10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
Newton's second law states that the rate of change of momentum of a body is directly proportional to the applied net force, and takes place in the direction of the force.
$F \propto \dfrac{dp}{dt} = \dfrac{d(mv)}{dt}$. For constant mass, $F \propto m\dfrac{dv}{dt} = ma$
Introducing a constant of proportionality $k$ (taken as 1 in SI units by defining the newton appropriately): $F = ma$
Newton's third law states that for every action, there is an equal and opposite reaction — the forces two bodies exert on each other are always equal in magnitude and opposite in direction.
Action-reaction pairs do not cancel because they act on different bodies, not the same body. Forces can only be added/cancelled when analysing the net force on a single object — since action acts on one body and reaction acts on the other, they never appear together in the same body's force equation.
On a frictionless banked road of angle $\theta$, the horizontal component of the normal reaction provides the centripetal force: $N\sin\theta = \dfrac{mv^2}{r}$, while the vertical component balances weight: $N\cos\theta = mg$.
Dividing the two equations: $\tan\theta = \dfrac{v^2}{rg}$
$\therefore v_{max} = \sqrt{rg\tan\theta}$
When pushing at a downward angle, the vertical component of the applied force adds to the object's weight, increasing the normal reaction $N$ — and since friction $f = \mu N$, this increases friction, making it harder to push.
When pulling at an upward angle, the vertical component of the applied force acts opposite to weight, reducing the normal reaction $N$ — this reduces friction, making it comparatively easier to pull the same object.
The law states that in the absence of an external force, the total linear momentum of a system of interacting bodies remains constant.
During any interaction (like a collision), by Newton's third law, the force one body exerts on another is equal and opposite to the force exerted back. These internal forces cause equal and opposite changes in momentum for the two bodies over the same time interval, so their combined momentum change sums to zero — hence total momentum is conserved, provided no external force acts on the system.
Forces on the person: weight $mg$ (downward) and normal reaction $N$ (upward, from the lift floor — this is the apparent weight).
Taking downward as positive, applying Newton's second law: $mg - N = ma$
$\therefore N = m(g - a)$
This is the apparent weight — it is less than actual weight $mg$ whenever the lift accelerates downward, and becomes zero if $a = g$ (free fall, giving a sensation of weightlessness).
Static friction acts between surfaces at rest relative to each other, opposing the tendency of relative motion — it adjusts itself up to a maximum limiting value.
Kinetic friction acts between surfaces already sliding relative to each other, and is generally slightly less than the maximum static friction.
Rolling friction acts when one surface rolls over another (like a wheel) — it is significantly smaller than both static and kinetic friction, since rolling avoids the continuous relative sliding present in kinetic friction.
Impulse is the product of force and the time interval for which it acts — a measure of the total effect of a force over time.
From Newton's second law: $F = \dfrac{dp}{dt} \Rightarrow F\,dt = dp$
Integrating over the time interval $t_1$ to $t_2$: $J = \displaystyle\int F\,dt = p_2 - p_1 = \Delta p$
Thus, impulse equals the total change in momentum — this is especially useful when force varies over a very short time interval, as in a collision.
By pulling their hands back while catching, the cricketer increases the time $\Delta t$ over which the ball's momentum changes to zero.
Since impulse $F\Delta t = \Delta p$ is fixed (the ball's momentum change is the same regardless of catching technique), increasing $\Delta t$ correspondingly decreases the average force $F$ experienced by the hands.
This reduces the risk of injury, compared to catching with a rigid hand (small $\Delta t$), which would produce a much larger, more painful impact force for the same momentum change.
A particle is in equilibrium under several concurrent forces (forces acting through a common point) when the vector sum of all forces acting on it is zero:
$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \cdots = 0$
Equivalently, resolving into components, the sum of components along any two mutually perpendicular directions (e.g., $x$ and $y$) must independently be zero: $\Sigma F_x = 0$ and $\Sigma F_y = 0$. Under these conditions, the particle remains at rest or continues moving with constant velocity.
10 questions · 5 marks each · Detailed step-by-step solutions
Let $m_1 = 6\text{ kg}$ (heavier, moves down) and $m_2 = 4\text{ kg}$ (lighter, moves up).
(a) Acceleration: $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2} = \dfrac{(6-4)(10)}{6+4} = \dfrac{20}{10} = 2\text{ m/s}^2$
(b) Tension: Using the lighter mass: $T - m_2g = m_2a \Rightarrow T = m_2(g+a) = 4(10+2) = 48\text{ N}$
Verification with heavier mass: $m_1g - T = m_1a \Rightarrow T = m_1(g-a) = 6(10-2) = 48\text{ N}$ ✓ Confirmed.
