Physics Question..If two gliders of equal mass and…?

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)

1/2 MaV^2a+1/2MbV^2b=1/2MaV’^2a+1/2MbV’^2b)

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

3 Answers

  • Physics Question..If two gliders of equal mass and…?

    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

    Since the masses are equal, let M = mass

    Since the velocities are equal, but in the opposite direction, Va is +, and Vb is -.

    The magnitude of Va = the magnitude of Vb, so, ½ * Va^2 = ½ * Vb^2

    So, let V^2 = Va^2 = Vb^2

    (½ * M * V^2) + (½ * M * V^2) = (½ * M * V’a^2) + (½ * M * V’b 2)

    Divide both sides by ½ * M

    (V^2) + (V^2) = (V’a^2) + (V’b 2)

    (V’a^2) + (V’b^2) = 2 * (V^2)

    The equation above can be true, if V’a and V’b are equal in magnitude, and in the same direction!

    AND

    The equation above can be true, if V’a and V’b are equal in magnitude, but in the opposite direction!

    After the collision, the gliders must move in opposite direction, or they will have to MOVE THROUGH each other !!!

    When a 100% elastic collision occurs, momentum and kinetic energy are conserved!

    Since the 2 gliders have equal mass, and initial velocity of both gliders are equal, but opposite,

    THIS PROBLEM IS PERFECTY SYMETRICAL.

    So, after the collision, each glider will be traveling at the same velocity as before the collision, but in the opposite direction.

    Before collision:

    Glider #1

    v = 20 m/s East

    Glider #2

    v = 20 m/s West

    After collision:

    Glider #1

    v = 20 m/s West

    Glider #2

    v = 20 m/s East

    2)If two gliders of equal mass and equal and opposite initial velocity collide and stick together, using given equation, what are the

    When the 2 gliders stick together, kinetic energy is not conserved, but momentum is conserved.

    MaVa + MbVb = (Ma + Mb) * Vfinal

    Vfinal =

    Since the masses are equal, let M = mass

    Since the velocities are equal, but in the opposite direction, Va is +, and Vb is -.

    Initial momentum = M * (Va – Vb)

    But the magnitude of Va = the magnitude of Vb, so (Va – Vb) = 0

    Initial momentum = 0, so Final momentum = 0.

    Final momentum = (Ma + Mb) * Vfinal = 0

    (Ma + Mb) is not 0, so, Vfinal = 0

    The 2 gliders collide, vibrate in place, and release all their initial kinetic energy as heat energy!!

  • MaVa+MbVb=MaV’a+MbV’b (momentum)

    As Va = – Vb

    and Ma=Mb

    => MaVa-MaVa=MaV’a+MaV’b

    =>MaV’a+MaV’b=0

    =>V’a+V’b=0

    =>V’a= – V’b……….(1)

    1/2 MaV^2a+1/2MbV^2b=1/2MaV’^2a+1/2MbV’^2b)

    As Magnitude of Va = mag. Vb

    and Ma=Mb

    1/2Ma(V’^2a+V’^2b)=MaV^2a

    V’^2a+V’^2b=2V^2a

    2V’^2a=2V^2a

    V’^2a=V^2a…………from(1)

    Therefore , after collision both gliders start moving in directions opposite to their initial directions and with same magnitude of velocity

  • i attempted fixing this using the gravity-centripetal tension dating. (GMm)/r^2 = (mv^2)/r (a million) M is the mass of the famous person so rearranging the above equation supplies: M = (v^2*r)/G (2) G=6.673E-11 using the circumfrence of the orbit and the time it took i found v and as a result v^2 = 5.728E12 r = a million.55E11 M = a million.33E34 kg after plugging values into equation 2

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