Asked by Anonymous
1) If two gliders of equal mass equal and opposite initial veloity collide perfectly elastically, using the given equation, what are the final velocities of the gliders in terms of the initial velocities
equations:
MaVa + MbVb = MaV'a + MbV'b (momentum)
1/2MaV^2a + 1/2MbV^2b = 1/2MaV'^2a
+ 1/2MbV'^2b (energy)
2)If two gliders of equal mass and equal and opposite initial velocity collide and stick together, using given equation, what are the final velocities of the gliders?
Equation:
MaVa + MbVb = (Ma+Mb)V'ab
equations:
MaVa + MbVb = MaV'a + MbV'b (momentum)
1/2MaV^2a + 1/2MbV^2b = 1/2MaV'^2a
+ 1/2MbV'^2b (energy)
2)If two gliders of equal mass and equal and opposite initial velocity collide and stick together, using given equation, what are the final velocities of the gliders?
Equation:
MaVa + MbVb = (Ma+Mb)V'ab
Answers
Answered by
bobpursley
Solve the equations, that is the point of all this.
On the second, it is trivial
v'ab=(MaVa+mbVb)/(Ma+Mb)
on the first, use the momentum equation to solve for V'a in terms of all the other. Go ahead and change the initial Vb to -Va, since it is the same, opposite direction, it makes the problem much easier (the left side of momentum is zero), and V'b=-V'a check that.
Now put all that into energy, and solve for either V'b or V'a, they are the same, except for sign. Your teacher is too easy on you making all the difficult and messy math disappear.
On the second, it is trivial
v'ab=(MaVa+mbVb)/(Ma+Mb)
on the first, use the momentum equation to solve for V'a in terms of all the other. Go ahead and change the initial Vb to -Va, since it is the same, opposite direction, it makes the problem much easier (the left side of momentum is zero), and V'b=-V'a check that.
Now put all that into energy, and solve for either V'b or V'a, they are the same, except for sign. Your teacher is too easy on you making all the difficult and messy math disappear.
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