Question: PAPER 2
ESSAY
[60 marks]
Answer eight questions in all; five questions from PART I and three questions from PART II.
PART I
[15 marks]
Answer any five (5) questions from this part. All questions carry equal marks (3 marks)
1. Define Projectile Motion and explain why a flying plane is not considered a projectile motion.
2. A body is projected vertically upwards with a velocity of 30 m/s. Calculate the
i. Maximum height reached
ii. Time taken to reach the maximum height
3. Explain the specific heat capacity of a material and state its unit.
4. A 1 kg block of pure ice at -10 °C is heated until the pure ice melts completely into pure water at its ice point. Calculate the total heat energy used in this process.
[specific heat capacity of ice = 2100 J/kgK and specific latent heat of fusion of ice = 3.36 × 10^5 J/kg]
5. Define these terms as applied in wave motion.
i. Frequency
ii. Period
iii. Wavelength
6. A progressive wave completes 20 oscillations in 0.5 s. Find the wave’s frequency and period.
7. Two equal charges placed 0.2 m apart experience a repulsive force of 2.7 N in vacuum. Calculate the magnitude of each charge. [k = 9 × 10^9 Nm^2/C^2]
8. Show that the work done W, by two equal charges Q, separated by a distance d in a vacuum, is given by:
W = k Q^2 / d
9. Describe the structure of the atom according to Neil Bohr’s model.
10. The de-Broglie wavelength of an electron accelerated through a potential difference is 1.0 × 10^-10 m.
Calculate the
i. Momentum of the electron
ii. Velocity of the electron
[mass of electron, m = 9.1 × 10^-31 kg and Planck’s constant, h = 6.63 × 10^-34 Js]
1. **Problem:** A body is projected vertically upwards with a velocity of $30$ m/s. Calculate:
i. Maximum height reached
ii. Time taken to reach the maximum height
2. **Formula and rules:**
- The maximum height $H$ reached by a projectile launched vertically upwards with initial velocity $u$ under gravity $g$ is given by:
$$H = \frac{u^2}{2g}$$
- The time $t$ to reach maximum height is:
$$t = \frac{u}{g}$$
- Here, $g = 10$ m/s$^2$ (approximate acceleration due to gravity).
3. **Calculations:**
i. Maximum height:
$$H = \frac{30^2}{2 \times 10} = \frac{900}{20} = 45 \text{ m}$$
ii. Time to reach maximum height:
$$t = \frac{30}{10} = 3 \text{ s}$$
4. **Explanation:**
- The body rises until its velocity becomes zero at the maximum height.
- The formulas come from the equations of motion under constant acceleration.
**Final answers:**
- Maximum height reached is $45$ m.
- Time taken to reach maximum height is $3$ s.