Question 1(d) Merit to Excellence

Tane has a toy car track: part is a horizontal banked curve of radius 0.750 m, part is a vertical loop of radius 0.250 m. The car has a mass of 0.120 kg. Assume sideways friction on the tyres is negligible.

The 0.120 kg car approaches the vertical loop of radius 0.250 m. Calculate the speed it must have at the bottom of the loop to go round so that it seems weightless at the top.

The 2023 paper, page 3 ↗ Marking schedule, page 2 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 What does “seems weightless at the top” tell you about the reaction force there?

   It is zero. Gravity alone supplies the centripetal force: mg = mv² / r.

2. 2 So how fast is the car going at the top?

   v_top² = gr = 9.81 × 0.250 = 2.4525 m² s⁻², so v_top = 1.566 m s⁻¹.

3. 3 Between the bottom and the top, where does the car’s kinetic energy go?

   Some becomes gravitational potential energy, because it rises 2r = 0.500 m, and the rest is the kinetic energy it still has at the top. With no friction: ½mv_bottom² = mg(2r) + ½mv_top².

4. 4 Work out the energy at the top.

   mg(2r) = 0.120 × 9.81 × 0.500 = 0.5886 J, and ½mv_top² = ½ × 0.120 × 2.4525 = 0.14715 J: 0.73575 J in all.

5. 5 All of that was kinetic energy at the bottom. Find the speed.

   ½ × 0.120 × v² = 0.73575, so v = √(2 × 0.73575 / 0.120) = √12.2625 = 3.50 m s⁻¹. (In symbols, v_bottom² = 5gr.)

What the marker wanted

• Achieved One of: the total energy at the top written as mgh + ½mv²; F_c = F_g at the top; v² = 2.4525; or v = 1.566 m s⁻¹.
• Merit The speed at the top used to find the kinetic energy there (0.14715 J), or the total energy at the top with one small error.
• Excellence The right speed at the bottom, 3.50 m s⁻¹.

In our words, from NZQA’s 2023 marking schedule.

Common mistake

How did it go?

Got it moves it further out. Review brings it back tomorrow.