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A 1kg ball moving at 12m/s collides head on with 2kg ball moving in opposite direction at 24m/s. If the coefficient of restitution is 2/3,then how much energy is lost in collision
A 1kg ball moving at 12m/s collides head on with 2kg ball moving in opposite direction at 24m/s. If the coefficient of restitution is 2/3,then how much energy is lost in collision

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2 years ago

```							we are givenfor ball 1m1 = 1 kgu1 = 12 m/sandfor ball 2m2 = 2 kgu2 = 24 m/snow,the coefficient of restitution is given ase = (v2 - v1) / (u1 - u2) = 2/3 = 0.66so,(v2 - v1) = 0.66 x (u1 - u2) = 0.66 x (12 - 24)or(v2 - v1) = -7.92orv2 = v1 - 7.92  ..............................(1)now, as e m1u1 + m2u2  = m1v1 + m2v2or1x12 + 2x24 = 1xv1 + 2xv2or60 = v1 + 2v2so,v2 =  (60 - v1) / 2  ...........................(2)by equating (1) and (2), we getv1 - 7.92 = (60 - v1) / 2or by rearranging 2v1 + v1 = 60 + 2x7.923 v1 = 75.48thus,v1 = 25.16 m/sandv2 = 25.16 - 7.92 = 17.24 m/sso,loss in kinetic energy will bedE = Ei - Ef = [(1/2)m1u12  +(1/2)m2u22] - [(1/2)m1v12  +(1/2)m1v22]which can be calculated...
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2 years ago
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