Motion & Forces Lesson 3 Action and Reaction Newton’s Third Law Momentum Conservation of Momentum

Newton’s Third Law Newton’s Third Law of Motion When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.

Newton’s Third Law Problem: How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force? Aren’t these “balanced forces” resulting in no acceleration? NO!!!

Newton’s Third Law Explanation: forces are equal and opposite but act on different objects they are not “balanced forces” the movement of the horse depends on the forces acting on the horse

Action and Reaction When a force is applied in nature, a reaction force occurs at the same time. When you jump on a trampoline, for example, you exert a downward force on the trampoline. Simultaneously, the trampoline exerts an equal force upward, sending you high into the air.

Action and Reaction Forces Don’t Cancel According to the third law of motion, action and reaction forces act on different objects. Thus, even though the forces are equal, they are not balanced because they act on different objects.

Action and Reaction Forces Don’t Cancel For example, a swimmer “acts” on the water, the “reaction” of the water pushes the swimmer forward. Thus, a net force, or unbalanced force, acts on the swimmer so a change in his or her motion occurs.

Rocket Propulsion In a rocket engine, burning fuel produces hot gases. The rocket engine exerts a force on these gases and causes them to escape out the back of the rocket. By Newton’s third law, the gases exert a force on the rocket and push it forward.

Newton’s Third Law Action-Reaction Pairs The hammer exerts a force on the nail to the right. The nail exerts an equal but opposite force on the hammer to the left.

Momentum A moving object has a property called momentum that is related to how much force is needed to change its motion. The momentum of an object is the product of its mass and velocity Momentum is given the symbol p and can be calculated with the following equation:

p = mv v Momentum p m Momentum quantity of motion p: momentum (kg·m/s) Use the unit kg·m/s m p v p: momentum (kg·m/s) m: mass (kg) v: velocity (m/s)

Force and Changing Momentum By combining these two relationships, Newton’s second law can be written in this way: In this equation mvf is the final momentum and mvi is the initial momentum

Law of Conservation of Momentum The momentum of an object doesn’t change unless its mass, velocity, or both change. Momentum, however, can be transferred from one object to another. The law of conservation of momentum states that if a group of objects exerts forces only on each other, their total momentum doesn’t change.

Law of Conservation of Momentum The results of a collision depend on the momentum of each object. When the first puck hits the second puck from behind, it gives the second puck momentum in the same direction.

When Objects Collide If the pucks are speeding toward each other with the same speed, the total momentum is zero.

Newton’s Third Law F m Action-Reaction Pairs Both objects accelerate. The amount of acceleration depends on the mass of the object. F m Small mass  more acceleration Large mass  less acceleration

Conservation of Momentum Law of Conservation of Momentum The total momentum in a group of objects doesn’t change unless outside forces act on the objects. pbefore = pafter

Conservation of Momentum Elastic Collision KE is conserved Inelastic Collision KE is not conserved

v Momentum p m p = ? p = mv m = 280 kg p = (280 kg)(3.2 m/s) Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s. GIVEN: p = ? m = 280 kg v = 3.2 m/s WORK: p = mv p = (280 kg)(3.2 m/s) p = 896 kg·m/s m p v

v Momentum p m p = 675 kg·m/s v = p ÷ m m = 300 kg The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg? GIVEN: p = 675 kg·m/s m = 300 kg v = ? WORK: v = p ÷ m v = (675 kg·m/s)÷(300 kg) v = 2.25 m/s m p v

Conservation of Momentum A 5-kg cart traveling at 1.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision. BEFORE Cart 1: m = 5 kg v = 4.2 m/s Cart 2 : m = 2 kg v = 0 m/s AFTER Cart 1 + 2: m = 7 kg v = ? p = 21 kg·m/s m p v p = 0 v = p ÷ m v = (21 kg·m/s) ÷ (7 kg) v = 3 m/s pbefore = 21 kg·m/s pafter = 21 kg·m/s

Conservation of Momentum A 50-kg clown is shot out of a 250-kg cannon at a speed of 20 m/s. What is the recoil speed of the cannon? BEFORE Clown: m = 50 kg v = 0 m/s Cannon: m = 250 kg AFTER Clown: m = 50 kg v = 20 m/s Cannon: m = 250 kg v = ? m/s p = 0 p = 1000 kg·m/s p = 0 p = -1000 kg·m/s pbefore = 0 pafter = 0

Conservation of Momentum So…now we can solve for velocity. GIVEN: p = -1000 kg·m/s m = 250 kg v = ? WORK: v = p ÷ m v = (-1000 kg·m/s)÷(250 kg) v = - 4 m/s (4 m/s backwards) m p v

Rocket Challenge After I check your fill in the blank notes. Come and get the instructions for the paper rockets. Depending on time you might have time to do all 3. If not you will complete them on Tuesday. You will use the best design and make it for your larger one we will do next week.