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26 Feb 2008 23:00:13 IST

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Collision Problem ??? Hint: Use Impulse Momentum Theorem

None

Normal 0 false false false MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-ansi-language:#0400; mso-fareast-language:#0400; mso-bidi-language:#0400;} Wedge M 2 is at rest (ie U 1 = 0m/s). Coefficient of restitution is e. Given U 1, U 2, M 1, M 2, e, beta, g

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Priyesh

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Joined: 18 Feb 2007

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26 Feb 2008 23:48:45 IST

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Click on figure to enlarge

velocity of mass m2 along the incline will not change let the velocity of mass m2 perpendicular to incline after collision be V

hence net velocity V2 after collision of mass M2 will be root(v^2 + (u2sin

)^2)

let velocity of mass M1 after collision be V1 towards left which is component of Va

so along common normal direction of collision i.e perp. to incline

we have

V + Va = e (U2cos

)--------------------------1

conserving momentum along this direction

M1Va - M2V = M2U2cos

------------------2

solve 1 &2 to get Va and V

hence V1 is Vasin

& V2 is root(v^2 + (u2sin

)^2)

The Sniper

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27 Feb 2008 09:24:22 IST

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$\text{Is the answer :}\\ \\ \text{Velocity of M1 in leftward direction }=\frac{U_2M_2sin\betacos\beta(1+e)}{M_2sin^2\beta+M_1}\\ \\ \text{Vel of M2 in x direction }=\frac{U_2M_1sin\betacos\beta(1+e)}{M_2sin^2\beta+M_1} \\ \\ \text{Vel of M2 in y direction }=\frac{U_2M_1cos^2\beta(1+e)}{M_2sin^2\beta+M_1}-U_2$

edison

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Joined: 19 Oct 2006

Posts: 7449

29 Feb 2008 20:22:48 IST

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I suppose the query is answered.

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