Collisions A + B C+D+

Conservation of Momentum Momentum in an isolated system in which a collision occurs is conserved An isolated system will not have external forces Specifically, the total momentum before the collision will equal the total momentum after the collision

Conservation of Momentum, cont. Mathematically: m v r + m v r = m v r + m v r 1 1i 2 2i 1 1f 2 2f Can be generalized to any number of objects Components of Momentum External forces? (short collision time)

Notes About A System Remember conservation of momentum applies to the system You must define the isolated system So its not the velocity which is conserved but the momentum mv.

HITT RF Remote Login Procedure: The radio channel number for this room is 07 (zero, seven). It is STRONGLY recommended to login your remote for every class just to be sure it is on the correct radio channel and working before class. 1. PRESS AND HOLD THE DOWN ARROW KEY until the GREEN light on the remote turns RED. 2. PRESS THE 0 KEY and you will see the RED light flash GREEN. 3. PRESS THE 7 KEY and you will see the RED light flash GREEN. 4. PRESS AND RELEASE THE DOWN ARROW KEY again and you will see the red light search for the receiver, if it BLINKS GREEN MULTIPLE TIMES you are logged in.

Two objects collide head on. Their masses and initial velocities are given. If the 10 kg mass has a final velocity of -0.1 m/s, what is the final velocity of the 1 kg mass? 1 m/s -1 m/s -0.1 m/s? m/s 10 kg 1 kg 10 kg 1 kg initial final

Why do the same number of balls move on the other side? A. Conservation of momentum B. Conservation of energy C. Newton s ghost D. All answers are correct (except maybe 3).

Types of Collisions Momentum is conserved in any collision Inelastic collisions Kinetic energy is not conserved Some of the kinetic energy is converted into other types of energy such as heat, sound, work to permanently deform an object (momentum is still conserved) Perfectly inelastic collisions occur when the objects stick together (momentum is still conserved) Not all of the KE is necessarily lost

Types of Collisions Elastic collision both momentum and kinetic energy are conserved Actual collisions Most collisions fall between elastic and perfectly inelastic collisions

73. A tennis ball of mass 57.0 g is held just above a basketball of mass 590 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m, as shown in Figure P6.69. (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. (b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. (b) To what height does the tennis ball rebound?

1.2 m 1.2 m v T v B

With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m. What are the velocities of the tennis ball and basketball when they strike the floor? A. v T = 4.85 m/s, v B = -4.85 m/s B. v T = -4.85 m/s, v B = -4.85 m/s C. v T = -4.85 m/s, v B = 4.85 m/s D. v T = 4.85 m/s, v B = 4.85 m/s

1.2 m 1.2 m 1.2 m v T v T v B v B '

m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f m 1 = m B, v 1i = v B m 2 = m T, v 2i = v T Elastic collision: (1/2)m 1 v 1i2 + (1/2)m 2 v 2i2 = (1/2)m 1 v 1f2 + (1/2)m 2 v 2f 2

Only for 1 dimensional elastic collisions m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f => m 1 v 1i - m 1 v 1f = -m 2 v 2i + m 2 v 2f => m 1 (v 1i - v 1f ) = m 2 (v 2f - v 2i )

Only for 1 dimensional elastic collisions Elastic collision: (1/2)m 1 v 1i2 + (1/2)m 2 v 2i2 = (1/2)m 1 v 1f2 + (1/2)m 2 v 2 2f => m 1 v 1i2 - m 1 v 1f2 = -m 2 v 2i2 + m 2 v 2 2f => m 1 (v 1i2 - v 1f2 ) = m 2 (v 2f2 - v 2i2 ) x 2 y 2 = (x y) (x + y) => m 1 (v 1i - v 1f ) (v 1i + v 1f ) = m 2 (v 2f - v 2i )(v 2f + v 2i ) Because m 1 (v 1i - v 1f ) = m 2 (v 2f - v 2i )

Only for 1 dimensional elastic collisions v 1i + v 1f = v 2i + v 2f or v v = (v 1i 2i 1f v 2f )

More About Elastic Collisions (1-d) Both momentum and kinetic energy are conserved Typically have two unknowns (1d) m1 v1 i + m2 v2i = m1 v1 f + m2 v2 f 1 2 2 1 2 1 2 m 1v1v i + 1 i + vm 1 2v f 2= i = v i + m1 v1 2 f + f 2 2 1 2 m 2 v 2 2 f Solve the equations simultaneously v 1I v 2I v 1F v 2F before after

Inelastic collisions Kinetic energy is not conserved Momentum is still conserved

A gun at rest of mass M shoots a bullet of mass m. If velocity of the bullet is +v b what is the velocity of the gun in terms of M, m and v b A. -(m/m)v b B. (m/m)v b C. -(M/m)v b D. (M/m)v b

What are p and KE in terms of m, M, and v b? A. 0 and -v b2 (1+M/m) B. 0 and (1/2)mv 2 b2 (1+m/M) C. -mv b and (v b M/m) 2 D. -mv b and (v b m/m) 2 KE i = 0 KE f = (1/2)mv b2 + (1/2)M(m/M) 2 v b 2

p f F p p = p f p i = F t p i

How can the stranded astronaut reach the shuttle? A. Kick her arms and legs like a swimmer B. Throw the wrench towards the shuttle C. Throw the wrench away from the shuttle

Rocket Propulsion v v = f i v e M ln( M i f )

Glancing Collisions For a general collision of two objects in threedimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved m v 1 m v 1 1ix 1iy + m v 2 + m v 2 2ix 2iy = = m v 1 m v 1 1fx 1fy + m v Use subscripts for identifying the object, initial and final velocities, and components 2 + m v 2 2fx 2fy and

Glancing Collisions The after velocities have x and y components Momentum is conserved in the x direction and in the y direction Apply conservation of momentum separately to each direction