Impulse and Momentum Some more extremely useful relations are the concepts known as impulse and momentum. Conservation of momentum underlies several important physical laws in a variety of scientific disciplines from quantum electrodynamics to fluid mechanics. A derivation for this fundamental quantity is shown below: We now define a quantity known as momentum, as , and the change in momentum over a discrete time interval is defined as the impulse, : Note: In comparison with potential and kinetic energy, which are scalar quantities, impulse and momentum are vector quantities and must be added vectorially.

Conservation of Linear Momentum Starting from a generalized expression for momentum We can take the derivative of both sides of the equation: And we arrive at another expression for Newton’s Second law. The force is equal to the rate of change of momentum. Thus, when the net force on an object is zero, then the momentum must be constant. This is the famous law of conservation of momentum.

Conservation of Linear Momentum Linear Momentum prior to impact Linear Momentum after impact Initial conditions: After impact:

Given the following information: Leads to 2 equations and 2 unknowns. These can be solved simultaneously, leading to the following result:

Relation between Force and Impulse As shown in previous slides, the impulse can be written as: If the force is constant over a discrete time interval, then it can be taken out of the integral and the impulse is re-written as: This equation is very useful for modeling collisions, because the force of impact is not usually known (or measurable), but the initial and final velocities are known and/or measurable.

Angular Momentum Like linear momentum, there is also an expression for angular momentum, which describes how much inertia an object carries in rotating about a coordinate center. While its derivation will be reserved for later in this course, the mathematical expression is given here as: Path of particle The cross product denotes that the angular momentum points in an entirely different direction from the instantaneous position vector, , and the instantaneous velocity vector . The direction can be found according to the right-hand rule. The O means that the angular momentum is taken with respect to the origin.

Conservation of Angular Momentum The rate of change of angular momentum can be derived as follows: (i.e. the cross product of a vector with itself is always zero (i.e. they point in the same direction) We define this new vector, , which is the rate of change of angular moment, and it is called the moment of force about a pivot. Note: If the net force (or moment in this case) is zero, then it leads directly to the law of conservation of angular momentum.

Momentum according to Isaac Asimov Isaac Asimov wrote in "Understanding Physics": “This tendency for motion (or for rest) to maintain itself steadily unless made to do otherwise by some interfering force can be viewed as a kind of "laziness," a kind of unwillingness to make a change. And indeed, Newton's first law of motion is referred to as the principle of inertia, from a Latin word meaning "idleness" or "laziness." He added a footnote: "In Aristotle's time the earth was considered a motionless body fixed at the center of the universe; the notion of 'rest' therefore had a literal meaning. What we ordinarily consider 'rest' nowadays is a state of being motionless with respect to the surface of the earth. But we know (and Newton did, too) that the earth itself is in motion about the sun and about its own axis. A body resting on the surface of the earth is therefore not really in a state of rest at all."

Central Forces In many forces, such as gravitational, electrical forces, etc., the magnitude of the force only depends on the distance away from the source (i.e. ), and the direction is always towards or away from the source . Using these assumptions, we can derive an expression for the rotation frequency:

Using this result, we can arrive at an expression for radial force written purely in terms of the radial distance Now the question becomes, what function f(r) satisfies the solution to this differential form? Or alternatively, what is the solution to this equation:

The trick is to write it in differential form: For example, consider the dot product of the velocity vector with itself. What is the time derivative of this quantity?

If you now multiply this function by ½ m, you will find that: The result is that the equation for radial motion is now reduced to: The next step is to write f(r) in a form that allows for solution of this differential equation.

Consider for the moment that the function f(r) is actually derivative of another function. After some manipulation, this formula becomes: So, now we can re-write the equation on the previous slide as:

This equation is easily solvable. It is simply a constant. But we also know that the expression for the square of velocity. This equation then becomes upon multiplication by ½ m…

This is the kinetic energy stored in the linear velocity This is the kinetic energy stored in the angular momentum This is the potential energy term When we make the substitution… The above equation becomes: Linear momentum Energy term Angular momentum Energy term

Problem 1 A 25-g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10-mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 10 mm 15o 20o B C D A

Problem 1 10 mm 15o 20o B C D A A 25-g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10-mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 1. Draw a momentum impulse diagram: The diagram shows the particle, its momentum at t1 and at t2, and the impulses of the forces exerted on the particle during the time interval t1 to t2.

S F is sum of the impulsive forces (the forces that are large Problem 1 10 mm 15o 20o B C D A A 25-g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10-mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 2. Apply the principle of impulse and momentum: The final momentum mv2 of the particle is obtained by adding its initial momentum mv1 and the impulse of the forces F acting on the particle during the time interval considered. mv1 +S F Dt = mv2 S F is sum of the impulsive forces (the forces that are large enough to produce a definite change in momentum).

