• Physics                                                                      Dr. Lu                         Dec.4- 5, 2018

    Unit 2: Mechanics

    Topic: Momentum, Impulse and change of Momentum

    Learning Targets:

    • I can calculate and determine the magnitude of a linear momentum by the definition of a linear momentum, p =mv, where m is the mass and v is the velocity of an object and P is the linear
    • I can calculate and determine the magnitude of an Impulse by the definition of the Impulse, J== Ft), where F is the force acting on an object and the t is the duration for the force and J is the impulse on an object.
    • I can connect the Impulse to the linear momentum by equating the impulse with the change of linear momentum, J= Ft) = F(t2 – t1) = P = ( P2 – P1) = m (V2 – V1)

          Unpacking the learning targets

          P=mv; two quantities  m (mass) and v (velocity) needed for p (momentum)

                      J= Ft),; two quantities F (force) and t (duration of force) for J (impulse)

     

    Mini-Lesson

     1. Show power point on momentum

         Video: https://www.youtube.com/watch?v=XFhntPxow0U  (momentum)

                     https://www.youtube.com/watch?v=ph48Xwj_eS8  

                     https://www.youtube.com/watch?v=E13h1E_Pc00

    2. Illustration

     (1)Linear momentum P

       P =mv;    where m is the mass of an object and the v is the velocity of the object and p is the

                       linear momentum.

      (2) Impulse J = Ft)

       where F is the force acting on the object and t is the duration of force acting on an object and

                  J is the impulse on the  object.

     (3)The unit

       - for momentum P is kg-m/s

      -  for impulse J is n-s

    -    Equivalent unit :  Newton-Second = kg-meter/sec

    1. P and J are vector quantities
    • The momentum has the same direction as the velocity
    • The impulse has the same direction as the force.
    1. Impulse = J = mat) = mt)/(t) [t]  = mt} = Impulse = change of momentum

                                   =P = ( P2 – P1) = m (V2 – V1)\

    3.Examples

    Example 1

    Calculate the magnitude of a linear momentum for an object with mass of 5 kg and moving at a speed of 2 m/s eastward.

     [Solution] P =mv= 5kg(2m/s) = 10 kg-m/s

     

    Example 2

    Find the magnitude of an impulse J when a force of 3 N acting on an object for a time duration of 0.2 sec.

    [Solution] Impulse J = F(t) = (3n)(0.2s) = 0.6 n-s

     

    Example 3

    A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a

    momentum of 30. kilogram•meters per second, strikes an obstacle. The obstacle exerts an

    impulse of 10. newton•seconds to the west on the block.The speed of the block after the

    collision is  (1) 1.7 m/s     (3) 5.0 m/s      (2) 3.3 m/s    (4) 20. m/s

    {Solution]  Let the direction along east is positive

    J = Impulse  = Ft) =  -  10 N-S = P = ( P2 – P1) = m (V2 – V1) =  x – ( 30 kg-m/s ) = - 10 N-S

    è x = P2 = (30 – 10) kg-m = 20 kg-m/s

    P2  =  20 kg-m /s = mV2 = (6 kg) V2=è V2 =  20/6  m/s =   3.3 m/s ç direction is pointing to east

     

    Check for understanding

    1.Calculate the magnitude of the momentum of a 5.0-kilogram object moving at 3.0 meters/sec

    2.A golf club hits a stationary 0.050-kilogram golfball with an average force of 5.0 × 103 newtons accelerating the ball to a speed of 44 meters per second. What is the magnitude of the impulse imparted to the ball by the golf club? ( # 8, June 2018)

    (1) 2.2 N•s             (2) 880 N•s           (3) 1.1 × 104  N•s      (4) 2.2 × 10 5  N•s

     

    Group work

    1. A 3.0-kilogram object is acted upon by an impulse having a magnitude of

    15 newton•seconds. What is the magnitude of the object’s change in momentum due to this

    impulse?    (1) 5.0 kg•m/s       (2) 15 kg•m/s          (3) 3.0 kg•m/s    (4) 45 kg•m/s

    2.A 1.5-kilogram cart initially moves at 2.0 meters per second. It is brought to rest by a constant net force in 0.30 second. What is the magnitude of the net force?

    (1) 0.40 N             (2) 0.90 N             (3) 10. N                       (4) 15 N

       (By the method of Newton’s 2nd law of Motion, F =ma)

    3. A 1.5-kilogram cart initially moves at 2.0 meters per second. It is brought to rest by a constant net force in 0.30 second. What is the magnitude of the net force?

     ( By the method of Impulse J  = Ft)

    4.A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have in order to hit the wall with the same magnitude of momentum as the ball?

    (1) 3.60 m/s      (2) 6.00 m/s        (3) 360. m/s      (4) 600. m/s

     

     

    Exit Ticket

    A 0.45-kilogram football traveling at a speed of 22 meters per second is caught by an

    84-kilogram stationary receiver. If the football comes to rest in the receiver’s arms, the

    magnitude of the impulse imparted to the receiver by the ball is

    • 1800 n-s    (2) 9.9 n-s     (3) 4.4  n-s    (4) 3.8 n-s

     

    Homework: Midterm homework for reviewing topics

    Review problem

    Newton’s Law of Universal Gravitation

       F= GM1M2/ r2  =è weight =M2(g) = GM1M2 / r2è  g = GM1 /r2; g is acceleration

       F=GM1M2 / r2 è     Centripetal force F = M2(V2/r )= GM1M2/ r2 è   V2/ r= GM1/r2

                                                                                           è  V2 = GM1/ r  = ( 2 r / T)2è GM1/ r = 4   r2 / T2

                                                                                         =è  T2 = 4   r2 (r) / [GM1] = 4   r3 / [GM1]

    1.The magnitude of the gravitational field strength near Earth’s surface is represented by (#14, June 2018)

    2.Which forces can be either attractive or repulsive? ( #16, June 2018)

    (1) gravitational and magnetic       (2) electrostatic and gravitational       (3) magnetic and electrostatic       (4) gravitational, magnetic, and electrostati

    3.What is the magnitude of the gravitational force of attraction between two 0.425-kilogram soccer balls when the distance between their centers is 0.500 meter?

    (#20, June 2018)

    (1) 2.41 × 10 -11 N      (2) (2) 4.82 × 10 -11 N      (3) 5.67 × 10 - 11 N     (4) 1.13 × 10 -10

    4.A 65.0-kilogram astronaut weighs 638 newtons at the surface of Earth. What is the mass of the astronaut at the surface of the Moon, where the acceleration due to gravity is 1.62 meters per second squared?

      (1) 10.7 kg   (2) 65.0 kg     (3) 105 N    (4) 638 N

    How far above the surface of the Earth is the astronanut

    5.The Hubble telescope’s orbit is 5.6 × 10 5 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth exerts a gravitational force of 9.1 × 104 newtons on the telescope. The magnitude of Earth’s gravitational field strength at this location is

      (1) 1.5 × 10 - 20  N/kg        (2) 0.12 N/kg         (3) 8.3 N/kg    (4) 9.8 N/kg  ( #11, 2015

    6.When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance between the point charges is increased to 3d, the magnitude of the electrostatic force between the two charges will be

      (1)  F/9     (2) 9 F    (3) 2F    (4) 4F

    7.Calculate the acceleration due to gravity at an altitude of 6.37 X 106 m above the earth's surface

    8.Two masses are attracted to each other by a force of 3.4 X 10-2 N. What would be the force of

        attraction if the masses of each were doubled and the distance between them kept constant?

    9.Referring to problem  8, what would be the force if one mass was doubled, the other mass

        tripled and the distance between them 1/2 the original separation distance?

    10.The mass of moon is about 7.35 x 1022 kg. The mass of the earth is about 5.98 x 1024

    1. If the centers of the two are 3.84 x 108 m apart, what is the gravitational force

        between them?

    11.The mass of an electron is 9.11 x 10-31 kg. The mass of a proton is 1.67 x 10-27 kg.

        They are about 1.0 x 10-10 m apart in a hydrogen atom. What is the gravitational force between these two particles in the hydrogen atom

    12.What is the acceleration due to gravity at an altitude of 3 x 106 m above the surface of the earth

    13.What velocity is needed to keep a satellite in orbit at an altitude of 4 000 km above the surface of the earth?

    14.A space probe is launched into space from Earth’s surface. Which graph represents the relationship between the magnitude of the gravitational force exerted on Earth by the space probe and the distance between the space probe and the center of Earth?

    Hooke’s Law:  F= -kX

    15.A spring has an unstretched length of 0.40 meter. The spring is stretched to a length of 0.60 meter when a 10.-newton weight is hung motionless from one end. The spring constant of this spring is (#32, June 2018)

    (1) 10. N/m         (2) 17 N/m         (3) 25 N/m             (4) 50. N/m

    16.The diagram below represents a 35-newton block hanging from a vertical spring, causing the spring to elongate from its original length. ( june 2017)

     Determine the spring constant of the spring.

    17.A vertical spring has a spring constant of 100. newtons per meter. When an object is

     attached to the bottom of the spring, the spring changes from its unstretched length of 0.50  meter to a length of 0.65 meter. The magnitude of the weight of the attached object is

      (1) 1.1 N          (2) 15 N           (3) 50. N               (4) 65 N

    18.

    Directions (61–63): Construct a graph following thedirections below.

    (A) Mark an appropriate scale on the axis labeled “Force (N).”

    (B)Plot the data points for force versus elongation.

    (C) Draw the best-fit line or curve.

    (D) Using your graph, calculate the spring constant of this spring. [Show all work, including the

    equation and substitution with units.]

      

    Uniform Circular Motion:  ac = V2/ r

                                                   Fc  = m V2 / r

                                                       

    19.Circular motion at a constant speed. Which diagram best represents the directions of both the car’s velocity, v, and acceleration, a?

    20.The diagram below represents a mass, m, being swung clockwise at constant speed in a horizontal circle.

     

      At the instant shown, the centripetal force acting on mass m is directed toward point

        (1) A       (2) B       (3) C       (4)

    21.An electron in a magnetic field travels at constant speed in the circular path represented

      in the diagram below.(June 2017)

      

     

    Which arrow represents the direction of the net force acting on the electron when the electron is at position A?

     

    22.Which diagram represents the directions of the velocity, v, and acceleration, a, of a toy car as it moves in a clockwise, horizontal, circular path at a constant speed? (#12, June 2018)

     23.If a 65-kilogram astronaut exerts a force with a magnitude of 50. newtons on a satellite that

        she applied, the magnitude of the force that the satellite exerts on her is

                (1) 0 N            (2) 50. N less than her weight             (3) 50. N more than her weight             (4) 50.

    24.On a flat, level road, a 1500-kilogram car travels around a curve having a constant radius of 45 meters. The centripetal acceleration of the car has a constant magnitude of 3.2 meters

        per second squared.(#76 to # 80 June 2018)

        (A) Calculate the car’s speed as it travels around the curve. [Show all work, including the equation and substitution with units.] [2]

        (B)  Determine the magnitude of the centripetal force acting on the car as it travels around the curve. [1]

       ( C ) What force provides the centripetal force needed for the car to travel around the curve?

        (D)Describe what happens to the magnitude of the centripetal force on the car as it travels around the curve if the speed of the car decreases. [1]

     

    Two Dimensional Motion of a projectile

       No force along X (horizontal direction)   V1x = Vfx = Vcos  ç no change for speed on X

                                                  X ( horizontal displacement) = [V cos(time of flight)

       Gravitational force along y (vertical direction), g is pointing downwards

                                                V1y = V sin                        

                                                 Vfy V sin  è Vfy = Viy – gt = V sin – gt

                                                 at peak of y, V iy  = 0è   t = V sin /g = time to peak     

                                                                        time of flight ( in air) = 2 V sin /g

                                          d  =  Viy (t) – (1/2) g t2

                                             Vfy2 = Viy2 – 2 g(d) = [V sin ]2  - 2(g)(d)

          Time of flight only depends on the initial y component of velocity (= Vsin )

                           Larger angle , longer time of flight, longest time of flight occurs at 900

                           Maximum X displacement occurs at 450

                           Maximum Y displacement occurs at 900

    25.A projectile with mass m is fired with initial horizontal velocity vx from height h above level ground. Which change would have resulted in a greater time of flight for the projectile?   

