Lecture 6.1.4 Momentum RCSI PDF

Introduction to General Physics 1 M 6.1.4 Momentum Dr Orlaith Brennan  Learning Outcomes On completion of this lecture students will be able to… Discuss and define momentum. Identify the quantities which effect momentum. Recognise the vector nature of momentum....

Introduction to General Physics 1 M 6.1.4 Momentum Dr Orlaith Brennan  Learning Outcomes On completion of this lecture students will be able to… Discuss and define momentum. Identify the quantities which effect momentum. Recognise the vector nature of momentum. Determine and calculate a momentum change and a total system momentum. State and explain the Principle of Conservation of Momentum. Utilise the principle of momentum conservation to solve collision problems for an unknown pre- or post-collision velocity or mass. 2 3 Why are moving objects difficult to stop?  Momentum The momentum (p) of an object is the resistance to a change in motion. p = mv The unit of momentum (p) = unit of mass x unit of velocity = kg m s-1 4 Momentum is a vector quantity → it has a direction.  Momentum Remember Newton’s First law! This idea applies to momentum as well. The momentum of an object will never change if it is left alone. p = mv The momentum of a body can change in any of the following three ways: Change in mass of object Change in speed of object velocity Change in direction of object 5  Momentum - Problem A train of mass 1200 kg moves west at 20 m s-1. What is its momentum? p = mv = 1200 (20) = 24,000 kg m s-1 West 6  Principle of Conservation of Momentum What happens when objects collide? Momentum Before Collision = Momentum After Collision 7  Principle of Conservation of Momentum In any interaction between two or more objects the total momentum of the objects before the interaction is equal to the total momentum after the interaction provided no external forces act on the system of objects. m1u1 + m2u2 = m1v1 + m2v2 8  Problem A 4kg block with an initial velocity of 10m/s is in collision with a 6kg block that is stationary. After the collision, the 6kg block is seen to be moving at 5m/s. (i) Sketch a diagram of the situation. 9  Problem (ii) Calculate the velocity of the 4kg block after the collision. Momentum Before = Momentum After (4)(10) + (6)(0) = (4)(v1’) + (6)(5) 40 + 0 = 4v1’ + 30 40 – 30 = 4v1’ 10/4 = v1’ 2.5 m/s = v1’ 10  Problem (iii) What is the velocity of the 4kg block if the 6kg block is moving with a velocity of 8 m/s after the collision? Momentum Before = Momentum After (4)(10) + (6)(0) = (4)(v1’) + (6)(8) 40 + 0 = 4v1’ + 48 40 – 48 = 4v1’ - sign indicates that 4 kg block -8/4 = v1’ is moving in the opposite 11 -2 m/s = v1’ direction  Problem A 10,000 kg railroad engine reversing at a velocity of 4 m/s strikes a resting railroad car of mass 5,000kg. If the cars lock together as a result of the collision, what is their common velocity afterwards? Before: 10,000 (4) + 5,000 (0) = 40,000 kg m/s 10,000 kg 5,000 kg After: 15,000 v Momentum Before = Momentum After 40000 = 15000 v → v = 2.66 m/s 12 15,000 kg  Momentum Momentum is conserved in a collision and in an explosion. or 13  Problem A gun of mass 4 kg fires a bullet of mass 15 grams with a velocity of 500m/s. Calculate the velocity of the gun after the bullet is fired? In these situations, the momentum before is always zero because both objects start with zero velocity. Momentum Before gun is fired = Momentum After gun is fired mgug + mbub = mgvg + mbvb 0+0 = 4vg + 15x10-3 (500) 0 = 4vg + 7.5 - 7.5 = 4vg -1.875 m/s = vg 14 Bullet and gun travel in opposite directions!  Problem (Past Exam) (a) Why are moving objects difficult to stop? Moving objects contain mass and velocity and as a result have momentum. Momentum is a resistance to the change in motion. (b) The image shows a 4.4 kg piñata being struck by a child with a stick of mass 0.54 kg. During an unsuccessful attempt to break the piñata, the child strikes it with the stick which is moving at 4.8 m/s. The stick stops but the piñata starts to move. Calculate the speed of the piñata immediately after being struck. Momentum Before = Momentum After 4.4 (0) + 0.54(4.8)= 0.54(0) + 4.4(v) 15 v = 0.59 m/s