Particles of masses 1.5kg and 3kg lie on a plane which is inclined at an angle of α to the horizontal, where tan α = 3/4. The section of the plane from A to B is smooth and the section of the plane from B to C is rough. The 1.5kg particle is held at rest at A and the 3 kg particle is in limiting equilibrium at B. The distance AB is xm and the distance BC is 4m (see diagram). [Figure X.X]
- (a)Show that the coefficient of friction between the particle at B and the plane is 0.75.[3]
- (b)The 1.5kg particle is released from rest. In the subsequent motion the two particles collide and coalesce. The time taken for the combined particle to travel from B to C is 2s. The coefficient of friction between the combined particle and the plane is still 0.75. Find x.[6]
- (c)Find the total loss of energy of the particles from the time the 1.5 kg particle is released until the combined particle reaches C.[3]
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- [B1] R=3gcosa = 3 × 10 × 0.8
- [M1] F = 3g sina = 3 × 10 × 0.6 Resolving parallel to plane.
- [A1] μ = 18/24 = 0.75 or μ = 3g sin a / 3g cos a = tan a = 0.75 Uses μ = F/R AG.
- [B1] a = g sina or PE loss=1.5gx sina for AB and a = 0 for BC 4.5g×sina-0.75×4.5g cosa = 4.5a leading to a = 0 Accelerations for AB and BC.
- [M1] v₁² = 2xgsinaxx ] or [1.5g×xsina=0.5×1.5×1₁² Uses 'suvat' or PE loss = KE gain for AB.
- [A1] v₁² = 20x sina=12x leading to v₁ = √12x
- [M1] 1.5×√12x+0=4.5×v₂ leading to v₂ = 1/3 √12x Conservation of momentum.
- [M1] 4= 2/3 x√12x Use of s = vt on BC since a = 0.
- [A1] x=3
- [B1] KE = 0.5×4.5×2² =9J KE gain for AC.
- [M1] PE loss = 15×(4+3)×3+30×4×3=135J Evaluates PE loss for AC.
- [A1] Loss of energy = 126 J
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