AP Physics
I am proud to say I did this one on my own. I was very surprised I got it right, and I just wanted to share it with everyone.
:P
AP Physics 29.(111)
Situation: Block A of mass M sits on top of a frictionless ramp that is 30cm high. The ramp is on a 90cm high table. At the bottom of the ramp there is a block B half the mass of M at rest. Block A is realeased from the top of the ramp so it slides and hits Block B. The collision is elastic. Determine where each block lands on the floor below the table after the collision. Ignore air resistance. Hint: Both blocks leave the ramp in a horizontal direction.
Pa+Pb=Pa'+Pb' : MAVA+MBVB=MAVA'+MBVB'
In order to determine the intital velocity of block A we must understand that at the top of the ramp block A will have max PE and at the bottom max KE. So...... mgh=(mv^2)/2. Then we will solve for v and arrive at sqrt(2gh).
therefore: M(sqrt(2gh)) + 0 (since block B begins at rest)= MVA' + M/2 (half the mass) VB'
M(sqrt(2gh)) = MVA' + (M/2)VB' Since we have two unknown variables we are forced to use the golf ball formula in order to identify VB'. The golf ball formula works in this situation since the larger object is hitting the smaller, at rest, object. The formula is : 2MAVA/ (MA+MB) = VB'.
The next step would substitution. (2M)(sqrt(2gh))/ M + (M/2). This can be simplified further since M and M/2 can be added by making a common denominator. (2M)(sqrt(2gh)) / (3M/2) or 4sqrt(2gh) / 3, (M's cancel).
Now that we have found our VB' we can plug it back into our elastic momentum formula.
M(sqrt(2gh)) = MVA' + ((M/2) ((4(sqrt(2gh)) / 3)))
Now it is pretty easy to solve for VA'
VA' = (M(sqrt(2gh)) - ((M/2) ((4(sqrt(2gh)) / 3))) / M
It is now important to understand that each of the respective V' is the vx for each of the blocks. this will remain constant and will determine how far something travelled when multiplied by time. In this case we will use Tfall = sqrt(2h/g).
Lets organize everything we know.
VAx = (M(sqrt(2gh)) - ((M/2) ((4(sqrt(2gh)) / 3))) / M
VBx = 4(sqrt(2gh)) / 3
Tfall = sqrt(2h/g), remains the same for both blocks since they are falling from the same height and horizontal velocity does not change hang time.
and (vx)(t) = dx which is what we are looking for for each block.
Now we actually use numbers and find out what each velocity is.
sqrt(2gh) = sqrt(2x10x(.3)) g= gravity h= height of ramp which 30 cm = .3m
sqrt(6) = 2.45
sqrt(2h/g) h is different in this one since it is talking about how high the blocks are falling from after it came down the ramp, which is 90cm or .9m.
so 2(.9)/10 = about .42
I will leave the rest of the calculator poking up to you. There should only be some M's left and a bunch of ugly constants. There might be some algebra involved but the concept is much simpler than what was shown above. Watch for the M's canceling, as I probably forgot to show that step.
dax = .34m
dbx = 1.37m
:P
AP Physics 29.(111)
Situation: Block A of mass M sits on top of a frictionless ramp that is 30cm high. The ramp is on a 90cm high table. At the bottom of the ramp there is a block B half the mass of M at rest. Block A is realeased from the top of the ramp so it slides and hits Block B. The collision is elastic. Determine where each block lands on the floor below the table after the collision. Ignore air resistance. Hint: Both blocks leave the ramp in a horizontal direction.
Pa+Pb=Pa'+Pb' : MAVA+MBVB=MAVA'+MBVB'
In order to determine the intital velocity of block A we must understand that at the top of the ramp block A will have max PE and at the bottom max KE. So...... mgh=(mv^2)/2. Then we will solve for v and arrive at sqrt(2gh).
therefore: M(sqrt(2gh)) + 0 (since block B begins at rest)= MVA' + M/2 (half the mass) VB'
M(sqrt(2gh)) = MVA' + (M/2)VB' Since we have two unknown variables we are forced to use the golf ball formula in order to identify VB'. The golf ball formula works in this situation since the larger object is hitting the smaller, at rest, object. The formula is : 2MAVA/ (MA+MB) = VB'.
The next step would substitution. (2M)(sqrt(2gh))/ M + (M/2). This can be simplified further since M and M/2 can be added by making a common denominator. (2M)(sqrt(2gh)) / (3M/2) or 4sqrt(2gh) / 3, (M's cancel).
Now that we have found our VB' we can plug it back into our elastic momentum formula.
M(sqrt(2gh)) = MVA' + ((M/2) ((4(sqrt(2gh)) / 3)))
Now it is pretty easy to solve for VA'
VA' = (M(sqrt(2gh)) - ((M/2) ((4(sqrt(2gh)) / 3))) / M
It is now important to understand that each of the respective V' is the vx for each of the blocks. this will remain constant and will determine how far something travelled when multiplied by time. In this case we will use Tfall = sqrt(2h/g).
Lets organize everything we know.
VAx = (M(sqrt(2gh)) - ((M/2) ((4(sqrt(2gh)) / 3))) / M
VBx = 4(sqrt(2gh)) / 3
Tfall = sqrt(2h/g), remains the same for both blocks since they are falling from the same height and horizontal velocity does not change hang time.
and (vx)(t) = dx which is what we are looking for for each block.
Now we actually use numbers and find out what each velocity is.
sqrt(2gh) = sqrt(2x10x(.3)) g= gravity h= height of ramp which 30 cm = .3m
sqrt(6) = 2.45
sqrt(2h/g) h is different in this one since it is talking about how high the blocks are falling from after it came down the ramp, which is 90cm or .9m.
so 2(.9)/10 = about .42
I will leave the rest of the calculator poking up to you. There should only be some M's left and a bunch of ugly constants. There might be some algebra involved but the concept is much simpler than what was shown above. Watch for the M's canceling, as I probably forgot to show that step.
dax = .34m
dbx = 1.37m