Mechanical Engineering Chapter 16 Problem The Masses The Surfacek Smooth Stationary The Spring Ismamb Unstretched And

Problem 16.41 The masses mA=mB. The surface

is smooth. At t=0,A is stationary, the spring is

unstretched, and Bis given a velocity v0toward the

right.

(a) In the subsequent motion, what is the velocity of

the common center of mass of Aand B?

(b) What are the velocities of Aand Bwhen the spring

is unstretched?

mA

k

mB

Problem 16.42 In Problem 16.41, mA=40 kg, mB=

30 kg, and k=400 N/m. The two masses are released

from rest on the smooth surface with the spring stretched

1 m. What are the magnitudes of the velocities of the

masses when the spring is unstretched?

274

Problem 16.43 A girl weighing 80 lb stands at rest on

a 325-lb ﬂoating platform. She starts running at 10 ft/s

relative to the platform and runs off the end. Neglect the

horizontal force exerted on the platform by the water.

(a) After she starts running, what is her velocity rela-

tive to the water?

(b) While she is running, what is the velocity of the

common center of mass of the girl and the platform

relative to the water? (See Active Example 16.4.)

Problem 16.44 Two railroad cars with weights WA=

120,000 lb and WB=70,000 lb collide and become cou-

pled together. Car Ais full, and car Bis half full, of

carbolic acid. When the cars collide, the acid in B sloshes

back and forth violently.

(a) Immediately after the impact, what is the velocity

of the common center of mass of the two cars?

(b) When the sloshing in B has subsided, what is the

velocity of the two cars?

2 ft/s

A1 ft/s

B

Solution:

vcenter of mass =(120,000 lb)(2 ft/s)+(70,000 lb)(1 ft/s)

Problem 16.45 Suppose that the railroad track in Prob-

lem 16.44 has a constant slope of 0.2 degrees upward

toward the right. If the cars are 6 ft apart at the instant

shown, what is the velocity of their common center of

mass immediately after they become coupled together?

32.2 ft/s2vcenter of mass

vcenter of mass =0.957 ft/s

Problem 16.46 The 400-kg satellite Straveling at

7 km/s is hit by a 1-kg meteor Mtraveling at 12 km/s.

The meteor is embedded in the satellite by the impact.

Determine the magnitude of the velocity of their

common center of mass after the impact and the angle β

between the path of the center of mass and the original

path of the satellite.

45

⬚

M

S7 km/s

b

Problem 16.47 The 400-kg satellite Straveling at

7 km/s is hit by a 1-kg meteor M. The meteor is

embedded in the satellite by the impact. What would

the magnitude of the velocity of the meteor need to

be to cause the angle βbetween the original path of

the satellite and the path of the center of mass of the

combined satellite and meteor after the impact to be

0.5◦? What is the magnitude of the velocity of the center

of mass after the impact?

vm=34.26 km/s : v=6.92 km/s.

7 km/s

β

= 0.5°

M

vM

276

c

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Problem 16.48 A 68-kg astronaut is initially station-

ary at the left side of an experiment module within an

orbiting space shuttle. The 105,000-kg shuttle’s center of

Solution: Consider the motion of the astronaut (A) and shuttle (S)

relative to a reference frame that is stationary with respect to their

common center of mass. During the astronaut’s motion,

Problem 16.49 An 80-lb boy sitting in a stationary

20-lb wagon wants to simulate rocket propulsion by

throwing bricks out of the wagon. Neglect horizontal

forces on the wagon’s wheels. If the boy has three bricks

weighing 10 lb each and throws them with a horizontal

velocity of 10 ft/s relative to the wagon, determine the

velocity he attains (a) if he throws the bricks one at a

time and (b) if he throws them all at once.

Solving,v

w1=0.769 ft/s.

Second brick:

(mB+mw+2mb)vw1=(mB+mw+mb)vw2+mbvb2,

vb2−vw2=−10.

Solving,v

w2=1.603 ft/s.

Third brick :

(mB+mw+mb)vw2=(mB+mw)vw3+mbvb3,

vb3−vw3=−10.

