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Nov 2010 p41 q6
3569
A smooth slide AB is fixed so that its highest point A is 3 m above horizontal ground. B is h m above the ground. A particle P of mass 0.2 kg is released from rest at a point on the slide. The particle moves down the slide and, after passing B, continues moving until it hits the ground (see diagram). The speed of P at B is vB and the speed at which P hits the ground is vG.
(i) In the case that P is released at A, it is given that the kinetic energy of P at B is 1.6 J. Find
the value of h,
the kinetic energy of the particle immediately before it reaches the ground,
the ratio vG : vB.
(ii) In the case that P is released at the point X of the slide, which is H m above the ground (see diagram), it is given that vG : vB = 2.55. Find the value of H correct to 2 significant figures.
Solution
(i) (a) The potential energy (PE) loss is equal to the kinetic energy (KE) gain. Therefore,
\(0.2g(3 - h) = 1.6\)
Solving for \(h\), we get:
\(h = 2.2 \text{ m}\)
(b) The kinetic energy of the particle immediately before it reaches the ground is given by the total potential energy loss from \(A\) to the ground:
\(\text{KE} = 0.2g \times 3 = 6 \text{ J}\)
(c) The ratio \(v_G : v_B\) can be found using the relationship: