Physics 11 Dynamics By: Bryon Long

Table of Contents Introduction to Dynamics Newton’s First Law of Motion Newton’s Second Law of Motion Law of Universal Gravitation Gravitational Field Strength Normal Force Frictional Force Elastic Force – Hooke’s Law Newton’s Third Law of Motion Momentum Impulse Elastic Collision Inelastic Collision Explosions Conclusion References Formulae

Introduction to Dynamics Dynamics is the study of forces. We will be focusing on the forces that explains why objects move. Sir Isaac Newton formulated three laws of motion and the universal law of gravitation in 1665. These laws are the foundation of Dynamics.

Newton’s First Law of Motion States that: an object will remain at constant velocity unless acted on by an unbalanced force. Also known as the Law of Inertia. Inertia is mass dependent. The larger the mass, the more resistant the object is to the change in motion.

Newton’s First Law of Motion Forces are not needed for an object to maintain constant velocity. Unbalanced forces cause the object to slow down and come to a rest. For example - when the platform an object is traveling on suddenly stops, the object will continue to travel in the direction of motion until an unbalanced force such as a brick wall stops it.

Newton’s First Law of Motion Object is traveling 100 m/s

Newton’s First Law of Motion Object is traveling 100 m/s

Newton’s First Law of Motion Object is traveling 100 m/s

Newton’s First Law of Motion Object is traveling 100 m/s

Newton’s First Law of Motion Object is traveling 100 m/s

Newton’s First Law of Motion Object is traveling 100 m/s

Newton’s First Law of Motion Object is traveling 0 m/s

Newton’s Second Law of Motion States that: the rate of change in an object’s velocity is directly proportional to the net force, and inversely proportional to the object’s mass. The net force is dependent on the change of velocity (acceleration) and mass of the object.

Newton’s Second Law of Motion F = ma F = Force m = mass a = acceleration The unit of force is newton (N)

Newton’s Second Law of Motion What is the force of a 15 kg object travelling with an acceleration of 2.5 m/s². F = ? m = 15 kg a = 2.5 m/s² F = ma = (15 kg)(2.5 m/s²) = 37.5 N

Newton’s Law of Universal Gravitation States that: the gravitational force between two masses is directly proportional to the product of their masses. Unit is in Newton (N)

Newton’s Law of Universal Gravitation Two objects are 2.20 m apart. One object has a mass of 20.0 kg and the other has a mass of 12.0 kg. What is the gravitational force between them?

Gravitational Field Strength Invisible fields that surround masses. Can be scalar or vector. Formula:

Gravitational Field Strength What is the mass of an object if it weighs 50.0 N near the earth’s surface? g = 9.8 m/s² Fg = 50.0 N m = ?

Normal Force On a flat surface, the normal force is equal to the force due to gravity (Fg). Fn = Fg = ma Fn Fg

Frictional Force Forces that oppose motion. Dependent on the normal force (Fn) and the nature of the two surfaces. Is NOT dependent on surface area.

Frictional Force Ff = μFn Ff = Frictional force Fn = Normal force μ = Coefficient of friction Ff Fn

Frictional Force A 25 kg object is pulled along a horizontal surface at a constant acceleration of 2.5 m/s². The coefficient between the two surfaces is 0.14. What is the frictional force? Ff = ? Fn = ma μ = 0.14 Ff = μFn = μma = (0.14)(25 kg)(2.5 m/s²) = 8.8 N

Elastic Force – Hooke’s Law States that: the elastic force is directly proportional to the distortion of the material. Elastic force is the force an object has to restore itself. The equilibrium position is the position where the material will restore itself to. The distortion is the distance from the equilibrium position to the stretched material.

Elastic Force – Hooke’s Law Equilibrium Position Distortion Clamp Elastic Material

Elastic Force – Hooke’s Law Formula: Fs = -kx Fs = elastic force k = spring constant – unit : N/m x = distortion - represents the direction of the elastic force. Unit is in newton(N)

Elastic Force – Hooke’s Law A 2.50 kg mass on a spring is 0.300m from its equilibrium position. If the spring constant is 32.0 N/m, what is the elastic force acting on the mass? Fs = ? k = 32.0 N/m x = 0.300 m Fs = -kx = -(32.0 N/m)(0.300 m) = -9.60 N

Newton’s Third Law of Motion States that: for every action (force) there is an equal but opposite reaction (force). For an example - a book is placed on a table. The book exerts a force (due to gravity) on the table. An equal force is also exerted from the table to the book. Both forces cancel out which explains why the book does not accelerate.

