Question

Bob throws a sticky piece of gum at a stationary cardboard box. The gum has a mass of 0.003 kg, and the box has a mass of 0.05 kg. The gum hits the box with a speed of 9 m/s, and they stick together and slide forward.

a. What is the total momentum of the system before the collision?

b. What is the total momentum of the system after the collision?

c. What is the velocity of the gum/box combination after the collision?

Answer 1

Same concept

Answer 2

So a

Answer 3

p=9*0.003
p=0.027kg * m/s

Answer 4

p=9*0.003
p=0.027kg * m/s

Answer 5

yup

and for b) it's the same answer since momentum is conserved

Answer 6

for c) use the final momentum to solve for velocity

keep in mind: the mass you must use is the mass of the gum + box b/c they're stuck together

Answer 7

0.027=0.053(v)

Answer 8

v=0.51 m/s

Answer 9

awesome, that' s it

Answer 10

A projectile launcher has a mass of 3 kg. It fires a projectile of mass 0.08 kg horizontally at a speed of 300 m/s.

a. What is the total momentum of the system before the projectile is fired?








b. What is the total momentum of the system after the projectile is fired?







c. What is the recoil velocity of the launcher?

Answer 11

a.
p=mv
p=3*300=900kg* m/s

b.
p=mv
p=0.08*300=24 kg * m/s

Answer 12

I can see where you're going but you have to be very careful with how the question is phrased -

before the projectile is fired, nothing is moving - so what's the total momentum?

Answer 13

Oh 0

Answer 14

good

as a general rule, the total momentum of a system before and after should be the same no matter what is happening

so a and b are both 0 kg*m/s

Answer 15

Got it

Answer 16

for c) since the total momentum is 0, the final momentum of the projectile and the final momentum of the launcher are equal, but opposite (so one of them is negative)

first calculate the momentum of the projectile (let's just use your calculations from earlier):

p=mv
p=0.08*300=24 kg * m/s for the projectile

Answer 17

the launcher has the opposite momentum -24 kg m/s

so -24 = 3v
v = ?

Answer 18

-8

Answer 19

good, so the launcher is -8 m/s [this is negative since it's in the opposite direction of the bullet]

Answer 20

and that's it ~

Answer 21

Cool ! )

Answer 22

Use Hooke's law to solve the following problems.

a. A spring with k = 11 N/m is stretched 0.25 m. What force does the spring apply?

b. What is the spring constant of a rubber band that exerts a force of 35 N when it is stretched 0.1 m?


c. An elastic band has a spring constant of 55 N/m. It will break when the force reaches 2000 N. How far do you need to stretch the band for it to break?


d. A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. How far will the spring stretch if the mass is motionless?

Answer 23

a) F = -kx

F = ?

Answer 24

F = -kx
F=-11(0.25)
F = -2.75

Answer 25

N

Answer 26

good (it's negative since it's acting in the opposite direction of the stretch)

for b) same thing, this time solving for k [just take the positive value of k since spring constants are always positive]

Answer 27

wait so its -2.75N

Answer 28

for a yes

Answer 29

okay

Answer 30

F=kx
35=0.1(k)
K=350 m/s

Answer 31

N/m actually as the unit

Answer 32

awesome, 350 N/m = your answer for b)

for c) same formula, this time you're solving for x

Answer 33

F = kx

2000 = 55x

x= ?

Answer 34

x=36.36

Answer 35

awesome, so 36.36 m = answer for c

for d) d. A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. How far will the spring stretch if the mass is motionless?

F is equal to gravity (spring is hanging down vertically)

so F = kx

mg = kx

solve for x

Answer 36

still there? having a little trouble? :S

Answer 37

Sorry I was afk

Answer 38

F = kx

mg = kx

2*9.8=8(x)

19.6=8(x)
x=2.45 m/s

Answer 39

good ^^ however, x is a distance so just m not m/s

Answer 40

A spring with spring constant 40 N/m is compressed 0.1 m past its natural length. A mass of 0.5 kg is attached to the spring.

a. What is the elastic potential energy stored in the spring?

b. The spring is released. What is the speed of the mass as it reaches the natural length of the spring?