+ = Problem 1 Solution D 10 mm 20o Draw a momentum impulse diagram. C x y Fx D t Fy D t m v1 15o m v2 20o Since the bullet leaves a 10-mm scratch and its average speed is 500 m/s, the time of contact D t is: D t = (0.010 m) / (500 m/s) = 2x10-5 s

+ = mv1 +S F Dt = mv2 + x components: Problem 1 Solution Apply the principle of impulse and momentum. + = x y Fx D t Fy D t m v1 15o m v2 20o mv1 +S F Dt = mv2 + x components: (0.025 kg)(600 m/s)cos15o+Fx(2x10-5s) = (0.025 kg)(400 m/s)cos20o Fx = - 254.6 kN -(0.025 kg)(600 m/s)sin15o+Fy(2x10-5s)=(0.025 kg)(400 m/s) sin20o Fy = 365.1 kN + y components:

Problem 1 Solution + = x y Fx D t Fy D t m v1 15o m v2 20o Fx = - 254.6 kN, Fy = 365.1 kN F = ( -254.6 kN )2 + ( 365.12 kN )2 = 445 kN F = 445 kN 40.1o

Problem 2 650 kg The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0 ), determine the average resistance of the ground to penetration.

Problem 2 140 kg 650 kg 1.2 m The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0 ), determine the average resistance of the ground to penetration. 1. Apply conservation of energy principle : When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T1 + V1 = T2 + V2 where 1 and 2 are two positions of the particle. 1a. Kinetic energy: The kinetic energy at each point on the path is given by: T = m v2 1 2

Problem 2 140 kg 650 kg 1.2 m The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0 ), determine the average resistance of the ground to penetration. 1b. Potential energy: The potential energy of a weight W close to the surface of the earth at a height y above a given datum is given by: Vg = W y

Problem 2 140 kg 650 kg 1.2 m The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0 ), determine the average resistance of the ground to penetration. 2. Apply conservation of momentum principle: During an impact of two bodies A and B, the total momentum of A and B is conserved if no impulsive external force is applied. mA vA + mB vB = mA v’A + mB v’B where vA and vB denote the velocities of the bodies before the impact and v’A and v’B denote their velocities after the impact. For perfectly plastic impact (e = 0), v’A = v’B = v’ and mA vA + mB vB = (mA + mB) v’

1 1 2 2 Problem 2 650 kg The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0 ), determine the average resistance of the ground to penetration. 3. Apply principle of work and energy: When a particle moves from position 1 to position 2 under the action of a force F, the work of the force F is equal to to the change in the kinetic energy of the particle. T1 + U1 2 = T2 where T1 = m v12 , T2 = m v22 and U 1 2 = F . dr 1 2 1 2

Apply conservation of energy principle. Problem 2 Solution Motion of the hammer during the drop just before impact. 650 kg v1 = 0 650 kg v2 y Position 1 Position 2 1.2 m 1.2 m T1 + V1 = T2 + V2 0 + mg (1.2 m) = m v22 + 0 v22 = 2 ( 9.81 m/s2 )(1.2 m) v2 = 4.85 m/s 1 2

(650 kg)( 4.85 m/s) + 0 = (650 kg + 140 kg) v’ v’ = 3.99 m/s Problem 2 Solution Apply conservation of momentum principle. Impact process: mH vH mP vP = 0 mH v’ mP v’ Before impact: After mH vH + mP vP = ( mH + mP ) v’ (650 kg)( 4.85 m/s) + 0 = (650 kg + 140 kg) v’ v’ = 3.99 m/s

( mH + mP ) v’2 + (WH + WP - R) y = 0 Apply principle of work and energy. Problem 2 Solution Hammer and pile move against ground resistance. v’ 110 mm v = 0 W R Position 1 Position 2 Work T1 + U1 2 = T2 ( mH + mP ) v’2 + (WH + WP - R) y = 0 (650 + 140) (3.99 m/s)2 + [(650 + 140)(9.81) - R](0.110) = 0 R = 65,000 N 1 2 1 2

Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 45o O B C C’ C’’ D A a

45o O B C C’ C’’ D A a Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 1. Apply conservation of energy principle : When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T1 + V1 = T2 + V2 where 1 and 2 are two positions of the particle.

45o O B C C’ C’’ D A a Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 1a. Kinetic energy: The kinetic energy at each end of the path is given by: T = m v2 1 2

45o O B C C’ C’’ D A a Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 1b. Potential energy: The potential energy of a weight W close to the surface of the earth at a height y above a given datum is given by: Vg = W y

S F is sum of the impulsive forces (the forces that are large B C C’ C’’ D A a Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 2. Apply the principle of impulse and momentum: The final momentum mv2 of the particle is obtained by adding its initial momentum mv1 and the impulse of the forces F acting on the particle during the time interval considered. mv1 +S F Dt = mv2 S F is sum of the impulsive forces (the forces that are large enough to produce a definite change in momentum).

