# Plz could you checkif i got the 2d collisions right?

1. Feb 6, 2013

### lioric

1. The problem statement, all variables and given/known data
A 0.30 kg puck A, initially at rest on a frictionless horizontal surface, is struck by a 0.20 kg puck B
that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20
kg puck has a speed of 1.0 m/s at an angle of θ = 530 to the positive x-axis.
a) Determine the velocity of the 0.30 kg puck after the collision.

2. Relevant equations

total mommentum before collision + total momentum after collision

3. The attempt at a solution

total momentum before collision in x

puck A

0.2 x 2 = 0.4

puck B

0.3 x 0

total = 0.7kgm/s

total momentum before collision in y = 0 for all

Total momentum after collision x

puck A

0.2 x 1 x cos 53=0.12

puck B
0.3 x V x cos 53

Total =0.7kgm/s

Total momentum after collision y

puck A

0.2 x 1 sin 53 = 0.1597

puck B

0.3 x V x sin53

Total =0

Resolve in x for puck B

0.3 x V x cos 53
Let Vcos 53 = Vx

0.3 x Vx + 0.12 = 0.7
Vx = 1.93333m/s

Resolve in y for puck B

0.3 x V x sin 53
Let Vsin 53 = Vy

0.3 x Vy + 0.1597 = 0
Vy = -0.53

V2=Vx2 + Vy2

V2 = 1.933332+(-0.53)2

V=√4.02

v=2.005m/s

Is this right???

2. Feb 6, 2013

### PeterO

Probably not correct - see red above.

3. Feb 6, 2013

### lioric

actually i knew there was something wrong
I did all of this all again since the pc froze.
previously i got it right
i knew there was a glitch since the answers looked different.

4. Feb 6, 2013

### lioric

total momentum before collision in x

puck A

0.2 x 2 = 0.4

puck B

0.3 x 0 = 0

total = 0.4kgm/s

total momentum before collision in y = 0 for all

Total momentum after collision x

puck A

0.2 x 1 x cos 53=0.12

puck B
0.3 x V x cos θ

Total =0.4kgm/s

Total momentum after collision y

puck A

0.2 x 1 sin 53 = 0.1597

puck B

0.3 x V x sinθ

Total =0

Resolve in x for puck B

0.3 x V x cos θ
Let Vcos θ = Vx

0.3 x Vx + 0.12 = 0.4
Vx = 0.93333m/s

Resolve in y for puck B

0.3 x V x sin θ
Let Vsin θ = Vy

0.3 x Vy + 0.1597 = 0
Vy = -0.53

V2=Vx2 + Vy2

V2 = 0.933332+(-0.53)2

V=√0.5902

v=0.768m/s

is this better?

5. Feb 6, 2013

### PeterO

I don't expect it to be, from the way you calculated.

Step 1: calculate initial momentum. check you have done that.

Step 2: calculate components of final momentum of the block you have been given. Check you did that.

Step 3: deduce the components of the momentum of the other block. - Not the way you did.

Step 4: Add those components (as vectors) to find the actual momentum of that other block.

Step 5: Calculate the velocity of that block.

eg: I am about to randomly select some values for example - so as not to solve your actual question for you.

Let's pretend the the momentum of the incoming block was 5 units in the x direction.

Lets also assume that the components after collision were 3.5 in the x direction, and 1.5 in the y direction.

The components of MOMENTUM of the other block would be
1.5 in the x direction [since 1.5 + 3.5 = 5; the original value was 5]
-1.5 in the y direction [since 1.5 + -1.5 = 0; the original value was 0]

So the net momentum = 1.5√2 at 45 degrees

Given the second mass was 0.3 kg, that meant the velocity must have been 5 m/s at an angle of 45 degrees with the positive x-axis.

Note: I chose values that were very easy to work with - even to havng equal components so I could make use of the 1,1,√2 Pythagorean triangle.

I expect your values to be a bit more messy

Note: the use of an angle of 53 degrees usually implies to use sin and cosine values of 0.8 and 0.6 rather that the complicated/inconvenient, accurate figures, as the angles in a 3,4,5 Pythagorean are just over 53 degrees and just under 37 degrees.

NOTE ALSO: there is a possibility that when you have found the final answer using momentum conservation, it may be inconsistent with kinetic energy calculations [there may appear to be an increase in total kinetic energy] The problem was that the question setter may have just chosen the speed and angle of the 0.2 kg mass unwisely.

6. Feb 6, 2013

### lioric

thank you very much that was very detail