Given final velocity v = 0 m/s (at rest at the highest point), time t = 4 s, and acceleration due to gravity a = -g = -10 m/s2 (taking upward as positive). We use the ki

Given final velocity $v = 0 \text{ m/s}$ (at rest at the highest point), time $t = 4 \text{ s}$, and acceleration due to gravity $a = -g = -10 \text{ m/s}^2$ (taking upward as positive). We use the kinematic equation $v = u + at$.

The velocity of projection is $\boxed{40 \text{ m/s}}$.

5 (a)
Explain the term forbidden gap.

The forbidden gap (or band gap) is an energy range in a solid where no electron energy states can exist. It separates the valence band (where electrons are bound) from the conduction band (where electrons are free to move).

5 (b)
Name the two charge carriers in semiconductors.

The two charge carriers in semiconductors are electrons and holes.

6 (a)
An electron of mass $9.1 \times 10^{-31} \text{ kg}$ has a velocity of $1 \times 10^6 \text{ m s}^{-1}$. Determine the de-Broglie wavelength of the electron. [$h = 6.6 \times 10^{-34} \text{ J s}$]

The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant, $m$ is mass, and $v$ is velocity.

The de-Broglie wavelength of the electron is $\boxed{7.25 \times 10^{-10} \text{ m}}$.

7 (a)
What is a magnetic material?

A magnetic material is a substance that can be magnetized or is strongly attracted to a magnet. These materials contain magnetic domains that can align in the presence of an external magnetic field.

7 (b)
Describe two ways of demagnetizing a magnet.

Two ways to demagnetize a magnet are:

• Heating: Heating the magnet above its Curie temperature causes the random thermal motion of atoms to disrupt the alignment of magnetic domains.
• Hammering/Mechanical shock: Repeatedly dropping or striking the magnet can disorient the magnetic domains, causing them to lose their organized alignment.

8 (a) (i)
Define work.

Work in physics is done when a force causes a displacement of an object in the direction of the force. It is a scalar quantity, calculated as the product of the force and the distance moved in the direction of the force.

8 (a) (ii)
A force F acts on a body of mass m, causing the body to accelerate at a m s⁻². The body travels through a distance s m in time t s in the direction of the force. Show that the work done is equal to change in the kinetic energy of the body.

Work done $W = F \times s$. From Newton's second law, $F = ma$.
So, $W = (ma)s$.

For constant acceleration, the kinematic equation relating final velocity ($v$), initial velocity ($u$), acceleration ($a$), and displacement ($s$) is $v^2 = u^2 + 2as$.
Rearranging for $s$: $s = \frac{v^2 - u^2}{2a}$.

Since $\frac{1}{2}mv^2$ is the final kinetic energy ($KE_f$) and $\frac{1}{2}mu^2$ is the initial kinetic energy ($KE_i$), we have:

This shows that the work done on the body is equal to the change in its kinetic energy.

8 (b) (i)
Define kinetic friction.

Kinetic friction is the force that opposes the relative motion between two surfaces that are sliding against each other. It acts parallel to the surfaces and in the direction opposite to the motion.

8 (b) (ii)
State two factors which affect kinetic friction.

Two factors that affect kinetic friction are:

• The nature of the surfaces in contact (e.g., their roughness or smoothness).
• The normal force pressing the two surfaces together.