Phys101 Lectures 14, 15, 16 and Collisions

Key points: • Momentum and • Condition for conservation of momentum and why • How to solve collision problems • Centre of

Ref: 9-1,2,3,4,5,6,7,8,9.

Page 1 Momentum is a vector:

It’s a quantity that represents the amount and direction of motion. It was once called “the quantity of motion”. Now we know that kinetic is another quantity of motion.   nd   dv dp Newton’s 2 Law F  ma  m  dt dt t f     Fdt  p  p  p  f i ti  t f    where J  Fdt  F t is called the impulse of F .  ave ti (area under the F-t curve) Impulse-momentum principle : The net impulse on an object is equal to the change in momentum of the object.   J  p Example 9-1: Force of a tennis serve. For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms (4 x 10-3 s), estimate the average force on the ball. [Solution]    x J  Favet  p p mv 0 F   ave t t 0.0655  0.004  800 (N) i-clicker question 14-1 A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s. What is the average force exerted by the ground on the ball during that time? A) 2.00 N   x B) 10.00 N F t  p C) 22.0 N Fx t  px D) 24.0 N px m( v fx  vix ) E) 12.0 N Fx   t t 0.1401.00 ( 1.20 )  0.014  22 ( N ) MasteringPhysics Assignment #6

Impulse-Momentum Theorem, Part I • Consider a particle of mass m=5.00kg moving along the x axis. It is at rest at t=0.00 sec. A varying force F(t)=6.00t2-4.00t+3.00 is acting on the particle between t=0.00 sec and t=5.00sec. Find the speed v of the particle at t=5.00 sec.        J  p  p f  pi  m( v f vi )  mv f x

 t2  J   Fdt t1

t2 5 5 J  F dt  6t 2  4t  3 dt  2t 3  2t 2  3t x  x     0 t1 0  2( 5 )3  2( 5 )2  3( 5 ) 2( 0 )3  2( 0 )2  3( 0 )  215 ( N  S ) J 215 v  x   43 ( m / s ) fx m 5 Many-body System  F Two students pulling each other on ice. 1 12 2  Internal : and F .  21   External forces:   F21 m1g, m2 g, FN1 and FN1      dp    dp Newton’s law: m g  F  F  1 , m g  F  F  2 1 N1 12 dt 2 N 2 21 dt Adding the two equations         dp dp m g  F  m g  F  F  F  1  2 1 N1 2 N 2 12 21 dt dt    dP   Fext   Fint  . P is the total momentum of the system dt    Since the internal forces always cancel, F21  F12, Fint  0   dP F  (only external forces can change the total momentum)  ext dt Demo Conservation of Momentum   dP  When F  0,  0, i.e., P  cons tant  ext dt

This is the law of conservation of linear momentum: When the net external force on a system of objects is zero, the total momentum of the system remains constant. Note 1: If one of the components of the net external force is zero, the corresponding component of the total momentum of the system is conserved (even though the total momentum vector may or may not be conserved). Note 2: For a one-object system, the condition for momentum conservation is that the net force acting on the object is zero. i-clicker question 14-2 The condition for mechanical energy to be conserved is (A) It’s a closed system. (B) The net external force is zero. (C) No nonconservative work. (D) The momentum is never conserved. (E) The momentum is always conserved. Example 9-4: Rifle recoil. Calculate the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s. [Solution] X-component of external Force is zero; Thus the x-component of the total momentum is conserved:

Pxf  Pxi  0

mBvBx  mRvRx  0

mBvBx 0.02620 vRx      2.5 ( m / s ) mR 5

vR  2.5 ( m / s ) Collisions

Momentum is conserved in all collisions. Why? Because the impulse of external forces can be ignored (much smaller than internal impulse).

Collisions in which is conserved as well are called elastic collisions, and those in which it is not are called inelastic. Elastic Collisions in One Dimension Here we have two objects colliding elastically. We know the and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds. x

Example 9-7: Billiard ball A of mass m moving with speed vA collides head-on with ball B of equal mass. What are the speeds of the two balls after the collision, assuming it is elastic? Assume ball B is initially at rest (vB = 0). (1)2  (2): 0  2v v Conservation of momentum: A B mv  mv  mv (x -comp) A A B Solution 1: vA  0

Conservation of kinetic energy: vB  vA

1 2 1 2 1 2  mv  mv  mv Solution 2: vB  0 2 A 2 A 2 B vA  vA v  v  v (1) A A B Solution 2 should be rejected 2 2 2 because it means no collision. vA  vA  vB (2) Inelastic Collisions

With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one in which the objects stick together afterward, so there is only one final velocity.   vA  vB Example 9-3: Railroad cars collide A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common speed immediately after the collision?