Problem 16.50 A catapult designed to throw a line to

ships in distress throws a 2-kg projectile. The mass of

the catapult is 36 kg, and it rests on a smooth surface.

It the velocity of the projectile relative to the earth as

it leaves the tube is 50 m/s at θ0=30◦relative to the

horizontal, what is the resulting velocity of the catapult

toward the left?

θ

0

278

Problem 16.51 The catapult, which has a mass of 36

as it leaves the tube is 50 m/s at θ0=30◦relative to the

horizontal. What is the resulting velocity of the catapult

toward the left?

Solution:

where

Problem 16.52 A bullet with a mass of 3.6 grams is

moving horizontally with velocity vand strikes a 5-kg

block of wood, becoming embedded in it. After the

impact, the bullet and block slide 24 mm across the ﬂoor.

The coefﬁcient of kinetic friction between the block and

the ﬂoor is µk=0.4. Determine the velocity v.

v

Solution: Momentum is conserved through the collision and then

Problem 16.53 A 0.12-ounce bullet hits a suspended

15-lb block of wood and becomes embedded in it. The

angle through which the wires supporting the block rotate

as a result of the impact is measured and determined to

be 7◦. What was the bullet’s velocity? 2 ft 7⬚

Problem 16.54 The overhead conveyor drops the 12-kg

package Ainto the 1.6-kg carton B. The package is tacky

and sticks to the bottom of the carton. If the coefﬁcient

of friction between the carton and the horizontal con-

veyor is µk=0.2, what distance does the carton slide

after impact?

1 m/s

A

26°

Let vbe the velocity of the combined object relative to the belt.

Use work and energy to determine the sliding distance d:

Problem 16.55 A 12,000-lb bus collides with a 2800-

lb car. The velocity of the bus before the collision

is vB=18i(ft/s) and the velocity of the car is vC=

33j(ft/s). The two vehicles become entangled and

remain together after the collision. The coefﬁcient of

kinetic friction between the vehicles’ tires and the road

is µk=0.6.

(a) What is the velocity of the common center of mass

of the two vehicles immediately after the collision?

(b) Determine the approximate ﬁnal position of the

common center of mass of the vehicles relative to

its position when the collision occurred. (Assume

that the tires skid, not roll, on the road.)

vC

y

x

vB

Solution:

(a) The collision (impulse – momentum).

280

Problem 16.56 The velocity of the 200-kg astronaut

Arelative to the space station is 40i+30j(mm/s). The

velocity of the 300-kg structural member Brelative to

the station is −20i+30j(mm/s). When they approach

each other, the astronaut grasps and clings to the struc-

tural member.

(a) What is the velocity of their common center of

mass when they arrive at the station?

(b) Determine the approximate position at which they

contact the station.

A

B

y

x

9 m

6 m

Solution:

Problem 16.57 The weights of the two objects are

WA=5 lb and WB=8 lb. Object Ais moving at vA=

2 ft/s and undergoes a perfectly elastic impact with the

stationary object B. Determine the velocities of the

objects after the impact.

vA

AB

Solution: Momentum is conserved and the coefﬁcient of restitu-

Problem 16.58 The weights of the two objects are

WA=5 lb and WB=8 lb. Object Ais moving at vA=

2 ft/s and undergoes a direct central impact with the sta-

tionary object B. The coefﬁcient of restitution is e=0.8.

Determine the velocities of the objects after the impact.

vA

AB

Solution: Momentum is conserved and the coefﬁcient of restitu-

Problem 16.59 The objects Aand Bwith velocities

vA=20 m/s and vB=4 m/s undergo a direct central

impact. Their masses are mA=8 kg and mB=12 kg.

After the impact, the object Bis moving to the right at

16 m/s. What is the coefﬁcient of restitution?

v

A

AB

vB

Problem 16.60 The 8-kg mass Aand the 12-kg mass

Bslide on the smooth horizontal bar with the velocities

shown. The coefﬁcient of restitution is e=0.2. Deter-

mine the velocities of the masses after they collide. (See

Active Example 16.5).

