Question about a collision

[thechairman]

Hi, I am having some trouble understanding a collision problem.


If two objects shown below collide and remain together without spinning, what will be their final velocity? Both objects are traveling at 10 m/s.

The answer given in the book is 5.8 m/s. But isn't this wrong? Shoudn't the final velocity of the clumped together objects be 10, as well, because of conservation of momentum.

Also, the solution goes on to say that "the horizontal momentum of the two object system equals zero so there will be no horizontal velocity. the vertical momentum is 87 kg m/s. We divide this by 15 kg to get 5.8 m/s."


Can anyone explain this? Thanks alot.

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[mrmilad]

Since the collision is taking place at an angle you must consider the collision in two directions. The X and the Y, and then at the end you can use vector addition to find the final velocity.


So in the X direction, the velocity of the second object is: sin 30 vi

Thus the momentum in X direction : m vi - m sin 30 vi = 2m vfx

The momentum in Y direction: 0 + m cos 30 vi = 2m vfy


To find the final velocity vector : vfx^2 + vfy^2 = vresultant^2.


Sorry I am too lazy to calculate it out at this moment, but thats how I would tackle this problem.

 

[misterex]

The horizontal momentum is equal to 5kg * 10m/s - 10kg * 10m/s*cos60=0. Since you are only taking the horizontal component the 10kg object, the momentum horizontally cancels.

Thus, there is only vertical momentum which is equal to 10kg * 10m/s * sin 60=87 kg m/s.

Now, since both stick together 87kg m/s= (10+50)kg * velocity

Remember, momentum is conserved in both the horizontal and vertical direction because it is a vector quantity. It is NOT like energy where you can just add everything because that is a scalar quantity.

 

[thechairman]

right, but the horizontal momentum shouldn't be 0, should it? because there is a non-zero sum from resulting from the x component vector additions.

 

[misterex]

No, it should be zero. The momentum in the right direction is 5*10=50. The momentum from the left direction is 10*10*cos60=50. So they both cancel.

 

[amikhchi]

I think you're a bit mixed up, that seems like the right way to explain it, only the masses aren't the same weight... here's how i did it:

if you break up the angled one into X,Y you get 5 (to the left) for the x, and 5rad3 (down) for the y

start with x:

p = mv => 10 * 5 = 50 (for the mass moving at an angle) that's the x momentum

same for the other one, mv => 5 * 10 = 50 (to the right)

m1v1 + m2v2 = (m1 +m2)v =>
(50) + (-50) = 0

long explanation for one part, but it takes long to explain not to do...
now all ur left with is the Y component (5rad3)

m1v1 + m2v2 = (m1 +m2)v

5*0 + 10(5rad3) = 15v

turns out to be about 5.8

hope that helps

 

[thechairman]

ok, but how do i explain the conservation of momentum?

if the final velocity is 5.8 m/s, and the final mass is 15 kg, doesn't that violate the conservation of momentum?

5 kg * 10 m/s + 10 kg * 10 m/s =/= 5.8 * 15 kg?

 

[misterex]

The final momentum is 5.8m/s * 15kg in the right direction (x axis)
The final momentum is 0 for the y axis.

Your last post is wrong because momentum is a VECTOR quantity. You can't just add them like that and get the final momentum. You have to add them seperataly in the x direction and seperately in the y direction.

Momentum in y cancels to 0, so there is no momentum in the y direction.

Momentum in the x is 5.8*15. This is really hard to explain by writing, sorry if this doesnt help.

 

[misterex]

My poor attempt at explaining it again.

Begining momentum=ending momentum.

X direction:
5*10-10*10*cos60=(5+10)Vx

Vx=0 when you solve. So no momentum in x direction

Y direction:
10*10*sin60=(5+10)Vy
Vy=5.8m/s

 

[thechairman]

so you are saying we can't think of momentum conservation in terms of scalar quantities?

 

[xlr8]

momentum is always conserved (in x and y) unless theres an external force (like gravity) acting on the body/bodies

 

[BlackSails]

so you are saying we can't think of momentum conservation in terms of scalar quantities?

Yes. Momentum is a vector quantity.

 

[misterex]

Right momentum is a vector and can't be thought of as scalar. Energy is scalar and you can just add that up.

 

[thechairman]

cripes, i find this out less than 4 weeks before the MCAT! is there any other conservation stuff that's like the conservation of momentum i should worry about for the MCAT?

 

[zhousilai]

no, Velocity in the x direction should be 0, because
5v1+10v2=15vf
5(10) + 10 (-10sin30) = 15 vf
0=15vf
0=vf

as for the y direction should be 5.8 going down

5v1+10v2=15vf
5(0) + 10 (-10cos30) = 15vf
10 (-10*0.866) = 15vf
10 (-8.66) = 15vf
-86.6 = 15vf
vf = -5.8