      [Neglect friction.]   (#8, June 2018)

      (1) decreasing the mass to m/2             (2) decreasing the height to h/2    (3) increasing the initial horizontal velocity to 2vx     (4) increasing the height to 2h

    26.A soccer ball is kicked into the air from level ground with an initial speed of 20. meters per second and returns to ground level. At which angle above the horizontal should the ball be

       kicked in order for the ball to travel the greatest total horizontal distance? [Neglect friction.]    (#5, June 2018)

        (1) 15°           (2) 30.°          (3) 45°        (4) 75

    27. What angle above the horizontal line will result in a longest the time of flight for a projectile

      with an initial speed of 50 m/s in the two-dimensional motion?

      (1) 200           (2) 300          (3)450           (4) 900

    1. What angle above the horizontal line will result in a highest height (y displacement) for a  projectile with an initial speed of 35 m/s in the two-dimensional motion?

      (1) 900           (2) 300          (3)450           (4) 600

     29.The diagram below represents a ball projected horizontally from a cliff at a speed of 10. meters per second. The ball travels the path shown and lands at time t and distance d

         from the base of the cliff. [Neglect friction.]

         

       A second, identical ball is projected horizontally from the cliff at 20. meters per second.

       Determine the distance the second ball lands from the base of the cliff in terms of d. [1]

    30.A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal component of the ball’s initial velocity is 13.0 meters per second. The magnitude

      of the vertical component of the ball’s initial velocity is 7.5 meters per second. [Neglect friction.] (#60-#62 June 2015)

       (A)On two perpendicular axes, draw a graph representing the relationship between the horizontal displacement of the football and the time the football is in the air. [1]

       (B)The football is caught at the same height from which it is thrown. Calculate the total time the football was in the air. [Show all work, including the equation and substitution with units.] [2]

     

    Friction force

           Ff  = u W = u (m g);          Fnet = ma = Fa  - Ff

             If Fa = 0 ( no applied force)è    Fnet = - Ff   and   a =u(g

    31.An object weighing 2.0 newtons is pushed across a horizontal, frictionless surface by a horizontal force of 4.0 newtons. The magnitude of the net force acting on the object is

      (1) 0.0 N                     (2) 2.0 N                 (3) 8.0 N                          (4) 4.0 N


    32. As represented in the diagram below, a constant 15-newton force, F, is applied to a 2.5-Kilogram box, accelerating the box to the right at 2.0 meters per second squared across

          a rough horizontal surface.

     

       (A)Calculate the magnitude of the net force acting on the box. [Show all work, including the equation annd substitution with units.] [2]

       (B) Determine the magnitude of the force of friction on the box

    33.A 150-newton force, applied to a wooden crate at an angle of 30.º above the horizontal, causes the crate to travel at constant velocity across a horizontal wooden floor, as

    represented below. (#71, to #75, June 2018)

      (A) Calculate the magnitude of the horizontal component of the 150-newton force. [Show all work,including the equation and substitution with units.] [2] 

      (B) Determine the magnitude of the frictional force acting on the crate. [1]

      (C ) Calculate the magnitude of the normal force exerted by the floor on the crate. [Show allwork, including the equation and substitution with units.] [2]

    34.An object is given an initial speed of 3.0 m/s on level ice and comes to rest in 30 m. What is the coefficient of friction

    35.A hockey puck weighing 1.1 N slides on the ice for 15 m before it stops. (a) If its initial speed was 6.0 m/s, what is the force of friction between the puck and ice? (b) What is the

         coefficient of friction?  (c) How long (time)  wll it take to stop?

     

    Motion along an inclined Plane:Weight perpendicular to inclined plane = W(cos)=mgcos

                                                           Acceleration along the inclined plane a = g(sin )

     

    36.(A)A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the component of the box’s weight perpendicular

         to the incline?   (1) 19 N   (2) 21 N  (3) 42 N   (4) 46 N  (#39, June 2016)

        (B)What is the magnitude of acceleration of the box along the inclined plane?

     

    Apparent weight inside an accelerated elevator

       Apparent weight >  your weight )mg) on surface of earth when elevator is being

                                     accelerated Upwards

      Apparent weight  <  your weight (Mg) on surface of earth when the elevator is being

                                     accelerated Downwards