3 m/s2 m/s

AB

Solution: Momentum is conserved, and the coefﬁcient of restitu-

282

Problem 16.61 In a study of the effects of an accident

on simulated occupants, the 1900-lb car with velocity

vA=30 mi/h collides with the 2800-lb car with velocity

vB=20 mi/h. The coefﬁcient of restitution of the

impact is e=0.15. What are the velocities of the cars

immediately after the collision?

vAvB

Problem 16.62 In a study of the effects of an accident

on simulated occupants, the 1900-lb car with velocity

vA=30 mi/h collides with the 2800-lb car with velocity

vB=20 mi/h. The coefﬁcient of restitution of the impact

is e=0.15. The duration of the collision is 0.22 s. Deter-

mine the magnitude of the average acceleration to which

the occupants of each car are subjected.

vAvB

Solution: The velocities before the collision are converted into

Problem 16.63 The balls are of equal mass m. Balls

Band Care connected by an unstretched spring and are

stationary. Ball Amoves toward ball Bwith velocity

vA. The impact of Awith Bis perfectly elastic (e =1).

(a) What is the velocity of the common center of mass

of Band Cimmediately after the impact?

(b) What is the velocity of the common center of mass

of Band Cat time tafter the impact?

A

ABC

k

x=mxB+mxC

m+m=xB+xC

2,

so

dx

dt =1

2dxB

dt +dxC

dt .

Immediately after the impact dxB/dt =vAand dxC/dt =0, so

dx

dt =1

2vA.

Problem 16.64 In Problem 16.63, what is the maxi-

mum compressive force in the spring as a result of the

impact?

284

Problem 16.65* The balls are of equal mass m. Balls

Band Care connected by an unstretched spring and are

stationary. Ball Amoves toward ball Bwith velocity

vA. The impact of Awith Bis perfectly elastic (e =1).

Suppose that you interpret this as an impact between ball

Aand an “object” Dconsisting of the connected balls

Band C.

(a) What is the coefﬁcient of restitution of the impact

between Aand D?

(b) If you consider the total energy after the impact

to be the sum of the kinetic energies, 1

2m(v′)2

A+

1

2(2m)(v′

D)2, where v′

Dis the velocity of the center

of mass of Dafter the impact, how much energy

is “lost” as a result of the impact?

(c) How much energy is actually lost as a result of the

impact? (This problem is an interesting model for

one of the mechanisms for energy loss in impacts

between objects. The energy “loss” calculated in

part (b) is transformed into “internal energy” — the

vibrational motions of Band Crelative to their

common center of mass.)

Solution: See the solution of Problem 16.135. Just after the impact

of A and B, A is stationary and the center of mass D of B and C is

AB D C

Problem 16.66 Suppose that you investigate an acci-

dent in which a 3400-lb car Astruck a parked 1960-lb

car B. All four of car B’s wheels were locked, and skid

marks indicate that it slid 20 ft after the impact. If you

estimate the coefﬁcient of kinetic friction between B’s

tires and the road to be µk=0.8 and the coefﬁcient of

restitution of the impact to be e=0.2, what was A’s

velocity vAjust before the impact? (Assume that only

one impact occurred.)

AB

vA

Solution: We can use work-energy to ﬁnd the velocity of car B

just after the impact. Then we use conservation of momentum and the

Problem 16.67 When the player releases the ball from

rest at a height of 5 ft above the ﬂoor, it bounces to a

height of 3.5 ft. If he throws the ball downward, releas-

ing it at 3 ft above the ﬂoor, how fast would he need to

throw it so that it would bounce to a height of 12 ft?

17.94 ft/s =0.837

0.837 =

33.23 ft/s before it hits the ﬂoor. To ﬁnd the original velocity when it

286

Problem 16.68 The 0.45-kg soccer ball is 1 m above

the ground when it is kicked upward at 12 m/s. If the

coefﬁcient of restitution between the ball and the ground

is e=0.6, what maximum height above the ground does

the ball reach on its ﬁrst bounce?

