2021 - JAMB Physics Past Questions and Answers - page 3
A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?
50J
100J
150J
200J
Given:
m\(_{1}\) = 0.05kg, u\(_{1}\) = 200ms\(^{-1}\), m\(_{2}\) = 0.95kg
K.E = \(\frac{1}{2}\)m\(_{r}\)v\(^{2}\)
m\(_{1}\)u\(_{1}\) = v(m\(_{1}\) + m\(_{2}\)) [law of conversation of momentum]
v = \(\frac{0.05 \times 200}{0.05 + 95}\) = 10ms\(^{-1}\)
K.E = \(\frac{1}{2}\)(1)10\(^{2}\) = 50J
A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms\(^{-1}\) from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is? (Take g = 10ms\(^{-2}\))
5J
10J
15J
20J
Given:
m = 0.1kg
u = 10ms\(^{-1}\)
h = 10m
g = 10ms\(^{-2}\)
At h\(_{max}\); v = 0
\(\not v\) = \(\not u\) = \(\not gt\)
\(\not 0\) = \(\not w\) - \(\not 10t\)
v\(^{2}\) = u\(^{2}\) - 2gh
0\(^{2}\) = 10\(^{2}\) - 2 x 10h
20h = 100
h = 5m
Total Height = 10 + 5
= 15m
Total Energy = mghr = 0.1 x 10 x1 5
= 15J
A car of mass 800kg attains a speed of 25m/s in 20secs. The power developed in the engine is?
1.25 x 10\(^{4}\)W
2.50x 10\(^{4}\)W
1.25 x 10\(^{6}\)W
2.50 x 10\(^{6}\)W
Power, P = \(\frac{mv^{2}}{t} = \frac{800 \times 25 \times 25}{20}\)
2.5 x 10\(^{4}\)W
When the breaks in a car are applied, the frictional force on the tyres is...?
a disadvantage because it is in the direction of the motion of the car
a disadvantage because it is in the opposite direction of the motion of the car
an advantage because it is in the direction of the motion of the car
an advantage because it is in the opposite direction of the motion of the car
The frictional force opposes the motion. Friction braking is the most commonly used braking method in modern vehicles. It involves the conversion of kinetic energy to thermal energy by applying friction to the moving parts of a system. The friction force resists motion and in turn generates heat, eventually bringing the velocity to zero. More.
If the stress on a wire is 10\(^{7}\)NM\(^{-2}\) and the wire is stretched from its original length of 10.00m to 10.05m. The young's modulus of the wire is?
5.0 x 10\(^{-4}\)Nm\(^{-2}\)
5.0 x 10\(^{-5}\)Nm\(^{-2}\)
2.0 x 10\(^{-8}\)Nm\(^{-2}\)
2.0 x 10\(^{9}\)Nm\(^{-2}\)
Young's modulus (E) is a property of the material that tells us how easy it can stretch and deform and is defined as the ratio of tensile stress (σ) to tensile strain (ε). Where stress is the amount of force applied per unit area (σ = F/A) and strain is extension per unit length (ε = dl/l). More.
E = \(\frac{stress}{strain}\)
strain = \(\frac{e}{l} = \frac{0.05}{10} = \frac{10^{7}}{0.005}\)
2 x 10\(^{9}\)Nm\(^{-2}\)
A solid weighs 10.00N in air, 6N when forcefully immersed in water, and 7.0N when fully immersed in a liquid, X. Calculate the relative density of the liquid X.
\(\frac{5}{3}\)
\(\frac{4}{3}\)
\(\frac{3}{4}\)
\(\frac{7}{10}\)
Relative Density = \(\frac{\text {weigh in air - weigh in liquid X}}{\text {weigh in air - weigh in water}}\)
= \(\frac{10 - 7}{10 - 6}
= \frac{3}{4}\)
When the temperature of a liquid increases, its surface tension
decreases
increases
remains constant
increases then decreases
An increase in temperature lowers the net force of attraction among molecules and hence decreases surface tension; soap decreases the pull of surface tension.
A gas at a volume of V\(_{0}\) in a container at pressure p\(_{0}\) is compressed to one-fifth of its volume. What will be its pressure if the magnitude of its original temperature T is constant?
\(\frac{p_{0}} {5}\)
\(\frac{4p_{0}}{5}\)
p\(_{0}\)
5P\(_{0}\)
p\(_{1}\) = P\(_{0}\), V\(_{1}\) = V\(_{0}\), V\(_{2}\) = \(\frac{1}{5}\)V\(_{0}\) P\(_{2}\) = ?
P\(_{1}\)V\(_{1}\) = P\(_{2}\)V\(_{2}\)
P\(_{2}\) = \(\frac{P_{1}V_{1}}{V_{1}}\)
= \(\frac{\not P_{o}V_{o}}{\frac{1}{5}P_{o}}\)
= 5\(\not P_{o}\)
A piece of substance of specific heat capacity 450Jkg\(^{-1}\)k\(^{1}\) falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms\(^{-2}\)]
\(\frac{2}{9}\)ºC
\(\frac{4}{9}\)ºC
\(\frac{9}{2}\)ºC
\(\frac{9}{4}\)ºC
According to the law of conservation of energy, energy states that energy can neither be created nor destroyed - only converted from one form of energy to another.
mc\(\bigtriangleup\theta\) = mgh
\(\bigtriangleup\theta\) = \(\frac{gh}{c}
= \frac{10 \times 20}{450}
= \frac{4}{9}\)ºC
I. A liquid boils when its saturated vapor pressure is equal to the external pressure
II. Dissolved substances in pure water lead to an increase in the boiling point.
III. When the external pressure is increased, the boiling point increases.
IV. Dissolved substances in pure water decreases the boiling point
Which of the above combinations are peculiarities of the boiling point of a liquid?
I, II and III only
I, II, III, and IV
I, II, and IV
II, III, and IV
The boiling point of a liquid depends on temperature, atmospheric pressure, and the vapour pressure of the liquid. Boiling begins when the atmospheric pressure is equal to the vapor pressure of the liquid.