T1 + V1 = T2 + V2 0 + 0 = m vC2 - m g (2 a sin 45o ) 1 vC = 1.682 g a Apply conservation of energy principle. Problem 3 Solution Motion of the sphere from point B to point C (just before the cord is taut). 45o O B D A a vB = 0 Position 1 C vC Position 2 y T1 + V1 = T2 + V2 0 + 0 = m vC2 - m g (2 a sin 45o ) vC = 1.682 g a 1 2

m vC cos 45o = m v’C v’C = vC cos 45o = 1.682 g a cos 45o Apply impulse and momentum principle. Problem 3 Solution Consider the sphere at point C as the cord becomes taut and the velocity of the sphere changes to be in direction normal to the cord. O D O D B B 45o 45o a a A A C C 45o v’C vC t t Momentum is conserved in the tangential direction since the external impulse (the cord on the sphere) is in the normal direction. m vC cos 45o = m v’C v’C = vC cos 45o = 1.682 g a cos 45o v’C = 1.1892 g a

m (v’C)2 - m g (2 a sin 45o ) = 0 - m g d Apply conservation of energy principle. Problem 3 Solution Motion of the sphere from point C to point C’’. y O D O D B B 45o 45o vC’’= 0 a a d A 2 a sin45o A C’’ Position 2 Position 3 v’C = 1.1892 g a C T2 + V2 = T3 + V3 m (v’C)2 - m g (2 a sin 45o ) = 0 - m g d m (1.1892)2 g a - m g (2 a sin 45o ) = 0 - m g d d = 0.707 a 1 2 1 2

Problem 4 A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle qB formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle qA , determine the required value of the ratio lB / lA of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. qA qB lA lB A B C D

qA qB lA lB A B C D Problem 4 A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle qB formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle qA , determine the required value of the ratio lB / lA of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 1. Apply principle of conservation of energy: When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T1 + V1 = T2 + V2 where 1 and 2 are two positions of the particle.

qA qB lA lB A B C D Problem 4 A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle qB formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle qA , determine the required value of the ratio lB / lA of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 1a. Kinetic energy: The kinetic energy at each point on the path is given by: T = m v2 1b. Potential energy: The potential energy of a weight W close to the surface of the earth at a height y above a given datum is given by: Vg = W y 1 2

qA qB lA lB A B C D Problem 4 A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle qB formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle qA , determine the required value of the ratio lB / lA of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 2. Apply conservation of momentum principle: During an impact of two bodies A and B, the total momentum of A and B is conserved if no impulsive external force is applied. mA vA + mB vB = mA v’A + mB v’B where vA and vB denote the velocities of the bodies before the impact and v’A and v’B denote their velocities after the impact.

qA qB lA lB A B C D Problem 4 A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle qB formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle qA , determine the required value of the ratio lB / lA of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 3. Apply the relationship for the coefficient of restitution: For impact of two particles A and B: v’B - v’A = e (vA - vB) where v’B - v’A and vA - vB are the relative velocities, normal to the impact plane, after and before the impact, respectively, and e is the coefficient of restitution.

qA qB lA lB A B C D Problem 4 Solution Motion of sphere A from its release until it hits sphere B. qA lA A C y Position 1 lA( 1 - cos qA ) vA1= 0 vA2 Position 2

0 + m g lA( 1 - cos qA ) = m (vA2)2 + 0 vA2 = 2 g lA( 1 - cos qA ) Position 1 Position 2 qA lA A C vA2 vA1= 0 lA( 1 - cos qA ) y Problem 4 Solution Apply principle of conservation of energy. Motion of sphere A from its release until it hits sphere B. T1 + V1 = T2 + V2 0 + m g lA( 1 - cos qA ) = m (vA2)2 + 0 vA2 = 2 g lA( 1 - cos qA ) 1 2

qA qB lA lB A B C D Problem 4 Solution Collision of balls A and B. B C D B C D v’A2 v’B2 After impact vA2 vB2 = 0 Before impact

m vA2 = m v’A2 + m v’B2 vA2 = v’A2 + v’B2 (1) C D Problem 4 Solution B C D Apply conservation of momentum principle. Apply the relationship for the coefficient of restitution. vA2 vB2 = 0 v’A2 v’B2 Before impact After impact m vA2 = m v’A2 + m v’B2 vA2 = v’A2 + v’B2 (1) ( v’B2 - v’A2 ) = e ( vA2 ) (2) Eliminating v’A2 from equations (1) and (2) gives: v’B2 = ( 1+ e ) 2 vA2

qA qB lA lB A B C D Problem 4 Solution Motion of sphere B following the collision. qB lB B D lB( 1 - cos qB ) Position 2 v’B2 vB3 Position 3

m ( v’B2 )2 + 0 = 0 + m g lB( 1 - cos qB ) D lB( 1 - cos qB ) Position 2 v’B2 vB3 Position 3 y Problem 4 Solution Apply principle of conservation of energy. Motion of sphere B following the collision. T2 + V2 = T3 + V3 m ( v’B2 )2 + 0 = 0 + m g lB( 1 - cos qB ) 1 2 v’B2 = ( 1+ e ) 2 vA2 Substituting and vA2 = 2 g lA( 1 - cos qA ) and qB = qA gives: lB lA = ( ) 1 + e 2