / s , the impulse applied to the golf ball by the floor is

02 cos 30.0 2(0.047 kg)(45 m/s)(cos 30.0 ) 3.7 N st mv = = = F

Pr. 1919. ssm A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 230 kg) that is

also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +3.6 m/s relative to the

shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance

between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off?

(b) Determine the velocity of the second log if the lumberjack comes to rest on it.

SOLUTION

a. The total linear momentum of the system before the lumberjack begins to move is

zero, since all parts of the system are at rest. Momentum conservation requires that

the total momentum remains zero during the motion of the lumberjack.

m1v

f1+ m

2v

f2

Total momentumjust before the jump

= 0

Initial momentum

Here the subscripts "1" and "2" refer to the first log and lumberjack, respectively. Let

the direction of motion of the lumberjack be the positive direction. Then, solving for

v1f gives

vm v

mf1

f2 (98 kg)(+3.6 m / s)

230 kg–1.5 m / s= = =– –2

1

The minus sign indicates that the first log recoils as the lumberjack jumps off.

b. Now the system is composed of the lumberjack, just before he lands on the second log,

and the second log. Gravity acts on the system, but for the short time under

consideration while the lumberjack lands, the effects of gravity in changing the linear

momentum of the system are negligible. Therefore, to a very good approximation, we

can say that the linear momentum of the system is very nearly conserved. In this case,

the initial momentum is not zero as it was in part (a); rather the initial momentum of

the system is the momentum of the lumberjack just before he lands on the second log.

Therefore,

m v m v m v m v1 2 1 2f1 f2

Total momentumjust after lumberjack lands

01 02

Initial momentum

+ = +

In this expression, the subscripts "1" and "2" now represent the second log and

lumberjack, respectively. Since the second log is initially at rest, v01= 0 . Furthermore,

since the lumberjack and the second log move with a common velocity, v v vf1 f2 f

= =

. The statement of momentum conservation then becomes

m v m v m v1 2 2f f 02

+ =

Solving for vf, we have

vm v

m mf

02 (98 kg)(+3.6 m / s)

230 kg 98 kg+ 1.1 m / s=

+=

+=2

1 2

The positive sign indicates that the system moves in the same direction as the original

direction of the lumberjack's motion.

Pr. 64*64. The drawing shows a human figure in a sitting position. For purposes of this problem, there

are three parts to the figure, and the center of mass of each one is shown in the drawing. These parts are: (1)

the torso, neck, and head (total mass = 41 kg) with a center of mass located on the y axis at a point 0.39 m

above the origin, (2) the upper legs (mass = 17 kg) with a center of mass located on the x axis at a point 0.17

m to the right of the origin, and (3) the lower legs and feet (total mass = 9.9 kg) with a center of mass

located 0.43 m to the right of and 0.26 m below the origin. Find the x and y coordinates of the center of mass

of the human figure. Note that the mass of the arms and hands (approximately 12% of the whole-body mass)

has been ignored to simplify the drawing.

Pr. 64 Solution64. REASONING The mass of each part of

the seated human figure will be treated as

if it were all located at the corresponding

center of mass point. See the drawing

given in the text. In effect, then, the

problem deals with the three-particle

system shown in the drawing at the right.

To determine the x and y coordinates of

the center of mass of this system, we will

employ equations analogous to Equation

7.10. The values for the masses are

m1 = 41 kg, m2 = 17 kg, and m3 = 9.9 kg.

SOLUTION The x and y coordinates of the center of mass for the three-particle

system in the drawing are

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

1 1 2 2 3 3cm

1 2 3

1 1 2 2 3 3cm

1 2 3

41 kg 0 m 17 kg 0.17 m 9.9 kg 0.43 m0.11 m

41 kg 17 kg 9.9 kg

41 kg 0.39 m 17 kg 0 m 9.9 kg 0.26 m0.20 m

41 kg 17 kg 9.9 kg

m x m x m xx

m m m

m y m y m yy

m m m

+ + + += = =

+ + + +

+ + + + −= = =

+ + + +