Physics 2014 Jamb Examination

The least possible error in a measurement, often referred to as the absolute error or precision of the instrument, is typically taken as half of the smallest graduation (the least count) of the instrument. A standard meter rule is graduated in millimeters (1.0mm). When taking a reading, the most a careful observer can be 'off' while rounding to the nearest mark is half of that smallest division. Therefore, the least possible error is:\[ \text{Error} = \frac{1}{2} \times 1.0\text{mm} = 0.5\text{mm} \]This accounts for the uncertainty in estimating the position between two millimeter marks.

Physical quantities are divided into scalars and vectors. A scalar quantity, such as temperature, distance, or mass, is completely described by its magnitude (size) alone. A vector quantity requires both magnitude and a specific direction for a complete description. Displacement is defined as the change in position of an object in a particular direction. Unlike distance, which only measures the path length, displacement must specify 'where' relative to the start, making it a vector quantity.

Let's evaluate each quantity:
I. Electrical potential is a scalar quantity; it represents work done per unit charge.
II. Torque is a vector quantity; it describes a rotational force and has a direction (clockwise or anticlockwise).
III. Kinetic Energy is a scalar quantity; energy does not have a direction.
IV. Momentum is a vector quantity; it is the product of mass (scalar) and velocity (vector).
Therefore, only torque and momentum are vectors.

The motion of the wheels of a moving car is a combination of two types of motion. As the wheels turn about their axles, they undergo rotational motion. Simultaneously, because the entire car is moving from one point to another along the road, the wheels also undergo translational motion. This combination is often referred to as 'rolling' motion.

We can use the equations of linear motion to find the final velocity. Given:
Initial velocity (\(u\)) = 0 (since it starts from rest)
Acceleration (\(a\)) = 3ms\(^{-2}\)
Distance (\(s\)) = 24m
We use the formula:\[ v^2 = u^2 + 2as \]\[ v^2 = 0^2 + 2(3)(24) \]\[ v^2 = 6 \times 24 \]\[ v^2 = 144 \]\[ v = \sqrt{144} = 12\text{ms}^{-1} \]

The escape velocity (\(v_e\)) is the minimum speed needed for an object to break free from the gravitational attraction of a massive body. It is calculated using the formula:\[ v_e = \sqrt{2gR} \]Where:
\(g = 9.8\text{ms}^{-2}\) (acceleration due to gravity)
\(R = 6.4 \times 10^{6}\text{m}\) (radius of the earth)\[ v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \]\[ v_e = \sqrt{125.44 \times 10^6} \]\[ v_e = 11.2 \times 10^3\text{ms}^{-1} \]\[ v_e = 11.2\text{kms}^{-1} \]The closest value in the options provided is 11.3kms\(^{-1}\).

First, we must correct a potential confusion: 'weight 8kg' usually implies a mass of 8kg. Weight is a force (\(W = mg\)).
Acceleration due to gravity on Earth (\(g_e\)) is approx 10ms\(^{-2}\).
Weight on Earth (\(W_e\)) = \(m \times g_e = 8 \times 10 = 80\text{N}\).
On the new planet, the gravity (\(g_p\)) is \(\frac{1}{3}g_e\).\[ W_p = m \times g_p = m \times \frac{1}{3}g_e \]\[ W_p = \frac{1}{3} \times (m \times g_e) = \frac{1}{3} \times 80\text{N} = 26.67\text{N} \]Rounding to the nearest whole number, we get 27N.

For an object to be in a state of static equilibrium, two conditions must be met. First, the resultant (vector sum) of all external forces acting on the body must be zero (translational equilibrium). Second, the sum of all torques (moments) acting on the body must be zero (rotational equilibrium). In the case of parallel forces, the first condition ensures that the object does not accelerate in any direction.

When three non-parallel, coplanar forces act on a body and keep it in equilibrium, they must be concurrent. This means that their lines of action must all pass through a single common point. If they did not meet at a point, they would create a resultant torque (moment), which would cause the object to rotate, thereby violating the condition for equilibrium.

According to the principle of conservation of mechanical energy, the potential energy at the maximum height is converted into kinetic energy as the object falls (ignoring air resistance). Just before hitting the ground, all the initial potential energy has been converted to kinetic energy.
Potential Energy (\(PE\)) = \(mgh\)
Given:
Mass (\(m\)) = 20kg
Gravity (\(g\)) = 10ms\(^{-2}\)
Height (\(h\)) = 10m\[ PE = 20 \times 10 \times 10 = 2000\text{J} \]Therefore, the Kinetic Energy (\(KE\)) just before impact is 2000J.