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You have not submitted your answer. Two identical pucks collide elastically on an air hockey table. Puck 1 was originally at

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Answer #1

The initial condition can be visualized by

9 m/s

The initial momentum of the system is therefore

pi= m x 9(-i) = -9m i

And the initial energy of the system is

E; = 5 mx mx 9281

Now, the final condition can be visualized by

The final momentum of the system is therefore

Ps = m(ū +ū) = m (u(- cos di + sin oj) +ul - cos 55°î – sin 55°)

And the final energy is

(2n + za)u=

The collision is elastic so the conservation of energy is applied here. i.e

E = E

= im(s + 1²) = Sim

合 := = 81 (1)

There is no external force applied so, the conservation of momentum is applied here. i.e

Pg = p

m(u( - cos oî+sin ) + ul- cos 55°î – sin 55°)) = -9m i

= (u(- cos i + sin 0j) +u(- cos 55°î – sin 55°)) = -9 i

This gives us

- cos - u cos 55º = -9

v cos 0 + u cos 55° = 9 .......

and

v sin A - u sin 55° = 0 ..........3

Squaring both sides of both the equations we get

v cose +ucos? 55° + 2vu cos 6 cos 55° = 81

v2 sino + usin55° – 2vu sin sin 55° = 0

Adding both the equations

v? (sin’e+cos? e)+u? (sinº 55° +cos? 55°)+(2vu cos cos 55°—2vu sin sin 55°) = 81

02+u? + (2vu cos cos 55° – 2vu sin sin 55°) = 81

Now using equation 1

81 + (2vu cos 6 cos 55° – 2vu sin sin 55°) = 81

2vu cos cos 55° – 2vu sin sin 55° = 0

cos 6 cos 550 = sin sin 55°

>>tan = tan 550

A=tan-1 35° tan 550 /

Putting this value in equation 3

v sin 350 – u sin 55° = 0

u = v sin 35° sin 550

And putting these values in equation 2

v cos 35° + v sin 350 sin 550 cos 55° = 9

v cos 35° + sin 350 tan 550)

= 7.37 m/s (cos 35° + sin35)

So, the final answers are

v = 7.37 m/s

0 = 180° - 35° = 1450 ccw from + acis

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