Question

@Abhisar A red car of mass m is heading north (direction 0°). It collides at an intersection with a yellow car of mass 1.3m heading east (direction 90°). Immediately after the collision, the cars lock together and travel at in direction 42°. What is the speed of the yellow car just before the collision?

24 m/s
28 m/s
11 m/s
17 m/s

Answer 1

The velocity after collision is 20m/s

Answer 2

Initial velocity of red car is not given ?

Answer 3

No

Answer 4

The red car mass is variable m and yellow is 1.3*m

Answer 5

@Astrophysics

Answer 6

Mass 1.3m lol mhm o_O, well do you have any idea on how to start it? Anything that comes in mind when it says collision?

Answer 7

Mass *v is the momentum

Answer 8

Yes this question deals with momentum, so first thing, draw a picture out :), and what for 1.3m mass mean? Oh I see the yellow is 1.3* the mass of red car hehe. The wording!

Anyways any attempt at the drawing?

Answer 9

Seems like there is some information missing when it says " the cars lock together and travel AT ... IN direction 42 degrees.

Answer 10

20m/S

Answer 11

\(\color{blue}{\text{Originally Posted by}}\) @akohl
The velocity after collision is 20m/s
\(\color{blue}{\text{End of Quote}}\)

Answer 12

So i think we will have to apply law of conservation of momentum ?

Answer 13

YEA mass times velocity

Answer 14

\[\sum p _{i}= \sum p _{f}\]

does this ring a bell?

(remember these are vector quantities)

Answer 15

But initial velocity of red ball is not given.

Answer 16

Newtons third law basically :P

Answer 17

Oh I see

Answer 18

cant add vectors without magnitude

Answer 19

Hey, do you have an image of the full question?

Answer 20

That's wr i got struck !

Answer 21

no there wasnt one

Answer 22

Or can you just post wherever you got your question from, I have doubts haha.

Answer 23

I cant its off a worksheet from my summer school teacher lol

Answer 24

*stuck

Answer 25

If we had an initial velocity, this would be cake :p

Answer 26

u will get \[V _{1}=2.3 V\]
\[V _{2}= \frac{ 2.3 }{ 1.3 }V\]
and then use the conservation of kinetic energy
and substitute with v1 and v2 to get V
\[\frac{ 1 }{ 2 }mV _{1}^{2}+\frac{ 1.3 }{ 2 }mV _{2}^{2}=\frac{ 2.3 }{ 2 }mV ^{2}\]

Answer 27

oh noo i forgot to write the values cos 45 write it and continue.

Answer 28

momentum is vector while energy isn't.