Fig. 5.1 shows a coin resting on a piece of stiff card placed over a glass. When the card is flicked sharply sideways, the coin drops into the glass. (a) St...

(a) Newton's first law of motion. [1]

Newton's first law states that an object remains at rest, or continues to move at constant velocity in a straight line, unless acted upon by a resultant external force. The coin is initially at rest, and if no horizontal force acts on it, it will stay at rest even when the card beneath it is removed.

(b) The coin is at rest before the flick. When the card is flicked sharply, the card moves so quickly that friction between the card and the coin has almost no time to act. [1] Without a significant horizontal force, Newton's first law tells us the coin tends to stay at rest in its original position. With the card now gone, the only force on the coin is its weight acting vertically downward, so it drops straight into the glass. [1]

(c) Using \( W = mg \):

\[ W = 0.012 \times 10 = 0.12 \text{ N} \]

[1]

(d) The resultant force is zero. [1]

The coin is stationary, so by Newton's first law, the forces must be balanced. The weight of the coin (0.12 N downward) is exactly balanced by the normal contact force from the card (0.12 N upward).

(e) Once the card is removed, the only force acting on the coin is its weight (gravity pulls it downward and air resistance is ignored):

\[ F_{\text{resultant}} = 0.12 \text{ N downward} \]

[1]

(f) Using \( F = ma \), rearranged to \( a = \frac{F}{m} \):

\[ a = \frac{0.12}{0.012} = 10 \text{ m/s}^2 \text{ downward} \]

This equals \( g \), as expected for free fall with no air resistance. [1]

(g) If the card is flicked slowly, it remains in contact with the coin for a longer time. During this longer contact time, friction between the card and the coin has time to act as a sustained horizontal force. [1] This friction accelerates the coin sideways along with the card, so the coin slides off to the side rather than dropping vertically into the glass. [1]