What Is the Speed of the Composite Object After Collision?

[cooney88]

Q. Two objects A and B are moving on a frictionless horizontal surface. A has a mass of 6kg and speed of 3m/s , B has a *** of 8kg and a speed of 5m/s. if they collide and stick together the speed in m/s of the composite object after the collision is:

diagram:

A----------------------> X
angle = 30->



B is traveling at an angle towards the x markso what i have been doing is i have been finding the before and after collision as x and y components. then putting the final xcomponent momentum and y component momentums into this formula :total momentum after = (x^2 + y^2)^.5

i would then have my total momentum before and then the total momentum after with v unknown. i got 4.02 for it. the question is multiple choice and these are the choices 3.23 16.2 2.71 4.02 or 3.76

 

[Doc Al]

so what i have been doing is i have been finding the before and after collision as x and y components. then putting the final xcomponent momentum and y component momentums into this formula :total momentum after = (x^2 + y^2)^.5

Sounds good.

i would then have my total momentum before and then the total momentum after with v unknown. i got 4.02 for it. the question is multiple choice and these are the choices 3.23 16.2 2.71 4.02 or 3.76

Looks like a match to me.

It's not clear exactly where that 30 degree angle fits in. For example: If A moves along the +x axis, does B move at an angle of 30 degrees above the +x axis? If so, then you're OK.

(Next time show every key step of your calculation--it makes it easier to follow.)

 

[cooney88]

pax= 6v1 pay = 0
pbx= 8v2cos30 pby = 8v2sin30

pcx= 6v1 +8v2cos30 = 12.9v pcy = 8v2sin30= 4v

(12.9v^2 + 4v^2)^.5 = 13.5v

so 58 = 13.5v

v= 4.2

a is moving along the x-axis and b is moving below a but is heading towards a. the angle between them is 30

 

[cooney88]

any ideas what I am doing wrong?

 

[cooney88]

sorry i meant i got 4.2 not 4.02 . so I am obviously doing something wrong

 

[Doc Al]

pax= 6v1 pay = 0
pbx= 8v2cos30 pby = 8v2sin30

OK.

pcx= 6v1 +8v2cos30 = 12.9v pcy = 8v2sin30= 4v

Careful: v1 and v2 are given, and they are not equal. Redo this step, putting in values for v1 and v2.

 

[cooney88]

yes only before the collision are v1 and v2 are different. however the 2 balls ''stick together'' therefore the velocity is the same for both of them after the collision

 

[Doc Al]

yes only before the collision are v1 and v2 are different. however the 2 balls ''stick together'' therefore the velocity is the same for both of them after the collision

That's true. But you're calculating the total momentum before the collision, so you must use the speeds they had before the collision.

 

[cooney88]

which i did but I am still getting 4.2 . thanks a million for helping me with this by the way. still not sure what I am doing wrong tho?

 

[Doc Al]

Redo the steps, using v1 and v2. Show me what you get.

 

[cooney88]

AW DUDE! couldn't we just use m1xv1+ m2xv2 = m1xv1+m2xv2 ? i think the angle might of been in the question to put u off?

 

[Doc Al]

couldnt we just use m1xv1+ m2xv2 = m1xv1+m2xv2 ?

No. That only works for collisions in one dimension.

i think the angle might of been in the question to put u off?

You were on the right track. Just redo it.