lioric said:
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/sis this better?
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.