Momentum and angular collision

In summary, the question is asking for the speed in m/s of the composite object after the collision of two objects, A and B, with masses of 6kg and 8kg respectively and speeds of 3m/s and 5m/s respectively on a frictionless horizontal surface. Using the formula for total momentum after a collision, (x^2 + y^2)^.5, and calculating the total momentum before and after the collision with the x and y components, the speed of the composite object is found to be 4.2 m/s.
  • #1
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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
 
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  • #2
cooney88 said:
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.)
 
  • #3
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
 
  • #4
any ideas what I am doing wrong?
 
  • #5
sorry i meant i got 4.2 not 4.02 . so I am obviously doing something wrong
 
  • #6
cooney88 said:
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.
 
  • #7
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
 
  • #8
cooney88 said:
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.
 
  • #9
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?
 
  • #10
Redo the steps, using v1 and v2. Show me what you get.
 
  • #11
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?
 
  • #12
cooney88 said:
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.
 

1. What is momentum in a collision?

Momentum is a physical quantity that describes the motion of an object. It is the product of an object's mass and velocity. In a collision, the total momentum of the system remains constant, meaning that the sum of the momenta of all the objects involved before the collision is equal to the sum of the momenta after the collision.

2. How is momentum conserved in a collision?

Momentum is conserved in a collision because of the law of conservation of momentum. This law states that in a closed system, the total momentum remains constant. This means that the total momentum before a collision is equal to the total momentum after the collision, even if the objects involved in the collision exchange momentum with each other.

3. What is an elastic collision?

An elastic collision is a type of collision where the total kinetic energy of the system is conserved. This means that in an elastic collision, the objects involved bounce off each other without any loss of energy. In other words, the objects involved in the collision have the same total kinetic energy before and after the collision.

4. How is angular momentum related to linear momentum in a collision?

Angular momentum is the rotational equivalent of linear momentum. In a collision, both linear and angular momentum are conserved. This means that the total angular momentum of the system before the collision is equal to the total angular momentum after the collision. In some collisions, angular momentum may be converted into linear momentum and vice versa.

5. What factors can affect the outcome of a collision?

Several factors can affect the outcome of a collision, including the masses and velocities of the objects involved, the angle of collision, and the coefficient of restitution. The coefficient of restitution is a measure of the elasticity of the collision and can affect how much kinetic energy is conserved. The shape and composition of the objects involved can also play a role in the outcome of the collision.

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