An object remains in a state of rest, or of uniform motion in a straight line, unless acted upon by a net external force. This is also called the law of inertia, since it essentially defines inertia — an object's inherent tendency to resist any change in its state of motion.
Inertia is directly related to mass — a more massive object has greater inertia, meaning it more strongly resists changes to its velocity (whether starting from rest, stopping, or changing direction). This is why a loaded truck is much harder to start moving or stop than an empty bicycle.
Explain, using the first law, why passengers standing in a bus lurch forward when the bus suddenly stops.
Before braking, both the bus and the passengers' bodies are moving forward at the same velocity. When the bus suddenly stops, an external force (friction from the brakes) acts on the bus itself, but the passenger's body has no such force acting on it directly — by the law of inertia, the passenger's body tends to continue moving forward at its original velocity, causing it to lurch forward relative to the now-stopped bus.
Momentum ($\vec{p} = m\vec{v}$) is a vector quantity representing the "quantity of motion" an object possesses, combining both its mass and velocity.
The rate of change of momentum of an object is directly proportional to the net external force applied, and this change occurs in the direction of the applied force.
For constant mass, this simplifies to the familiar form: $\vec{F} = m\vec{a}$.
A force of 20 N acts on a mass of 4 kg, initially at rest. Calculate the acceleration produced and the momentum after 5 seconds.
$a = F/m = 20/4 = 5\ \text{m/s}^2$
$v = u + at = 0 + (5)(5) = 25\ \text{m/s}$; Momentum $= mv = (4)(25) = 100\ \text{kg m/s}$
Two masses, 3 kg and 5 kg, experience the same net force of 15 N. Compare their resulting accelerations.
$a_1 = F/m_1 = 15/3 = 5\ \text{m/s}^2$; $a_2 = F/m_2 = 15/5 = 3\ \text{m/s}^2$
The lighter mass (3 kg) experiences a greater acceleration than the heavier mass (5 kg) under the same applied force — directly illustrating the inverse relationship between mass and acceleration for a fixed force.
Impulse is the product of force and the (typically short) time interval over which it acts: $\vec{J} = \vec{F} \times \Delta t$. From the second law, impulse equals the total change in momentum produced: $\vec{J} = \Delta \vec{p}$ — this is especially useful for analysing collisions and impacts, where force varies rapidly and isn't easily measured directly, but the change in momentum can be calculated.
A ball of mass 0.5 kg moving at 10 m/s is brought to rest by a wall in 0.02 s. Calculate the impulse delivered and the average force exerted by the wall.
$J = \Delta p = m(v_f - v_i) = 0.5(0 - 10) = -5\ \text{kg m/s}$ (magnitude 5 kg m/s, directed opposite to the ball's initial motion)
$F_{avg} = J/\Delta t = 5/0.02 = 250\ \text{N}$
Pulling the hands back while catching increases the time over which the ball's momentum changes to zero. Since impulse (= change in momentum) is fixed for a given catch, increasing $\Delta t$ correspondingly decreases the average force experienced by the hands ($F = J/\Delta t$), making the catch far less painful and reducing injury risk.
For every action, there is an equal and opposite reaction — when object A exerts a force on object B, object B simultaneously exerts an equal-magnitude, oppositely-directed force back on object A.
Action and reaction forces always act on two different objects, and therefore never cancel each other out for the purpose of analysing either object's individual motion — a common misconception is thinking they cancel and "nothing should ever move," but each force must be considered only in the context of the specific object it acts upon.
Explain, using Newton's third law, how a rocket propels itself forward in the vacuum of space, where there's no air to "push against."
The rocket engine expels hot exhaust gases backward at high speed (the "action" force, exerted by the rocket on the gas). By Newton's third law, the expelled gas simultaneously exerts an equal and opposite "reaction" force on the rocket, pushing it forward. This process doesn't require air or any external medium to push against — it works purely through the mutual interaction between the rocket and its own expelled exhaust, which is why rockets can accelerate even in the vacuum of space.
In the absence of any net external force, the total momentum of a system of interacting objects remains constant — this follows directly from Newton's third law, since internal forces between objects in the system always occur in equal-and-opposite pairs, cancelling out when the total momentum of the whole system is considered.
where $u_1, u_2$ are initial velocities and $v_1, v_2$ are final velocities of two colliding objects.
A gun of mass 5 kg fires a bullet of mass 0.02 kg with a velocity of 400 m/s. Calculate the recoil velocity of the gun.
Initial momentum of the system (gun + bullet, at rest) $= 0$.
By conservation of momentum: $m_{gun}v_{gun} + m_{bullet}v_{bullet} = 0$
$(5)v_{gun} + (0.02)(400) = 0 \Rightarrow v_{gun} = -\dfrac{8}{5} = -1.6\ \text{m/s}$
The negative sign shows the gun recoils in the direction opposite to the bullet's motion, with a speed of 1.6 m/s.
In a perfectly elastic collision, both momentum and kinetic energy are conserved — the objects separate after collision without any permanent deformation or heat loss. In an inelastic collision, momentum is still conserved, but kinetic energy is not — some is converted to heat, sound, or deformation energy. A perfectly inelastic collision is the extreme case where the colliding objects stick together and move with a common final velocity.
A 4 kg ball moving at 6 m/s collides elastically with a stationary 4 kg ball. Given that for an elastic collision between equal masses, velocities are simply exchanged, find the final velocity of each ball.
