Physics 151: Lecture 17 Today’s Agenda

Physics 151: Lecture 17Today’s Agenda • Today’s Topics : • Momentum (Chapter 9) • Conservation of Momentum • Introduce Collisions (Elastic and Inelastic)

ACT- 2 • Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value . At the highest point, • K A > K B . • K A = K B . • K A < K B . • The work done by F on A is greater than the work done on B. • The work done by F on A is less than the work done on B.

. . Wf + . . Lecture 16, ExampleSkateboard • Let’s suppose that the surface is not frictionless and the same skateboarder reach the speed of 7.0 m/s at bottom of the hill. What was the work done by friction on the skateboarder ? Conservation of Total Energy : K1 + U1 = K2 + U2 m = 25 kg Wf + 0 + mgR = 1/2mv2 + 0 Wf = 1/2mv 2 - mgR R=3 m Wf = (1/2 x25 kg x (7.0 m/s2)2 - - 25 kg x 10m/s2 3 m) Wf = 613 - 735 J = - 122 J Total mechanical energy decreased by 122 J !

Definition: For a single particle, the momentump is defined as: • (p is a vector since v is a vector). p= mv • So px = mvxetc. • Newton’s 2nd Law: F = ma dv p v Chapter 9Linear Momentum • Units of linear momentum are kg m/s.

Momentum Conservation • The concept of momentum conservation is one of the most fundamental principles in physics. • This is a component (vector) equation. • We can apply it to any direction in which there is no external force applied. • You will see that we often have momentum conservation even when (mechanical) energy is not conserved.

vi • A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter • Car crashes, collisions where objects stick together, etc. Elastic vs. Inelastic Collisions • A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. Kbefore = Kafter • Carts colliding with a spring in between, billiard balls, etc.

Lecture 17, ACT 1Collision in 1-D Winter in Storrs ice (no friction)

M = 2m V0 vf = ? Lecture 17, ACT 1Collision in 1-D initially m ice v = 0 (no friction) finally Vf = A) 0 B) Vo/2 C) 2Vo/3 D) 3Vo/2 E) 2Vo

Lecture 17, Review problem: numerical • High-speed stroboscopic photographs show that the head of a golf club of mass 200 grams is traveling at 55 m/s just before it strikes a 46-gram golf ball at rest on a tee. After the collision, the clubhead travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after

v Inelastic collision in 1-D: Example 1 • A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, andV : • What is the initial speed of the bullet v ? • What is the initial energy of the system ? • What is the final energy of the system ? • Is energy conserved ? x V before after See example 12-6

initial v final v+Dv Lecture 19, ACT 2Let’s do some rocket engineering ! • A rocket engine consumes 450 kg of fuel per minute. If the exhaust speed of the ejected fuel is 5.2 km/s, what is the thrust of the rocket ? • a. 48 kN • b. 39 kN • c. 55 kN • d. 32 kN M+m M vg m

Lecture 17, ACT 3Momentum Conservation • Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. • The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. • Which box ends up moving fastest ? (a)Box 1(b)Box 2(c)same 2 1

Lecture 17Review problem: more involved • A 3.0-kg mass is sliding on a horizontal frictionless surface with a speed of 3.0 m/s when it collides with a 1.0-kg mass initially at rest as shown in the figure. The masses stick together and slide up a frictionless circular track of radius 0.40 m. To what maximum height, h, above the horizontal surface will the masses slide?

Lecture 17Ballistic Pendulum

Lecture 17 ACT 4 The law of conservation of momentum applies to a collision between two bodies if: they exert forces on each other respectively proportional to their masses. they exert forces on each other respectively proportional to their velocities. their accelerations are proportional to their masses. they exert equal and opposite forces on each other.

V Inelastic collision in 2-D • Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). v1 m1 + m2 m1 m2 v2 before after

Inelastic collision in 2-D... • There are no net external forces acting. • Use momentum conservation for both components. v1 V = (Vx,Vy) m1 + m2 m1 m2 v2

P P p2  p1 p2 p1 Inelastic collision in 2-D... • So we know all about the motion after the collision ! V = (Vx,Vy) Vy  Vx

Comment on Energy Conservation • We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. • Energy is lost: • Heat • Bending of metal (crashing cars) • Kinetic energy is not conserved since work is done during the collision ! • Momentum along a certain direction is conserved when there are no external forces acting in this direction. • In general, easier to satisfy than energy conservation.

Recap of today’s lecture • Momentum and Collisions Ch. 9.1-9.4 (part of 9.4)

Physics 151: Lecture 17 Today’s Agenda