Question 5.2:
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
a) during its upward motion,
b) during its downward motion,
c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms-2 = 0.5 N.
(b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards).
(c) The pebble is not at rest at the highest point but has a horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at 45° because no other acceleration except ‘g’ is acting on the pebble.
Question 5.3:
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