Momentum Law of Conservation of Momentum: momentum will remain constant unless an unbalanced force acts upon it. Momentum is the product of mass and velocity. P = mv Unit is in kg · m/s.

Momentum Calculate the moment of a 15.3 kg object traveling and a velocity of 30.0 m/s east. P = ? m = 15.3 kg v = 30.0 m/s P = mv = (15.3 kg)(30.0 m/s) = 459 kg · m/s

Impulse The Change in momentum. Δp = mΔv Unit is in kg · m/s

Impulse A net force of 31.6 N south is used to accelerate a 15.0 kg object uniformly from rest to 10.0 m/s. What is the change in momentum? Δp = ? m = 20.0 kg Δv = 10.0 m/s Δp = mΔv = (20.0 kg)(10.0 m/s) = 300 kg · m/s

Elastic Collision Collision of two objects. Do not stick. A 0.50 kg ball is travelling east at a velocity of 9.5 m/s when it collides with a 0.35 kg object travelling west at a velocity of 7.0 m/s. After the collision the 0.50 kg ball is travelling west at a velocity of 4.0 m/s. What is the velocity of the 0.35 kg ball after collision?

Elastic Collision Before collision 1 2 m = 0.50 kg v = 9.5 m/s p = 4.75 kg·m/s m = 0.35 kg v = -7.0 m/s p = -2.45 kg·m/s Pbefore = 2.30 kg·m/s

Elastic Collision After collision 1 2 m = 0.50 kg v = -4.0 m/s p = -2.0 kg·m/s m = 0.35 kg v = ? p = 4.3 kg·m/s Pafter = 2.30 kg·m/s

Elastic Collision After collision 1 2 m = 0.35 kg v = ? p = 4.3 kg·m/s Pafter = 2.30 kg·m/s

Inelastic Collision Two objects collide. Both objects stick to each other (masses are added). A 2.1 x 10² kg car is travelling east at a velocity of 52 km/h collides with a 3.1 x 10² kg car travelling west at a velocity of 20 km/h. During collision, both cars lock together. What is the velocity of the locked cars after the collision?

Inelastic Collision Before collision 1 2 m = 2.1 x 10² kg v = 52 km/h p = 10920 kg · km/h m = 3.1 x 10² kg v = -20 km/h p = -6200 kg · km/h Pbefore = 4720 kg · km/h

Inelastic Collision After collision 1+2 m = 5.2 x 10² kg (add both masses) v = ? km/h p = 4720 kg · km/h Pafter = 4720 kg · km/h

Explosions Objects are combined before explosion and are separated after explosion. A 0.125 kg bullet is fired from a 5.5 kg gun. If the velocity of the bullet is 325 m/s, what is the recoil velocity of the gun?

Explosions Before collision 1+2 m = 5.625 kg v = 0 m/s p = 0 kg · m/s Pbefore = 0 kg · m/s

Explosions After explosion 1 2 m = 5.5 kg v = ? m/s p = -40.625 kg · km/h m = 0.125 kg v = 325 m/s p = 40.625 kg · m/s Pafter = 40.625 kg · m/s

Explosions After explosion 1 2 m = 5.5 kg v = ? m/s p = -40.625 kg · km/h Pafter = 40.625 kg · m/s

Conclusion 1st law of motion: if the body is at rest, it will remain at rest. If the body is moving with a constant velocity, it will continue to do so. 2nd law of motion: ΣF = ma 3rd law of motion: For every reaction there is an equal and opposite reaction. Hooke’s Law: Fs = -kx

Conclusion Normal force: Fn = Fg Frictional force: Ff = μFn Law of universal gravitation: Gravitational field strength:

Conclusion Momentum: P = mv, momentum is conserved. Impulse: Δp = mΔv = Fnett, change in momentum Elastic: Collisions that do not stick. Inelastic: Collisions that stick. Explosions: Separate after explosion.

References Physics 11 Student Notes and Problems (SNAP) Fundamentals of Physics, Halliday Resnick, Third Edition