Question

Consider a rock mass that fails on a mountainside at an elevation of 2300 metres. This rock

mass crashes onto the valley floor at an elevation of 1600 metres and comes to an abrupt halt

   on the opposite valley wall at an elevation of 1800 metres. Calculate the maximum velocity

   (in km/hr) at which the rock mass was moving when it first arrived on the valley floor.

   Calculate the minimum velocity (in km/hr) the rock mass was travelling as it began its ascent

   up the opposite valley wall.

The rock mass travelled for a distance of 3 kilometres as it crossed the valley floor. Given

      the range of rock mass velocities, compute how much time citizens living on the valley floor

      would have to get out of the path of this rapidly moving rock mass if they were fortunate

      enough to have witnessed the initial rock failure.

                                        v = d/t  ; where d = 3 km

The computations in Q.1 and Q.2 relate to the physics of the Frank Slide which occurred

    in the morning of April 29, 1903. Eyewitness accounts of this event indicate that the rock mass

    took 90 seconds to cross the valley floor permitting geologists to estimate that the rock mass

    was travelling at a velocity of 110 km/hr (30.5 m/s).

      Account for the differences between your calculated velocities and travel times and those

      based on direct observations of this event.

Answer 1

1. v^2 = u^2 + 2as
= 0 + 2 x 9.81 x 2300
v = 212.4 m/s
= 212.4 x3600 / 1000 = 764.7 kph

PE gain = remaining KE after impact
mgh = 1/2 mv^2 (m cancels out)
9.81 x 200 = 0.5 x v^2
v^2 = 3924
v = 62.64 m/s = 225.5 kph

2. Using s = 1/2 (u+v)t
Average velocity as it moves across the valley = (62.64 + 0) /2 = 31.32 m/s
Time taken = distance / velocity = 1500 / 31.32 = 47.9 seconds
Velocity = distance/ time