For a perfectly elastic collision between two equal masses where one is initially at rest, the moving ball comes to rest and the previously stationary ball moves off with the first ball's original velocity.
Final velocity of first ball $= 0\ \text{m/s}$; Final velocity of second ball $= 6\ \text{m/s}$ (momentum conserved: $4(6)+4(0) = 4(0)+4(6) = 24$, consistent on both sides).
Friction is the force that opposes relative motion (or the tendency of relative motion) between two surfaces in contact. Static friction acts on an object at rest, preventing it from starting to move, and adjusts itself (up to a maximum limit) to exactly match any applied force. Kinetic friction acts on an object already in relative motion, and has an essentially constant value (for given surfaces) regardless of speed.
where $\mu_s$, $\mu_k$ are the coefficients of static and kinetic friction respectively, and $N$ is the normal force. Generally, $\mu_s > \mu_k$ for the same pair of surfaces — meaning it takes more force to start an object moving than to keep it moving.
A 10 kg block rests on a horizontal floor, with a coefficient of static friction of 0.4. Calculate the maximum static friction force and determine if a 30 N horizontal push will move the block. (g = 10 m/s²)
$N = mg = (10)(10) = 100\ \text{N}$
$f_{s,max} = \mu_s N = (0.4)(100) = 40\ \text{N}$
Since the applied force (30 N) is less than the maximum static friction (40 N), static friction adjusts to exactly balance the push (also 30 N), and the block does not move.
Rolling friction occurs when an object rolls (rather than slides) over a surface, and is typically much smaller than both static and kinetic (sliding) friction for the same surfaces. This is precisely why wheels, ball bearings, and rollers are so widely used in machinery and vehicles — converting sliding motion into rolling motion drastically reduces frictional energy loss.
For an object on an inclined plane at angle $\theta$, the object's weight ($mg$) is resolved into two components: $mg\cos\theta$ (perpendicular to the incline, balanced by the normal force) and $mg\sin\theta$ (along the incline, tending to slide the object down).
The angle of repose is the maximum incline angle at which an object placed on the surface just remains at rest without sliding, satisfying $\tan\theta = \mu_s$ — beyond this angle, the component of gravity along the incline exceeds the maximum available static friction, and the object begins to slide.
A block just begins to slide on an inclined plane when the angle of inclination is 30°. Calculate the coefficient of static friction between the block and the incline.
At the angle of repose: $\mu_s = \tan\theta = \tan30° = \dfrac{1}{\sqrt{3}} \approx 0.577$
A block of mass 5 kg slides down a frictionless incline at 37° (sin37° ≈ 0.6, cos37° ≈ 0.8). Calculate its acceleration down the slope. (g = 10 m/s²)
On a frictionless incline, the only force component along the slope is $mg\sin\theta$, giving acceleration:
$a = g\sin\theta = (10)(0.6) = 6\ \text{m/s}^2$
On a flat (unbanked) road, friction alone must supply the entire centripetal force needed for a vehicle to turn — this limits the maximum safe speed and becomes unreliable in wet or slippery conditions. Banking the road (tilting it at an angle $\theta$) allows a component of the normal force to also contribute towards the required centripetal force, reducing dependence on friction.
This gives the ideal speed $v$ for which a vehicle can safely take a banked turn of radius $r$ with no reliance on friction at all — at exactly this speed, the horizontal component of the normal force alone provides the necessary centripetal force.
A circular road of radius 100 m is banked at an angle such that a car can safely travel at 20 m/s without relying on friction. Calculate the banking angle. (g = 10 m/s²)
$\tan\theta = \dfrac{v^2}{rg} = \dfrac{(20)^2}{(100)(10)} = \dfrac{400}{1000} = 0.4$
$\theta = \tan^{-1}(0.4) \approx 21.8°$
An inertial frame is one that is not accelerating (either at rest or moving at constant velocity) — Newton's laws hold true directly within such a frame. A non-inertial frame is one that is itself accelerating (e.g., a braking car, a rotating platform); Newton's laws don't directly apply here unless a pseudo (fictitious) force is introduced.
A pseudo force is an imaginary, "as-if" force introduced in a non-inertial (accelerating) reference frame, equal in magnitude to $ma$ (where $a$ is the frame's own acceleration) but acting in the direction opposite to the frame's acceleration — adding this force allows Newton's second law to be applied consistently even within the accelerating frame, even though this force doesn't correspond to any real physical interaction.
A ball hangs from the ceiling of a car accelerating forward at 5 m/s². Explain, from the perspective of an observer inside the car, why the ball appears to tilt backward, using the concept of pseudo force.
From inside the car (a non-inertial, accelerating frame), a pseudo force of magnitude $ma$ acts on the ball, directed backward (opposite to the car's forward acceleration). This apparent backward force, combined with gravity, causes the ball's string to tilt backward, appearing to an observer inside the car as if some real force is pushing the ball backward — even though an observer standing outside the car (in an inertial frame) would correctly explain the same tilt simply as the string providing the net forward force needed to accelerate the ball along with the car.
The centrifugal force commonly referred to in everyday language (the outward "force" felt when going around a curve) is itself a pseudo force — it only appears when analysing motion from within a rotating (and therefore non-inertial) reference frame, such as sitting inside a car taking a sharp turn. From an inertial (outside, stationary) frame, there is no actual outward force at all — only the real, inward-directed centripetal force is acting, and the sensation of being "pushed outward" is simply the body's inertia resisting the inward centripetal acceleration.