Ramp-to-Trailer Transfer
A trainee at a vocational engineering workshop tests a transfer chute. A 0.50 kg block leaves the chute horizontally at 4.0 m/s from a height of 1.25 m above the floor of a low trailer. The trailer's near edge is initially 1.80 m horizontally from the chute exit, and its flat bed is 2.40 m long. In trial A the trailer is stationary. In trial B it moves directly away from the chute at 1.5 m/s. In trial C it moves toward the chute at 1.5 m/s. The block is small, air resistance is negligible, and g = 9.8 m/s2. A safety rail sits at the far end of the bed, 2.40 m from the near edge. The trainee must decide whether the block lands on the bed in each trial and what launch speed would be needed in trial B. The chute can be adjusted only to speeds between 3.0 and 6.0 m/s.
1. For trial A, calculate the fall time and horizontal distance travelled by the block. Does it land on the trailer bed? Show the comparison with the near and far edges.
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Use vertical motion to find the fall time:
h = 1/2 g t^2, so t = sqrt(2h/g) = sqrt(2 x 1.25 / 9.8) = 0.505 s.
Horizontal distance at 4.0 m/s: x = v t = 4.0 x 0.505 = 2.02 m.
The bed runs from 1.80 m to 1.80 + 2.40 = 4.20 m from the chute exit. Since 1.80 < 2.02 < 4.20, the block lands on the bed, about 0.22 m past the near edge. Trial A succeeds.
2. In trial B, derive the range of launch speeds that would put the block on the bed. Can the existing 4.0 m/s setting work? Respect the chute limit.
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The fall time is still 0.505 s because the height does not change. In that time the trailer moves away by 1.5 x 0.505 = 0.758 m.
Near edge position at landing: 1.80 + 0.758 = 2.558 m. Far edge position at landing: 4.20 + 0.758 = 4.958 m.
Required horizontal speed is distance divided by 0.505 s: minimum = 2.558 / 0.505 = 5.07 m/s. maximum = 4.958 / 0.505 = 9.82 m/s.
So the block lands on the bed for speeds from about 5.07 m/s to 9.82 m/s. The chute is limited to 6.0 m/s, so the usable interval is 5.07 m/s to 6.00 m/s. The existing 4.0 m/s setting fails: it gives only 2.02 m, less than the 2.558 m needed to reach the receding near edge.
3. In trial C, calculate where the block lands relative to the bed. If the chute is reset to a speed that works for trial B, is the safety rail at risk in trial C?
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In trial C the trailer moves toward the chute, so the near edge moves closer by 0.758 m: near edge = 1.80 - 0.758 = 1.042 m. far edge = 4.20 - 0.758 = 3.442 m.
At 4.0 m/s the block travels 2.02 m, so it lands 2.02 - 1.042 = 0.98 m past the near edge, well inside the bed. The safety rail at the far end is not reached.
If the speed is raised to about 5.1 m/s to satisfy trial B, the range becomes 5.1 x 0.505 = 2.58 m. This is still less than the 3.442 m far-edge position, so the rail is not at risk. At the chute maximum of 6.0 m/s, the range is 3.03 m, still short of 3.442 m. A single setting near 5.1-6.0 m/s can therefore work for both moving trials, but it leaves little margin in trial C if the trailer speed increases.