Question

a 3.0 kg mass moving to the right at 1.4 m/s collides in a perfectly inelastic collision with a 2.0 kg mass initially at rest. what will the vleocity of the combined mass be after collision? show your work.

Answer 1

@airjordan200

Answer 2

Do you have any idea how to work this problem?

Answer 3

not really

Answer 4

The total momentum of the moving mass (before collision) and the stationary mass will be equal to the momentum of the combined mass (after collision).

Answer 5

?

Answer 6

What is the momentum of the moving mass before the collision?

Answer 7

1.4 m/s

Answer 8

No, that's the velocity. What is the momentum?

Answer 9

3.0 kg to the right?

Answer 10

im really horrible at physics. math and history are my strong points

Answer 11

if you're good at math, physics should be fun :-)

\[p=mv\]\(p\) is momentum, \(m\) is mass, \(v\) is velocity

Answer 12

\(3.0 \text{ kg}\) is the mass moving to the right, with velocity \(1.4 \text { m/s}\)

momentum of the moving object (before collision) is thus
\[p=mv = (3.0\text{ kg})(1.4\text{ m/s}) = \]

Answer 13

so 4.2 m/s?

Answer 14

No, did you keep all of the stuff you multiplied together? What happened to the \(\text{kg}\)?

Answer 15

4.2kg?

Answer 16

treat units exactly like algebraic variables

Answer 17

4.2kg m/s lol

Answer 18

bingo!

okay, what about the momentum of the stationary object? What is that?

Answer 19

2.0kg with no velocity

Answer 20

well, okay, the mass is \(2.0 \text{ kg}\) and velocity is \(0 \text{ m/s}\), what is the momentum? \[p=mv\]

Answer 21

2.0kg m/s
so 4.2kg m/s *2.0kg m/s?

Answer 22

8.4kg m/s

Answer 23

\[m =2.0\text{ kg}\]\[v = 0\text{ m/s}\]\[\text{momentum} = p = mv = \]

Answer 24

sso i was wrong?

Answer 25

yes, in many ways :-)

For starters, \[2.0 \text{ kg m/s} * 4.2\text{ kg m/s} = 2.0*4.2*\text{kg}*\text{kg}*\text{m}*\text{m}/(\text{s}*\text{s}) = 8.4 \text{kg}^2\text{m}^2/\text{s}^2\]

Answer 26

remember, treat units just like you would \(x\) when doing algebra.

Answer 27

oooh
o forgot the units
that explains a lot of stuff

Answer 28

\[4x*2x=8x^2\]

Answer 29

so is 8.4kg^2 m^2/s^2 the answer?

Answer 30

and if you take the units along through the calculation, they provide a valuable error check on unit conversions, formula setup, etc. For example, the units of momentum have the dimension [mass][length]/[time], for example, \(\text{kg m}/\text{s}\) but your answer has different dimensions, does it not? That is a big red flag that something is rotten in Denmark, so to speak.

Answer 31

As I said, momentum (\(p\)) is given by
\[p = mv\]and we have \(m = 2.0 \text{ kg}\) and \(v = 0\text{ m/s}\)
so the momentum of the stationary object (before the collision) is
\[p = m v = (2.0\text{ kg})(0 \text{ m/s}) = \]

Answer 32

so what is the bloody answer then?

Answer 33

im more confused than when i first asked

Answer 34

lol don't worry about the answer, worry about knowing how to get there

Answer 35

most of the time answer is only worth 2 of 10 points anyways...

Answer 36

I'm trying to get you to do it :-)

Look, you find the momentum of all of the pieces before the collision. The momentum of an object is simply the mass * velocity. We found the momentum of the moving object. Now you need to find the momentum of the stationary object. I did everything except punch the buttons for you...

\[p = mv = (2.0\text{ kg})(0 \text{ m/s}) = \]

Answer 37

You know how to multiply by 0, don't you?

Answer 38

the question is worth 3 points

Answer 39

so the answer is 0?
oh my god, this is confusing!

Answer 40

@Comrad , @whpalmer4 is trying to guide you thru the process of learning how to do this problem... You seem to be a little too eager to get the answer.

Answer 41

Yes, the momentum of the stationary object is 0. Objects at rest have no momentum.

So, the total momentum of the system before collision is the sum of the momentums of the moving object and the stationary object.

\[4.2\text{ kg m/s} + 0 \text{ kg m/s} = \]

Answer 42

0kg m2/s^2

Answer 43

no.... :-(

Let's try something else.

\[4.2x + 0x = \]

Answer 44

0x so 0
there is no value that could make the answer something else other than 0.

