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Two billiard balls of equal mass undergo a perfectly elastic head

By Barry Long,2014-04-01 19:08
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Two billiard balls of equal mass undergo a perfectly elastic head

PHY 201 Practice test 2 Summer 05

     Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one 1.

    ball’s initial speed was and the other’s was in the opposite direction, 2.00 ms,3.00 ms

    what will be their speeds after the collision?

Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B

    represent the ball moving at 3.00 m/s in the opposite direction. So v2.00ms and A

    . Use Eq. 7-7 to obtain a relationship between the velocities. v!,3.00msB

    ???? vvvvvv,!,,?!? 5.00ms;?ABABBA

    Substitute this relationship into the momentum conservation equation for the collision,

    mmnoting that . AB

    ????mvmvmvmvvvvv?!???!?? AABBAABBABAB

    ???? ,!???!,?!,1.00ms5.00ms 26.00ms 3.00msvvvv;?AAAA

    ??vv!?!5.00ms2.00msBA

    The two balls have exchanged velocities. This will always be true for 1-D elastic collisions

    of objects of equal mass.

2. A 95-kg fullback is running at to the east and is stopped in 0.75 s by a head-on 4.0 ms

    tackle by a tackler running due west. Calculate (a) the original momentum of the

    fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler, and (d) the average force exerted on the tackler.

Call east the positive direction.

    2 (a) pmv!!!?95 kg4.0ms3.810kgms ;?;?originaloriginalfullbackfullback

     (b) The impulse on the fullback is the change in the fullback’s momentum.

    )?2 ,!,!,!,?pmvv95 kg04.0ms3.810kgms;?;?fullbackfinalfinal?:fullbackfullback,,

     (c) The impulse on the tackler is the opposite of the impulse on the fullback, so

    2 3.810kgms?

     (d) The average force on the tackler is the impulse on the tackler divided by the time of interaction.

    2,?p3.810kgms2 F!!!?5.110Nt0.75 s

     1

     A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a 3.

    distance of 115 m. (a) What is the total number of revolutions the wheel rotates in

    coming to rest? (b) What is the angular acceleration? (c) How long does it take for the

    wheel to come to the stop?

See Example 8-7 p. 202-203

4. A hollow cylinder (hoop) is rolling on a horizontal surface at speed when it v3.3 ms

    reaches a 15º incline. (a) How far up the incline will it go? (b) How long will it be on the

    incline before it arrives back at the bottom?

    (a) Assuming that there are no dissipative forces doing work,

    conservation of energy may be used to find the final

    height h of the hoop. Take the bottom of the incline to be

    the zero level of gravitational potential energy. We

    assume that the hoop is rolling without sliding, so that h

    . Relate the conditions at the bottom of the ~vR;

    incline to the conditions at the top by conservation of

    energy. The hoop has both translational and rotational

    2 kinetic energy at the bottom, and the rotational inertia of the hoop is given by . ImR

    2v22221111EEmvImghmvmRmgh!??!??!? ~bottomtop22222R 22;?3.3msvh!!!1.111 m2g9.8ms

    h1.111 m!!!? The distance along the plane is given by d4.293 m4.3 mosinsin15

(b) The time can be found from the constant accelerated linear motion. Use the relationship

    24.293 m2x;?1. ,!??!!!xvvtt 2.602 s;?o2vv??03.3mso

    This is the time to go up the plane. The time to come back down the plane is the same,

    and so the total time is . 5.2 s

     2

     A 140-kg horizontal beam is supported at each end. A 320-kg piano rests a quarter of 5.

    the way from one end. What is the vertical force on each of the supports?

    L Let m be the mass of the beam, and M be the mass of the piano. L/4 Calculate torques about the left end of the beam, with

    counterclockwise torques positive. The conditions of

    equilibrium for the beam are used to find the forces that the

    support exerts on the beam. FRFMg11mgL !,,!FLmgLMgL0;?;?(R24

    231111FmMg!?!?!?140 kg320 kg9.80ms1.4710N;?;?;?(;;?R2424

     FFFmgMg!?,,!0(yLR

    233FmMgF!?,!,?!?460 kg9.80ms1.4710N3.0410N;?;?;?LR

     The forces on the supports are equal in magnitude and opposite in direction to the above

    two results.

    33 F!?1.510N downF!?3.010N downRL

6. (a) An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water.

    What is the specific gravity of the object?

     3327.84 N 6.86 N = g V ------> 0.98 N = (1x10 kg/m) (9.80 m/s) V water

     -432V = 1x10 m Mass of the object M = 7.84 N/(9.8 m/s) = 0.8 Kg

     -4333Density of object is = M/V = 0.8 Kg/(1x10 m ) = 8x10 kg/m. object

Specific gravity = / = 8 objectwater

     (b) How much pressure does it take for a pump to supply a drinking fountain with

    300 kPa, if the fountain is 30.0 m above the pump?

     1122 PvgyPvgy11122222 33With v = v , y = 0 , y = 30.0 m , P = 300 kPa , = 1x10 kg/m 12122

You calculate P = 594 kPa 1

     3

    7. A mass sitting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. 3.0 J of work is required to compress the spring by 0.12 m. If the mass is released from rest with the spring compressed, the mass experiences a

    2 Find the value of (a) the spring stiffness constant and maximum acceleration of 15 ms.

    (b) the mass.

(a) The work done to compress a spring is stored as potential energy.

    23.0 J2W;?221 Wkxk!?!!!?? 416.7Nm4.210Nm222x0.12 m;?

    (b) The distance that the spring was compressed becomes the amplitude of its motion. The

    kmaximum acceleration is given by . Solve this for the mass. aAmaxm

    2kk)?4.16710Nm? aAmA!?!!!? 0.12 m3.333 kg3.3 kg;?max?:2ma15ms,,max

     4

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