Answer 41

a) potential energy = U = (1/2)kx^2 = ?

Answer 42

U = (1/2)kx^2 = ?
(1/2)(40)(0.1)

Answer 43

2

Answer 44

missed the exponent

(1/2)(40)(0.1)^2 = ?

Answer 45

0.2

Answer 46

awesome, so a) = 0.2J

b) initial potential energy = final potential energy

so 0.2 = (1/2)mv^2 solve for v

Answer 47

0.2 = (1/2)mv^2 solve for v
0.2 = (1/2)0.5v^2
v=0.89

Answer 48

good, so 0.89 m/s

and that's it ~

Answer 49

A pendulum has a mass of 3 kg and is lifted to a height of 0.3 m. What is the maximum speed of the pendulum?

Answer 50

potential energy = U = mgh

all of the initial PE is converted to KE

so mgh = (1/2)mv^2

solve for v

Answer 51

mgh = (1/2)mv^2
3*9.8*0.3=(1/2)3v^2
v=2.42 N/m

Answer 52

good but v is velocity so m/s

Answer 53

Use the universal law of gravitation to solve the following problems.

a. Two cars of mass 3000 kg are 2 m apart. What is the force of gravity between them?

b. The force of gravity between a planet and its moon is 371 N. If the planet has a mass of 4 × 1022 kg and the moon has a mass of 5 × 105 kg, what is the distance between their centers?

c. The force of gravity between Tim and John is 1.85 × 10–9 N. If they are 18 m apart and Tim has a mass of 120 kg, what is John's mass?

Answer 54

G = (6.67 * 10^-11)

for a) solve for F

Answer 55

F=(6.67 * 10^-11)(3000/2^2)

Answer 56

LIke that

Answer 57

almost, there are two masses so it needs to be 3000*3000

Answer 58

F=(6.67 * 10^-11)(9000000/2^2)

Answer 59

yup then just chuck that into mathway or whatever you're using

Answer 60

1.650825 x 10^8

Answer 61

hm. that's actually not what I got (that's too big for two trucks anyway)

Answer 62

can I see what the input looks like on your screen?

Answer 63

that's really weird can you try entering

(6.67 * 10^-11)(3000*3000) * (1/4) = ?

Answer 64

Got the same thing

Answer 65

0.00015 for f though right?

Answer 66

yeah (I think the problem might be mathway not interpreting the exponent correctly)

Answer 67

force is N as usual

anyway for b)

b. The force of gravity between a planet and its moon is 371 N. If the planet has a mass of 4 × 10^22 kg and the moon has a mass of 5 × 10^5 kg, what is the distance between their centers?

same equation, this time we're finding r

Answer 68

still there?

Answer 69

Yess sorry sorry I was doing something

Answer 70

371=

Answer 71

would i combine the 2 masses

Answer 72

nah it's just m1 * m2

Answer 73

4 × 10^22* 5 x 10^5

Answer 74

371= (6.67*10^-11)*( 4 * 10^22 *5 * 10^5)/r^2 solve for r

Answer 75

r=6

Answer 76

don't forget the exponent

6 * 10^7 m = your ans

Answer 77

c. The force of gravity between Tim and John is 1.85 × 10–9 N. If they are 18 m apart and Tim has a mass of 120 kg, what is John's mass

same formula, this time you are solving for one of the masses

Answer 78

371= (120*18)

Answer 79

we don't have both masses here, just 1 mass so

120*m

18 is the distance (r)

Answer 80

371=120*m/18

Answer 81

like that

Answer 82

almost there

18 needs to be squared, and then G needs to be added to the formula

Answer 83

(also the force isn't 371 anymore)

Answer 84

1.85 × 10–9 =120*m/18^2

Answer 85

good, now solve for m.

Answer 86

Hm :S the software is interpreting m as meters


try this

1.85 × 10^–9=(6.67 * 10^-11)(120x)/18^2

Answer 87

wolframalpha can be really stupid if you use a variable other than x :S

Answer 88

74.89

Answer 89

good so 74.89 kg = ans