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Maths Demo Worksheet

Maths Demo Worksheet

A wedge of mass MM rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle α\alpha to the horizontal. A particle of mass mm slides down the face of the wedge, starting from rest. At a later time tt, the speed VV of the wedge, the speed vv of the particle and the angle β\beta of the velocity of the particle below the horizontal are as shown in the diagram.

A particle sliding down a wedge inclined at alpha, showing wedge velocity V, particle velocity v and angle beta below the horizontal.

Let yy be the vertical distance descended by the particle. Derive the following results, stating in (ii) and (iii) the mechanical principles you use:

1(i) Vsin⁡α=vsin⁡(β−α)V \sin \alpha = v \sin(\beta - \alpha);

1(ii) tan⁡β=(1+mM)tan⁡α\tan \beta = \left(1 + \frac{m}{M}\right) \tan \alpha;

1(iii) 2gy=v2(M+mcos⁡2β)M2gy = \frac{v^2\left(M + m \cos^2 \beta\right)}{M}.

Write down a differential equation for yy and hence show that

y=gMt2sin⁡2β2(M+mcos⁡2β)y = \frac{gMt^2 \sin^2 \beta}{2\left(M + m \cos^2 \beta\right)}

Question sourced from STEP Database, University of Cambridge

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Maths Demo Worksheet

Maths Demo Worksheet

A wedge of mass MM rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle α\alpha to the horizontal. A particle of mass mm slides down the face of the wedge, starting from rest. At a later time tt, the speed VV of the wedge, the speed vv of the particle and the angle β\beta of the velocity of the particle below the horizontal are as shown in the diagram.

A particle sliding down a wedge inclined at alpha, showing wedge velocity V, particle velocity v and angle beta below the horizontal.

Let yy be the vertical distance descended by the particle. Derive the following results, stating in (ii) and (iii) the mechanical principles you use:

1(i) Vsin⁡α=vsin⁡(β−α)V \sin \alpha = v \sin(\beta - \alpha);

1(ii) tan⁡β=(1+mM)tan⁡α\tan \beta = \left(1 + \frac{m}{M}\right) \tan \alpha;

1(iii) 2gy=v2(M+mcos⁡2β)M2gy = \frac{v^2\left(M + m \cos^2 \beta\right)}{M}.

Write down a differential equation for yy and hence show that

y=gMt2sin⁡2β2(M+mcos⁡2β)y = \frac{gMt^2 \sin^2 \beta}{2\left(M + m \cos^2 \beta\right)}

Question sourced from STEP Database, University of Cambridge

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