Conservation of momentum (x-comp.): mv A  m v A  m v B

Hit and stick (perfectly inelastic): v A  v B v Solve: v  v  A A B 2 Example 9-11: . The ballistic pendulum is a device used to measure the speed of a projectile, such as a bullet. The projectile, of mass m, is fired into a large block of mass M, which is suspended like a pendulum. As a result of the collision, the pendulum and projectile together swing up to a maximum height h. Determine the relationship between the initial horizontal speed of the projectile, v, and the maximum height h. [Solution] Two events: 1. Hit and stick; mv  ( m M )v 1 2. Swing . ( m  M ) v  2  ( m  M ) gh 2

1 m2 m  M Eliminate v: v2  ( m  M )gh , v  2gh 2 m  M m Collisions in Two or Three Dimensions Example 9-13: Proton-proton collision. A proton traveling with speed 8.2 x 105 m/s collides elastically with a stationary proton in a hydrogen target. One of the protons is observed to be scattered at a 60°angle. At what angle will the second proton be observed, and what will be the velocities of the two protons after the collision? Elastic collision: mA  mB  m 1 1 1 mv2  mv2  mv2 2 A 2 A 2 B 2 2 2 i.e., vA  vA  vB (1)

Conservation of momentum:       mvA  mvA  mvB , i.e., vA  vA  vB , 3 unknowns: vA ,vB ,B . x : vA  vA cos A  vB cosB (2) 3 equations : (1), (2), (3). y : 0  vA sin A  vB sinB (3) i-clicker question 14-3 A car with a mass of 1200 kg and a speed of 12 m/s heading north approaches am intersection. At the same time, a minivan with a mass of 1300 kg and speed of 24 m/s heading east is also approaching the intersection. The car and the minivan collide and stick together. Consider the total momentum and the total kinetic energy of the two vehicles before and after the collision. WhatA. is theBoth velocity the total of momentumthe wrecked and vehicles total kinetic just after energy the collision? are    Hit and stick:conserved. v   v   v  C V R y (N) B. The total momentum is conserved  but the total kinetic Conservationenergy of momentum:is not conserved. pC  pV  PR C. Neither the total momentum nor the total kinetic energy is P  conserved.m v 2  m v 2  34400 ( kgm / s ) R D. TheV Vtotal kineticC C energy is conserved but the total x (E)  momentumPR is34400 not conserved.  vR E. The change in total momentum13.7 ( m equals/ s ) the change in total m  m 1300 1200 pV VkineticC energy.  p 1 pV 1 mV vV C     tan  tan  65 PR pC mCvC Center of Mass (CM) In (a), the diver’s motion is pure translation; in (b) it is translation plus rotation. There is one point that moves in the same path a particle would take if subjected to the same force as the diver. This point is called the center of mass (CM). Center of Mass (CM)

The general motion of an object can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other forms of motion about the CM. Center of Mass (CM) For two particles, the center of mass lies closer to the one with the most mass:

where M is the total mass.

In general,

mAxA  mB xB mA yA  mB yB xCM  , yCM  , mA  mB mA  mB Exercise 9-15: Three particles in 2-D. Three particles, each of mass 2.50 kg, are located at the corners of a right triangle whose sides are 2.00 m and 1.50 m long, as shown. Locate the center of mass.

[Solution]

mA xA  mB xB  mC xC mA yA  mB yB  mC yC xCM  yCM  mA  mB  mC mA  mB  mC 1 1  x  x  x  ( m  m  m )  y  y  y  3 A B C A B C 3 A B C 1 1  0  2.0  2.0  0  0 1.5 3 3 1.33( m )  0.50 ( m ) Example 9-17: CM of L-shaped flat object. Determine the CM of the uniform thin L-shaped object shown. [Solution] The object consists of two rectangular parts: A and B, whose centres of mass are (xA, yA) and (xB, yB).

xA 1.03m, yA  0.10m

xB 1.96m, yB  0.74m

mA  t( 2.060.20 )  0.412t (  density, t  thickness )

mB  t(1.480.20 )  0.296t

mA xA  mB xB 0.4121.03 0.2961.96 xCM   1.42 ( m ) mA  mB 0.412  0.296

mA yA  mB yB 0.4120.10  0.296( 0.74 ) yCM    0.25 ( m ) mA  mB 0.412  0.296 Center of Mass and Translational Motion

The total momentum of a system of particles is equal to the product of the total mass and the velocity of the center of mass.   PTotal  MvC

The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the of the center of mass:

Therefore, the center of mass of a system of particles (or objects) with total mass M moves like a single particle of mass M acted upon by the same net external force. Conceptual Example 9-18: A two-stage rocket. A rocket is shot into the air as shown. At the moment it reaches its highest point, a horizontal distance d from its starting point, a prearranged explosion separates it into two parts of equal mass. Part I is stopped in midair by the explosion and falls vertically to Earth. Where does part II land? Assume g = constant.