Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Nov 2009 p41 q2
3574
A smooth narrow tube AE has two straight parts, AB and DE, and a curved part BCD. The part AB is vertical with A above B, and DE is horizontal. C is the lowest point of the tube and is 0.65 m below the level of DE. A particle is released from rest at A and travels through the tube, leaving it at E with speed 6 m/s (see diagram). Find
the height of A above the level of DE,
the maximum speed of the particle.
Solution
(i) Using conservation of energy, the potential energy at A is converted to kinetic energy at E. The equation is:
\(mgh = \frac{1}{2} m (6)^2\)
Solving for \(h\):
\(gh = \frac{1}{2} (6)^2\)
\(h = \frac{18}{g}\)
Assuming \(g = 9.8 \text{ m/s}^2\),
\(h = \frac{18}{9.8} \approx 1.8 \text{ m}\)
(ii) The maximum speed occurs at the lowest point C. Using energy conservation: