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Nov 2012 p41 q6
3554
The diagram shows the vertical cross-section ABCD of a surface. BC is a circular arc, and AB and CD are tangents to BC at B and C respectively. A and D are at the same horizontal level, and B and C are at heights 2.7 m and 3.0 m respectively above the level of A and D. A particle P of mass 0.2 kg is given a velocity of 8 m s-1 at A, in the direction of AB (see diagram). The parts of the surface containing AB and BC are smooth.
Find the decrease in the speed of P as P moves along the surface from B to C.
The part of the surface containing CD exerts a constant frictional force on P, as it moves from C to D, and P comes to rest as it reaches D.
Find the speed of P when it is at the mid-point of CD.
Solution
(i) Using the principle of conservation of energy from A to B and A to C:
\(\frac{1}{2} m v_B^2 = \frac{1}{2} m v_A^2 - mg \times 2.7\)
\(\frac{1}{2} m v_C^2 = \frac{1}{2} m v_A^2 - mg \times 3\)
Substituting for \(v_A = 8\) m s-1:
\(v_B^2 = 8^2 - 20 \times 2.7\)
\(v_C^2 = 8^2 - 20 \times 3\)
Loss of speed = \(\sqrt{v_B^2} - \sqrt{v_C^2} = 10^{1/2} - 2 = 1.16\) m s-1