Normal reaction (horizontal surface, no vertical acceleration): $N = mg = 10 \times 10 = 100\text{ N}$
Maximum static friction (limiting friction): $f_{max} = \mu_s N = 0.4 \times 100 = 40\text{ N}$
To just start the block moving, the applied force must equal (or marginally exceed) this maximum static friction.
Minimum force required = 40 N.
On an unbanked road, friction alone provides the centripetal force: $\mu mg = \dfrac{mv^2}{r}$
Simplifying (mass cancels): $v^2 = \mu rg = 0.4 \times 30 \times 10 = 120$
$v = \sqrt{120} \approx 10.95\text{ m/s}$
The maximum safe speed is approximately 10.95 m/s (about 39.4 km/h). Note that the car's mass does not affect this maximum speed.
This is a perfectly inelastic collision — momentum is conserved (no external horizontal force during the brief collision).
$m_1u_1 + m_2u_2 = (m_1+m_2)v$
With $m_1 = 0.02\text{ kg}$, $u_1 = 300\text{ m/s}$, $m_2 = 2\text{ kg}$, $u_2 = 0$:
$(0.02)(300) + 0 = (0.02+2)v$
$6 = 2.02v \Rightarrow v = \dfrac{6}{2.02} \approx 2.97\text{ m/s}$
The common velocity immediately after impact is approximately 2.97 m/s.
Forces along the incline: component of gravity down the incline = $mg\sin\theta$; friction opposes motion, acting up the incline = $\mu mg\cos\theta$ (since $N = mg\cos\theta$).
Net force down the incline: $F = mg\sin\theta - \mu mg\cos\theta = mg(\sin\theta - \mu\cos\theta)$
Acceleration: $a = g(\sin\theta - \mu\cos\theta) = 10(\sin30° - 0.2\cos30°)$
$a = 10(0.5 - 0.2\times0.866) = 10(0.5 - 0.173) = 10(0.327) \approx 3.27\text{ m/s}^2$
The block accelerates down the incline at approximately 3.27 m/s².
Force = rate of change of momentum = $n \times m \times v$, where $n$ = bullets fired per second.
$180 = n \times 0.05 \times 400$
$180 = 20n \Rightarrow n = 9$ bullets per second
Bullets per minute = $9 \times 60 = \mathbf{540\text{ bullets/minute}}$
Normal reaction: $N = mg = 2\times10 = 20\text{ N}$
Kinetic friction: $f = \mu_kN = 0.3\times20 = 6\text{ N}$ (opposing motion)
Net force: $F_{net} = F_{applied} - f = 20 - 6 = 14\text{ N}$
Acceleration: $a = \dfrac{F_{net}}{m} = \dfrac{14}{2} = \mathbf{7\text{ m/s}^2}$
For the hanging mass (3 kg): $m_1g - T = m_1a$
For the mass on the table (5 kg, horizontal, only tension acts horizontally): $T = m_2a$
Adding both equations: $m_1g = (m_1+m_2)a$
$a = \dfrac{m_1g}{m_1+m_2} = \dfrac{3\times10}{3+5} = \dfrac{30}{8} = 3.75\text{ m/s}^2$
Tension: $T = m_2a = 5\times3.75 = \mathbf{18.75\text{ N}}$
For a perfectly elastic collision between equal masses, with one initially at rest, a well-known result is that the velocities are exchanged: the moving ball comes to rest, and the stationary ball moves off with the initial velocity of the first.
After collision: Ball 1 (initially moving) → $v_1' = 0$; Ball 2 (initially at rest) → $v_2' = 10\text{ m/s}$
Momentum check: Before: $0.5(10) + 0.5(0) = 5\text{ kg·m/s}$. After: $0.5(0) + 0.5(10) = 5\text{ kg·m/s}$ ✓
Kinetic energy check: Before: $\frac{1}{2}(0.5)(10)^2 = 25\text{ J}$. After: $\frac{1}{2}(0.5)(10)^2 = 25\text{ J}$ ✓ Both conserved, confirming an elastic collision.
As the monkey climbs upward with acceleration $a$, applying Newton's second law (taking upward as positive):
$T - mg = ma$
At the maximum permissible tension, $T = 250\text{ N}$:
$250 - (20)(10) = 20a$
$250 - 200 = 20a \Rightarrow 50 = 20a \Rightarrow a = 2.5\text{ m/s}^2$
The monkey can climb with a maximum acceleration of 2.5 m/s² without breaking the rope.