Momentum, Impulse, and Collisions

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Goals for Chapter 8 • To understand the concept of momentum Chapter 8 • To study the relationship of impulse and momentum “Impulse-momentum Theorem” Momentum, Impulse, • To know when momentum is conserved and examine the implications of conservation of momentum in a and Collisions variety of physics problems - e.g. collision • To understand the concept of center of mass and how

PowerPoint® Lectures for to use it to analyse physics problems University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman

Lectures by James Pazun Modified by P. Lam 6_6_2011

Compare kinetic energy and momentum Newton’s second law & Impulse-momentum Theorem Kinetic enegy (K) is a scalar (a number) ! ! ! ! dv d(mv) 1 F = ma = m = K ! mv2 dt dt 2 ! ! ! dp Momentum (p) is a vector ! F = (another form of Newton's second law) ! ! dt p=mv Both describe the properties of a mass and its velocity. ! t f ! t f dp ! ! Kinetic energy is related to the magnitude of the momentum ! " Fdt = " dt = p f # pi (Impulse-momentum Theorem) ! ti ti dt 1 m2v2 |p|2 K= = t f ! ! 2 m 2m " Fdt $ I is called the "implulse" % Faverage&t t Both kinetic energy and i momentum are useful concepts compare with Work-kinetic energy Theorem in analyzing collisions. f ! ! " F•d" = K f # Ki (Work-kinetic energy Theorem) i

1 Compare momentum and kinetic energy Examples of impulse-momentum theorem

Example: Batting a baseball 20 • To change the Dropping a wine glass on a hard floor In this case, there is a momentum of an vs. change in momentum but no object, one needs to Dropping a wine glass on a soft carpet change in KE apply a force on the Hard floor Δt small => Faverage is large object. W-K.E. Theorem: ! ! • The amount of force K ! K = W = F •s f i depends on the change (The bat does negative work on the ball to slow it down to zero velocity, in momentum and the then it does positive work to speed it up to +20 m/sˆi time the force acts on " zero net work and no change in kenetic energy) the object Impulse-momentum Theorem: • Impulse-momentum ! ! ! ! p mv (0.4)( 20)iˆ; p mv (0.4)( 20)iˆ Theorem i = i = ! f = f = + ! ! ! ! ! ! ˆ ˆ ˆ p f " p i = Faverage #t p f ! pi = (8i )! (!8i ) =16i = Faverage"t ! If "t = 0.01s # Faverage =1600N

Examples of momentum change - quantified Like energy, momentum also has conservation rules ! ! ! (a) A 0.2kg wine glass is dropped on • When the net “external” force is zero, the total momentum of the system p f " p i = Faverage #t ! ! ! ! a hard floor. Let the initial velocity is conserved (m v + m v ) = (m v + m v ) Identify the initial and final of the glass right before it hits the 1 1 2 2 initial 1 1 2 2 final momentum vectors ground is 5 m/s and it took 0.01 and deduce Faverage second to stop the glass. ! (both direction and magnitude) Will go over this in class ! (b)

Internal forces (force of A on B and force of B on A) are equal and opposite => produce equal and opposite momenta thus the internal forces do not affect the total momentum; only the external forces do.

2 An application of conservation of total momentum Another example of conservation of total momentum More correct analysis: Typically a rifle is specified by its • Consider the muzzle velocity, i.e. the bullet's velocity collision in wrt the gun. Suppose the muzzle is 300 m/s. Now find the rifle recoil velocity and the Example 8.5. bullet's velocity wrt the ground. ! ! (ptotal )i = (ptotal ) f • Find v . ! ! ! A2x 0 = (mB (vBR + vRG ) + mRvRG ) ! ! 0 = (0.005)(300iˆ + v ) + (3)v The bullet’s velocity is 300 m/s ( RG RG ) ! 1.5 m m wrt the ground ! v = " iˆ # "0.499 iˆ RG 3.005 s s ! ! Due to recoil, the bullet's velocity wrt ground (ptotal )i = (ptotal ) f ! ! is a little bit smaller than the muzzle velocity. 0 = (mBv B + mRv R ) ! ! ! ˆ ˆ vBG = vBR + vRG = 300i " 0.499i ˆ ! The correction in this case is very small due to 0 = (0.005)(300i ) + (3)v R ( ) the small mass of the bullet compared to the rifle. ! m " v = #0.5iˆ Correction is not small for cannons R firing heavy cannon balls. s ** Don’t forget momentum is a vector

Compare elastic and inelastic collisions Completely inelastic collisions

• Cars are designed to crumple and absorb as much energy as possible so the passengers do not need to.

• In any inelastic collisions, total momentum is conserved but total kinetic energy is not conserved (some of the kinetic energy went into deforming the cars!)

Find the final velocity, the initial and final total kinetic energies.

A collision where two object stick together is called completely inelastic collision because it loses the most kinetic energy. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

3 Another completely inelastic collision - automobile accidents 1-D elastic collision - 2 equations and 2 unknowns

• Refer to Example 8.9 and Figure 8.19, find v and θ.

First equation : Conservation of total momentum (x - direction because net Fx = 0) Second equation : Conservation o kinetic energy (elastic collision)

Elastic collisions—Figure 8.23 and 8.22 Gravitational “sling-shot” effect

• There are collisions where you can guess the solutions.

• Two equal masses • Two very different masses

What happen to the ping-pong ball’s final velocity if the bowling ball is also initially moving toward Can be analyzed effectively as 1-D collision the ping-pong ball? This is the concept behind gravitational sling-shot effect.

4 2-D elastic collision - 3 equations and 4 unknowns. Center of mass motion Center of mass position : • Consider Example 8.12. ! ! ! m1r1 + m2r2 + ... Defn : rcm " m1 + m2 + ... ! ! ! # Mtotal rcm . = m1r1 + m2r2 + ...

Center of mass velocity : ! ! ! Mtotalv cm. = m1v1 + m2v 2 + ...= total momentum

Center of mass acceleration: ! ! ! ! ! Mtotal acm . = m1a1 + m2a2 + ...= net Fexternal + net Finternal ! net Finternal = 0 according to Newton's third law. Conservation of total momentum x & y - directions " 2 equation # Center of mass motion is governed by external forces only Conservation of kinetic energy " 1 equation ! ! ! Furthermore, if net Fexternal = 0, then a cm. = 0 # Mtotalv cm = constant 4 unknowns : | vA2 |,| vB2 |,#,$ ! This is a re - statement of conservation of total momentum!! (Not enough equation because we treated masses as points)

Another example of center of mass motion when a =0 Example of center of mass motion when acm=0 cm • Refer to Example 8.14.

Find the center of mass

A wrench spins around and slides across a very smooth table. The net external force ~0, hence the center of mass (white dot) exhibits a nearly constant velocity=> conservation of momentum.

5 Example of center of mass motion is governed by external force Rocket propulsion-example of conservation of total momentum only • Look at Figure 8.32 in section 8.6.

! ! Net external force = Mtotalg " a cm = g " center of mass motion follows a parabola path even after the shell exploded (because internal forces which cause the explosion do not affect the center of mass Set up the conservation of total momentum equation motion.) Note : Total momentum is NOTconserved because there is an external force.