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
To solve both questions, you can use the principles of conservation of momentum and conservation of energy. Let's break down the steps for each scenario:
1) Perfectly Elastic Collision:
a) Conservation of Momentum:
- Start with the equation: MaVa + MbVb = MaV'a + MbV'b
- Where Ma and Mb are masses of the two gliders, Va and Vb are their initial velocities, and V'a and V'b are their final velocities.
- Since the initial velocities are equal and opposite, Va = -Vb. Substituting this into the equation, we get: Ma(-Vb) + MbVb = MaV'a + MbV'b.
- Simplifying, we have: -MbVb + MbVb = MaV'a + MbV'b. The opposite terms cancel out: 0 = MaV'a + MbV'b.
- Rearranging the equation, we find the final velocities in terms of initial velocities: V'a = -MbVb / Ma and V'b = MaVb / Mb.
b) Conservation of Energy:
- Use the equation: 1/2MaV^2a + 1/2MbV^2b = 1/2MaV'^2a + 1/2MbV'^2b.
- Since the given problem states that the collision is perfectly elastic, the total kinetic energy before and after the collision remains the same.
- Substitute Va = -Vb (from the initial velocities being equal and opposite) into the equation.
- Rearrange the equation as needed to solve for V'a and V'b in terms of Va and Vb.
2) Completely Inelastic Collision:
a) Conservation of Momentum:
- Use the equation: MaVa + MbVb = (Ma+Mb)V'ab.
- This scenario assumes that the two gliders stick together after the collision, forming a single combined glider.
- Substituting Va = -Vb (from the initial velocities being equal and opposite) into the equation.
- Rearrange the equation as needed to solve for V'ab in terms of Va and Vb.
Note: It's important to remember that in the given equations, Ma, Mb, Va, and Vb refer to the masses and velocities of the gliders, and V'a, V'b, and V'ab denote their final velocities.
1) In a perfectly elastic collision, the kinetic energy is conserved. Therefore, we can equate the initial and final kinetic energy equations:
1/2MaV^2a + 1/2MbV^2b = 1/2MaV'^2a + 1/2MbV'^2b
Since the gliders have equal mass and equal and opposite initial velocities, Ma = Mb and Va = -Vb. Let's substitute these values into the equation:
1/2Ma(-Vb)^2 + 1/2MbVb^2 = 1/2MaV'^2a + 1/2MbV'^2b
Simplifying, we get:
1/2MaVb^2 + 1/2MaVb^2 = 1/2MaV'^2a + 1/2MbV'^2b
Since Ma = Mb, we can substitute:
MaVb^2 + MaVb^2 = MaV'^2a + MbV'^2b
2MaVb^2 = MaV'^2a + MbV'^2b
Since Ma = Mb, we can further simplify:
2MaVb^2 = MaV'^2a + MaV'^2b
Now, substitute the condition that Va = -Vb:
2MaVb^2 = MaV'^2a + Ma(-V'^2a)
Simplifying further:
2MaVb^2 = MaV'^2a - MaV'^2a
2MaVb^2 = 0
This implies that Vb = 0, which means the glider with initial velocity Vb comes to a complete stop, and the glider with initial velocity Va remains unchanged.
Therefore, the final velocities of the gliders in terms of the initial velocities are:
Va = Va
Vb = 0
2) In an inelastic collision where the two gliders stick together, we can use the equation:
MaVa + MbVb = (Ma + Mb)V'ab
Since the gliders have equal masses and equal and opposite initial velocities, Ma = Mb and Va = -Vb. Let's substitute these values into the equation:
Ma(-Vb) + MbVb = (Ma + Mb)V'ab
Simplifying, we get:
-MaVb + MaVb = (Ma + Ma)V'ab
0 = 2MaV'ab
This implies that V'ab = 0, which means that the gliders stick together and come to a complete stop.
Therefore, the final velocities of the gliders in this case are:
V'ab = 0
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.