equal and opposite initial veloity collide perfectly elastically, making use of the given equation, what room the last velocities of the gliders in terms of the early 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 early velocity collide and stick together, using given equation, what space the final velocities of the gliders?

Equation:

Physics Question..If two gliders of same mass and...?

equal and also opposite initial veloity collide perfectly elastically, utilizing the offered equation, what are the final velocities of the gliders in regards to the early stage velocities

Since the masses room equal, permit M = mass

Since the velocities space equal, yet in the opposite direction, Va is +, and also Vb is -.

You are watching: If two gliders of equal mass and equal and opposite initial velocity collide

The magnitude of Va = the size 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 political parties by ½ * M

(V^2) + (V^2) = (V"a^2) + (V"b 2)

(V"a^2) + (V"b^2) = 2 * (V^2)

The equation over can it is in true, if V"a and also V"b space equal in magnitude, and in the exact same direction!

AND

The equation above can be true, if V"a and also V"b room equal in magnitude, however in the contrary direction!

After the collision, the gliders must relocate in the opposite direction, or lock will need to MOVE through each other !!!

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

Since the 2 gliders have actually equal mass, and initial velocity that both gliders space equal, but opposite,

THIS trouble IS PERFECTY SYMETRICAL.

So, after ~ the collision, every glider will certainly be travel at the same velocity as prior to the collision, but in opposing 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 early stage velocity collide and stick together, using given equation, what room the

When the 2 gliders rod together, kinetic energy is no conserved, yet momentum is conserved.

MaVa + MbVb = (Ma + Mb) * Vfinal

Vfinal =

Since the masses space equal, allow M = mass

Since the velocities room equal, yet in opposing direction, Va is +, and also Vb is -.

Initial momentum = M * (Va – Vb)

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

Initial inert = 0, so last momentum = 0.

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

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

The 2 gliders collide, vibrate in place, and release all your initial kinetic power as warmth 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 size 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 relocating in directions opposite to their initial directions and also with exact same magnitude that velocity

i attempted fixing this utilizing the gravity-centripetal tension dating. (GMm)/r^2 = (mv^2)/r (a million) M is the mass of the famed person therefore rearranging the above equation supplies: M = (v^2*r)/G (2) G=6.673E-11 making use of the circumfrence that the orbit and also the time it take it i discovered v and as a an outcome v^2 = 5.728E12 r = a million.55E11 M = a million.33E34 kg after ~ plugging values right into equation 2

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