State the law of conservation of linear momentum.

Gravel falls vertically onto a moving horizontal conveyor belt.

(i) The gravel falls at a constant rate of 13 kg s^–1 through a height of 1.9 m. Show that the vertical speed of the gravel as it lands on the conveyor belt is about 6 m s^–1.

(ii) The gravel lands on the conveyor belt without rebounding. Calculate the rate of change of the vertical momentum of the gravel.

(iii) Gravel first reaches the belt at t = 0.0 s and continues to fall. Determine the total vertical force that the gravel exerts on the conveyor belt at t = 5.0 s.

The conveyor belt moves with a constant horizontal speed of 1.5 m s^–1. As the gravel lands on the belt, it has no horizontal speed.

(i) Calculate the rate of change of the kinetic energy of the gravel due to its change in horizontal speed.

(ii) Determine the power required to move the conveyor belt at constant speed.

(iii) Outline why the answers to (c)(i) and (ii) are different.

if no external forces act / isolated system;
momentum is constant / (total) momentum before=(total) momentum after;

(i) use of \(v = \sqrt {2gh} \);
6.11ms^–1; (must show calculation to better than 1 sf)

(ii) rate of change of vertical momentum=13×6.11;
79N; (accept answers in the range of 78N to 80N)

(iii) mass accrued=5.0×13=65kg;
weight of this mass (=65×9.8)=637N; (650 from g=10ms^–2)
total force (637+79=)716N; } (allow ECF from (b)(ii) and from incorrect weight)

(i) 14.6Js^–1;

(ii) horizontal momentum gain per second =13×1.5( =19.5kgms^–1);
power required=29.3W;

(iii) additional energy/power required to accelerate gravel (through friction at the surface of the belt) / the gravel has to slip to gain horizontal speed / OWTTE;