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Physics - Forces and Motion (Kinematics and Newton's Laws)

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Kinematics is the study of motion without considering its causes, focusing on parameters like displacement (ss), velocity (vv), and acceleration (aa).

Scalars are quantities with magnitude only (e.g., distance, speed), whereas Vectors have both magnitude and direction (e.g., displacement, velocity, force).

Instantaneous velocity is the velocity at a specific moment in time, defined as the derivative of displacement with respect to time: v=dsdtv = \frac{ds}{dt}.

Acceleration is the rate of change of velocity over time: a=ΔvΔta = \frac{\Delta v}{\Delta t}.

Newton's First Law (Inertia): An object will remain at rest or move with constant velocity unless acted upon by a non-zero resultant force (Fnet=0F_{net} = 0).

Newton's Second Law: The resultant force acting on an object is equal to the rate of change of momentum, commonly expressed as F=maF = ma when mass is constant.

Newton's Third Law: If body A exerts a force on body B, body B exerts a force of equal magnitude and opposite direction on body A (FAB=FBAF_{AB} = -F_{BA}).

Weight (WW) is the force of gravity acting on an object: W=mgW = mg, where g9.81 m/s2g \approx 9.81 \text{ m/s}^2 on Earth.

📐Formulae

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2=u2+2asv^2 = u^2 + 2as

s=(u+v)2ts = \frac{(u + v)}{2}t

Fnet=maF_{net} = m \cdot a

p=mvp = m \cdot v

W=mgW = m \cdot g

💡Examples

Problem 1:

A car travelling at 20 m/s20 \text{ m/s} comes to a complete stop over a distance of 50 m50 \text{ m} under uniform deceleration. Calculate the acceleration of the car.

Solution:

a=4 m/s2a = -4 \text{ m/s}^2

Explanation:

We use the kinematic equation v2=u2+2asv^2 = u^2 + 2as. Here, u=20 m/su = 20 \text{ m/s}, v=0 m/sv = 0 \text{ m/s} (stop), and s=50 ms = 50 \text{ m}. Rearranging for aa: 0=202+2(a)(50)0=400+100a100a=400a=4 m/s20 = 20^2 + 2(a)(50) \Rightarrow 0 = 400 + 100a \Rightarrow 100a = -400 \Rightarrow a = -4 \text{ m/s}^2.

Problem 2:

A wooden crate of mass 15 kg15 \text{ kg} is pushed across a floor with a horizontal force of 50 N50 \text{ N}. If the force of friction is 20 N20 \text{ N}, find the acceleration of the crate.

Solution:

a=2 m/s2a = 2 \text{ m/s}^2

Explanation:

First, calculate the net force: Fnet=FappliedFfriction=50 N20 N=30 NF_{net} = F_{applied} - F_{friction} = 50 \text{ N} - 20 \text{ N} = 30 \text{ N}. Using Newton's Second Law (F=maF = ma), we find a=Fnetm=30 N15 kg=2 m/s2a = \frac{F_{net}}{m} = \frac{30 \text{ N}}{15 \text{ kg}} = 2 \text{ m/s}^2.