12 m/s

1 m

Solution: We must ﬁrst ﬁnd the velocity with which the ball strikes

Post Impact:

Problem 16.69 The 0.45-kg soccer ball is stationary

just before it is kicked upward at 12 m/s. If the impact

lasts 0.02 s, what average force is exerted on the ball by

the player’s foot?

Problem 16.70 By making measurements directly

from the photograph of the bouncing golf ball, estimate

the coefﬁcient of restitution.

Solution: For impact on a stationary surface, the coefﬁcient of

Problem 16.71 If you throw the golf ball in Prob-

lem 16.70 horizontally at 2 ft/s and release it 4 ft above

the surface, what is the distance between the ﬁrst two

bounces?

Solution: The normal velocity at impact is vAn =−

√2g(4)=

Problem 16.72 In a forging operation, the 100-lb

weight is lifted into position 1 and released from rest. It

falls and strikes a workpiece in position 2. If the weight

is moving at 15 ft/s immediately before the impact and

the coefﬁcient of restitution is e=0.3, what is the

velocity of the weight immediately after impact?

1

288

Problem 16.73 The 100-lb weight is released from

rest in position 1. The spring constant is k=120 lb/ft,

and the springs are unstretched in position 2. If the

coefﬁcient of restitution of the impact of the weight

with the workpiece in position 2 is e=0.6, what is

the magnitude of the velocity of the weight immediately

after the impact? 16 in

2

1

Problem 16.74* A bioengineer studying helmet design

uses an experimental apparatus that launches a 2.4-kg

helmet containing a 2-kg model of the human head

against a rigid surface at 6 m/s. The head, suspended

within the helmet, is not immediately affected by the

impact of the helmet with the surface and continues to

move to the right at 6 m/s, so the head then undergoes

an impact with the helmet. If the coefﬁcient of restitu-

tion of the helmet’s impact with the surface is 0.85 and

the coefﬁcient of restitution of the subsequent impact

of the head with the helmet is 0.15, what is the velocity

of the head after its initial impact with the helmet?

6 m/s

Problem 16.75*

Solution: See the solution to Problem 16.74

Problem 16.76 Two small balls, each of 1-lb weight,

hang from strings of length L=3 ft. The left ball is

released from rest with θ=35◦. The coefﬁcient of resti-

tution of the impact is e=0.9. Through what maximum

angle does the right ball swing? L

L

m

u

2[32.2 ft/s2][3 ft] =33.2◦.

290

Problem 16.77 In Example 16.6, if the Apollo

command-service module approaches the Soyuz space-

craft with velocity 0.25i+0.04j+0.01k(m/s) and the

docking is successful, what is the velocity of the center

of mass of the combined vehicles afterward?

x

B

A

v

A

z

y

Problem 16.78 The 3-kg object Aand 8-kg object B

undergo an oblique central impact. The coefﬁcient of

restitution is e=0.8. Before the impact, vA=10i+

4j+8k(m/s) and vB=−2i−6j+5k(m/s). What are

the velocities of Aand Bafter the impact?

AB

y

x

Problem 16.79 A baseball bat (shown with the bat’s

axis perpendicular to the page) strikes a thrown base-

ball. Before their impact, the velocity of the baseball is

vb=132(cos 45◦i+cos 45◦j)(ft/s) and the velocity of

the bat is vB=60(−cos 45◦i−cos 45◦j)(ft/s). Neglect

the change in the velocity of the bat due to the direct

central impact. The coefﬁcient of restitution is e=0.2.

What is the ball’s velocity after the impact? Assume that

the baseball and the bat are moving horizontally. Does

the batter achieve a potential hit or a foul ball?

baseball

Bat

y

vb

vB

Problem 16.80 The cue gives the cue ball Aa velocity

parallel to the yaxis. The cue ball hits the eight ball

Band knocks it straight into the corner pocket. If the

magnitude of the velocity of the cue ball just before the

impact is 2 m/s and the coefﬁcient of restitution is e=1,

what are the velocity vectors of the two balls just after

the impact? (The balls are of equal mass.)

x

y

45°

A

B

292

Problem 16.81 In Problem 16.80, what are the veloc-

ity vectors of the two balls just after impact if the coef-

ﬁcient of restitution is e=0.9?