Vectors and Motion in Two Dimensions: Practice Questions (Physics 1 1.5)

Answer: (A). The angle is measured from the +x (east) axis, so the east component is next to it: v_x = 12 cos 30° = +10.4 m/s. The north–south component is opposite the angle: 12 sin 30° = 6.0 m/s, and it points south, so v_y = −6.0 m/s. Check: √(10.4² + 6.0²) = 12 m/s.

• (B) swaps sin and cos, as if the angle were measured from the y-axis.
• (C) gets the sizes right but ignores that south is the −y direction.
• (D) splits the 12 m/s equally between the axes. Components only add to the magnitude through Pythagoras, never by simple sharing.

Answer: (C). At the top, only the vertical component v_y is zero. The horizontal component never changes (a_x = 0), so the ball is still moving sideways at v_x0. Gravity still acts, so a_y = −9.8 m/s² throughout the flight.

• (A) makes both classic errors: it forgets v_x, and it confuses “v_y = 0” with “no acceleration”.
• (B) forgets the horizontal motion. Only a ball thrown straight up is momentarily at rest at the top.
• (D) has the velocity right, but if the acceleration were zero at the top, v_y would stay zero and the ball would never come down.

Answer: (B). The fall time depends only on the height: h = ½ g t², so t = √(2h / g) and t is proportional to √h. Four times the height gives √4 = 2 times the time. The horizontal distance is x = v_x0 t, so it is multiplied by 2 (speed) × 2 (time) = 4.

• (A) accounts for the doubled speed but forgets that the higher table also gives a longer fall time.
• (C) treats the fall time as proportional to h (×4) instead of √h (×2).
• (D) squares both factors, as if distance depended on v² and on h. That is not what x = v_x0 √(2h / g) says.

(a)

1. Move 1: A_x = +3.0 m, A_y = 0.
2. Move 2: the angle is measured from north (the y-axis), so the y-component is next to it. B_x = −4.0 sin 30° = −2.0 m (west). B_y = +4.0 cos 30° = +3.46 m.
3. Resultant: R_x = 3.0 − 2.0 = +1.0 m; R_y = 0 + 3.46 = +3.46 m.
4. Size: R = √(1.0² + 3.46²) = 3.6 m. Both components are positive, so the direction is tan⁻¹(3.46 ÷ 1.0) = 74° north of east.

(b) Average velocity = 3.61 m ÷ 12 s = 0.30 m/s at 74° north of east. Average speed = (3.0 + 4.0) m ÷ 12 s = 0.58 m/s.

Suggested mark points (4): 1 for the components of move 2 with correct signs (cos for the north component); 1 for adding components; 1 for 3.6 m at 74° north of east; 1 for both averages, with the average velocity given a direction.

Common error: adding the lengths, 3.0 + 4.0 = 7.0 m. That is the distance travelled, not the displacement.

(a) Vertical motion: v_y0 = 0, y₀ = 25 m, a_y = −9.8 m/s². It lands when y = 0: 0 = 25 − 4.9 t², so t = √(25 ÷ 4.9) = 2.26 s (2.3 s to 2 s.f.).

(b) Horizontal motion at constant velocity: x = 8.0 m/s × 2.259 s = 18 m.

(c) v_x = 8.0 m/s (unchanged). v_y = −9.8 × 2.259 = −22.1 m/s. Size = √(8.0² + 22.1²) = 24 m/s (23.5 m/s). Direction: tan⁻¹(22.1 ÷ 8.0) = 70° below the horizontal.

Suggested mark points (5): 1 for using only the vertical motion to find t; 1 for 2.3 s; 1 for 18 m; 1 for v_y = −22 m/s with v_x unchanged; 1 for 24 m/s at 70° below the horizontal.

Common error: dividing height by horizontal speed, 25 ÷ 8.0 = 3.1 s. The horizontal speed has nothing to do with how long the stone takes to fall.

(a) Release the ball from rest at the same marked point on the ramp each time, without pushing it, and keep the ramp fixed to the table so its angle and exit edge do not change. Measure h with a metre rule held vertically. Use a plumb line to mark the point on the floor directly below the edge, and carbon paper to record each landing spot. Repeat each height several times and average d.

(b) From the vertical motion, h = ½ g t², so t = √(2h / g). Horizontally, d = v_x0 t = v_x0 √(2 / g) × √h. If v_x0 is constant, d is proportional to √h. So plot d (vertical axis, m) against √h (horizontal axis, m^½). The points should lie on a straight line through the origin with slope v_x0 √(2 / g).

(c)

Slope from the first and last points = (0.678 − 0.303) ÷ (1.000 − 0.447) = 0.678 m^½. Since √(2 / g) = √(2 ÷ 9.8) = 0.452 s/m^½, the launch speed is v_x0 = 0.678 ÷ 0.452 = 1.5 m/s.

(d) The data support the claim. The ratio d ÷ √h is 0.678 m^½ for every row (to within rounding), so the points lie on one straight line through the origin. A constant ratio means a constant launch speed. By contrast, d ÷ h falls from 1.5 to 0.68, so d is not proportional to h.

Alternative method. Plotting d² against h also gives a straight line through the origin, with slope 2v_x0² / g = 0.46 m. This earns full credit if the slope is linked correctly to v_x0. A slope from a best-fit line through all five points (0.678 m^½) is better practice than a two-point slope.

(a) Components: v_x0 = v₀ cos θ and v_y0 = v₀ sin θ. Vertically, the ball returns to y = 0 when 0 = v_y0 t − ½ g t², so t = 2v₀ sin θ / g. Horizontally, R = v_x0 t, so R = 2v₀² sin θ cos θ / g.

(b)

(c) The claim is incorrect. Ball Q does stay in the air longer, 1.73 times as long as P. But it moves across more slowly: its horizontal component is 7.50 m/s, not 13.0 m/s. Range depends on the product of time and horizontal speed. In R = 2v₀² sin θ cos θ / g, swapping sin θ and cos θ (30° and 60°) leaves the product unchanged, so both balls land 19.9 m away. Q goes three times higher (8.61 m against 2.87 m), but not farther.

Using the identity 2 sin θ cos θ = sin 2θ is fine but not needed. An answer that reaches the conclusion from the numbers in (b) alone, with the trade-off explained, earns the last point.