Answer 45

no value for x should i say

Answer 46

No, \[4.2x + 0x = (4.2+0) x = 4.2x\]!!!

Answer 47

4.2 miles + 0 miles = how many miles? 4.2 miles, or 0 miles?

Answer 48

my bad. the plus came through as a multiplication symbol. sorry!!!!

Answer 49

4.2 miles

Answer 50

*facepalm*

Answer 51

Okay, so our total momentum is
\[4.2 \text{ kg m/s} + 0 \text{ kg m/s} = 4.2 \text{ kg m/s}\]

And after the collision, the two objects are now one object, with that same momentum, and the mass of the object is \[3.0\text{ kg}+2.0\text{ kg} = 5.0\text{ kg}\]

Answer 52

Using \[p = mv\]find \(v\) such that \(p = mv\) is true when \(m = 5.0\text{ kg}\) and \(p = 4.2 \text{ kg m/s}\)

Answer 53

\[p = mv\]\[4.2 \text{ kg m/s} = (5.0\text{ kg}) v\]

Answer 54

.84kg m/s^2

Answer 55

You got the number right, the units are wrong, which means the answer is wrong.

\[4.2 \text{ kg m/s} = (5.0 \text{ kg}) v\]
\[v = \frac{4.2}{5.0} \frac{\text{ kg m/s}}{\text{kg}} = 0.84 \frac{\cancel{\text{ kg}}\text{ m/s}}{\cancel{\text{kg}}} = 0.84 \text{ m/s}\]

Answer 56

k

Answer 57

so it only square when multiplying 2 velocities?

Answer 58

*squared

Answer 59

Here's what we did. We had two objects, with masses \(m_1 = 3.0 \text{ kg}\) and \(m_2 = 2.0 \text{ kg}\) and velocities before collision of \(v_1 = 1.4\text{ m/s}\) and \(v_2 = 0\text{ m/s}\)

momentum is conserved in an inelastic collision, so:

\[p_1 + p_2 = p_{final}\]\[m_1*v_1 + m_2*v_2 = (m_1+m_2)v_{final}\]
We just plugged in the numbers and solved for \(v_{final}\)


Yes. Treat units exactly like you would algebraic variables. \[2x+3x = 5x\]\[2x*3x=6x^2\]

Answer 60

Because our \(v_2=0\), our equation simplified to
\[m_1*v_1 = (m_1+m_2)v_{final}\]

You can see that if the mass of the stationary object is much larger, the collision will essentially result in the first object stopping and not much else. Think about firing a rifle bullet into the ground. The Earth doesn't exactly get knocked out of its orbit :-)

If the situation is reversed, where a bit object hits a small one, think dump truck hitting some poor schmuck on his bicycle, the big object continues along pretty much at the same speed, with a new hood ornament. Ouch.

Answer 61

ok so can you put all of our equations and answers on 1 things? its kinda confusing

Answer 62

i hate physics

Answer 63

Okay.

Moving object:
\[m_1 = 3.0 \text{ kg}\]\[v_1 = 1.4\text{ m/s}\]
Stationary object:
\[m_2 = 2.0 \text{ kg}\]\[v_2 = 0\text{ m/s}\]

Combined objects after collision:
\[m_{final} = m_1+m_2\]

\[p_1 + p_2 = p_{final}\]\[m_1 * v_1 + m_2*v_2 = (m_1+m_2)v_{final}\]\[v_{final} = \frac{m_1*v_1 + m_2*v_2}{m_1+m_2}\]\[v_{final} = \frac{(3.0\text{ kg})(1.4\text{ m/s}) + (2.0\text{ kg})(0\text{ m/s})}{(3.0\text{ kg}+2.0\text{ kg})} = \frac{4.2\text{ kg m/s}}{5.0\text{ kg}}\]\[v_{final} = 0.84\text{ m/s}\]

Answer 64

so thats the final answer? this is just the most confusing thing ever? i mean it makes since mathematicaly but i still dont get it

Answer 65

There ya go!

Answer 66

"what will the vleocity of the combined mass be after collision?"
the velocity of the combined mass after collision will be 0.84 m/s.

Answer 67

thanks man. ur a life saver.

Answer 68

We roughly double the mass, so the velocity is roughly cut in half.

Answer 69

wait, so this isn't the answer?

Answer 70

No, you have the answer, I'm just trying to help you get a feel for what's going on.

Answer 71

ooooh ok.

Answer 72

Momentum before = momentum after. That's all there is to it.

Answer 73

Well thanks!

Answer 74

you're welcome! time for me to get back to my real work...