- #26

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so i add 1200 and 1400 then x 9.8 = 25480N

then divide .6 = 15288N

? im not doin this right am i?

then divide .6 = 15288N

? im not doin this right am i?

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- Thread starter brightesthalo
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- #26

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so i add 1200 and 1400 then x 9.8 = 25480N

then divide .6 = 15288N

? im not doin this right am i?

then divide .6 = 15288N

? im not doin this right am i?

- #27

andrevdh

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- #28

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a = F/m

= 15288/2600

= 5.88m/s

as for initial speed how do i calculate that?

= 15288/2600

= 5.88m/s

as for initial speed how do i calculate that?

- #29

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would that be using v2=u2 +2as somehow?

- #30

andrevdh

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That is correct.

- #31

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do i have to assume the distance of the suburu seeing as it hasn't been given

- #32

andrevdh

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The cars are mangled together and are sliding as a unit.

- #33

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still wont help me decide whether the suburu was going faster than 80 km/h would it?

- #34

andrevdh

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The answer need to be appoached in several steps. This is one of them before you will be able to answer that question.

The combined wrecks start with some speed v and are decelerated to zero by the frictional force. What do you get their initial speed?

The combined wrecks start with some speed v and are decelerated to zero by the frictional force. What do you get their initial speed?

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- #35

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i tried that and got 12.26 m/s ---> 44.136km/h

that right?

- #36

andrevdh

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Now you can calculate the momentum of the combined wrecks after the collision. What will the direction and magnitude of this vector be?

- #37

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so it would be p=mv

= 2600 x 12.26

= 31876

??

= 2600 x 12.26

= 31876

??

- #38

andrevdh

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Units of the calculated momentum? What would the direction of the vector be?

- #39

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? ok now im lost here im gussin 7 deg

- #40

andrevdh

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What will the components of this momentum vector be in the north-south and east-west directions?

- #41

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would this by any chance be wher sin tan cos would be used? if not then i dont really know

- #42

andrevdh

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- #43

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- #44

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actually ill stay a little longer i really need this done

- #45

andrevdh

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It is time for me to go home also. I might be online in about 16 hours time.

What you need to do is calculate the components of this vector and compare it with the momentum vectors before the collision. Since momentum is conserved the components need to be the same in both directions before and after collision.

What you need to do is calculate the components of this vector and compare it with the momentum vectors before the collision. Since momentum is conserved the components need to be the same in both directions before and after collision.

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- #46

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how did u work that out?

- #47

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thank u for your help anyhow

i appreciate it very much

- #48

andrevdh

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You need to calculate the components of the momentum after the collision, [itex]p_a[/itex]. These two components are such vectors that when added together will produce the vector [itex]p_a[/itex]. They also form the sides of a right angle (its base and perpendicular) as indicated in the attachment, while the vector [itex]p_a[/itex] are the hypotenuse of the triangle.

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