Physics 40a Final Exam Review

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  • 8/12/2019 Physics 40a Final Exam Review

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    Physics 40A - Final Exam Solutions - Winter 2013

    Problem 1[25 points]

    Two boxes of masses m1= 1.65 kg and m

    2= 3.30 kg slide down an inclined plane (with = 30)

    while attached by a massless rod as shown. The coefficient of kinetic friction between box 1 and

    the incline is 1= 0.226; that between box 2 and the incline is 2= 0.113. (a) Draw free bodydiagrams for box 1 and box 2. Compute (b) the tension in the rod, and (c) the commonacceleration of the two boxes.

    Solution:(a) See diagrams above. Tis the magnitude of the force in the rod,N

    1and N

    2are the normal

    forces on boxes 1 and 2,f1andf

    2are the kinetic friction forces on boxes 1 and 2.

    (b) For each block we take +xdown the incline and +yin the direction of the normal force.Applying Newton's second law to thexandydirections we obtain two equations for each box:

    m1 g sin f 1 T = m1 a 1

    N1

    m1g cos = 0 2

    m2

    g sin f2

    T = m2

    a 3

    N2 m2g cos = 0 4Dividing Eq. (1) by m

    1and dividing Eq. (3) by m

    2gives

    g sin f 1/ m1 T/ m1 = a and g sin f 2 / m2 T/ m2 = aSubtracting gives

    f

    1

    m1

    f2

    m2

    T

    m1

    T

    m2= 0 Tm1 m2m1m2 =

    m2

    f1

    m1

    f2

    m1 m2

    T = m

    2 f

    1 m

    1 f

    2

    m1 m2Using f1 = 1N1 = 1 m1 g cos and f 2 = 2N2 = 2 m2 g cos we get

    T = m2 1 m1g cos m1 2 m2 g cos

    m1

    m2

    = m1m2g cos 1 2

    m1

    m2

    = 1.65kg 3.30 kg9.8m/s

    2cos 30

    0.26 0.113

    1.65kg 3.30 kg T = 1.05 N

    (c) Substituting forf1in Eq. (1) we ge

    a = g sin 1 g cos T

    m1= 9.8m/s2sin30 0.2269.8m/s2 cos30

    1.05N

    1.65kg

    a = 3.62 m/s2

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    Physics 40A - Final Exam Solutions - Winter 2013

    Problem 2[25 points]

    A pulley of rotational inertia 0.385 kgm2and radius 33 cm is mounted on a horizontal axle. Abucket of mass 1.53 kg hangs from a massless cord which is wrapped around the rim of thepulley. A frictional torque of 1.1 Nm acts on the pulley at the axle. Assume that the cord does

    not slip. (a) Calculate the angular acceleration of the pulley and the linear acceleration of thefalling bucket. (b) Determine the angular velocity of the pulley and the linear velocity of thebucket at time t= 3.0 s if the pulley (and bucket) start from rest at t= 0.

    Solution:

    (a) Suppose the pulley rotates counterclockwise due to the torque caused by the tension in thecord. Then the frictional torque would tend to oppose the motion, i.e. it would tend to make itrotate clockwise. Therefore, Newton's second law applied to the rotating pulley gives

    T R f = I

    Newton's second law applied to the bucket gives

    T m g = ma T = m g a

    Because the cord does not slip, the tangential component of the acceleration of a point P at therim of the pulley is equal to the acceleration of the bucket. Therefore, since Pmoves in a circle,

    a = RNow we have

    I = T R f = m g aR f = m g R R fSolving for gives

    = m g R f

    I mR2

    = 1.53 kg 9.8 m/s20.33m 1.1 Nm

    0.385 kgm2 1.53kg 0.33 m2 = 6.98 rad/s2

    Now we obtain the linear acceleration

    a = R = 0.33m 6.98rad/s2 = 2.30 m/s2

    (b) Since the angular acceleration is constant, after 3.0 s

    = 0 t = 0 6.98rad/s 3.0s = 20.9rad/s

    The velocity of the bucket is the same as that of point Pon the rim of the pulley:

    v = R = 20.9rad/s 0.33 m = 6.90 m

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    Physics 40A - Final Exam Solutions - Winter 2013

    Problem 3[25 points]In the Figure below, a 2.5 kg block slides head on into a spring with a spring constant of 320

    N/m.When the block stops it has compressed the spring by 7.5 cm. The coefficient of kineticfriction between the block and the horizontal surface is 0.25. While the block is in contact with

    the spring and being brought to rest, what are (a) the work done by the spring force, and (b) theincrease in thermal energy of the block-floor system? (c) What is the block's speed just as theblock reaches the spring?

    Solution:

    (a) The work done by the spring force is

    W =xi

    xfF dx =

    xi

    xfkx dx =

    [

    1

    2

    k x2

    ]x

    i

    xf

    = 1

    2

    kxi2 x f2

    With x i = 0 and x f = 7.5 cm = 0.075 m we find

    W = 1

    2320 N/m 0 0.075 m2

    W = 0.9 J

    (b) The increase in thermal energy of the block-floor system is

    Eth = f kd

    wherefkis the frictional force and d = 7.5 cm is the distance moved during the time the block isin contact with the spring and being brought to rest.. The frictional force on the block from thefloor is

    f k = kN

    and Newton's second law applied to the block in the vertical direction isN mg = 0Therefore,

    f k = kmg

    and we find

    Eth = f kd = km g d = 0.25 2.5 kg 9.8 m/s2 0.075 m

    Eth = 0.46J

    (c) Consider the motion of the block from the moment it makes contact with the spring (when itsspeed is v) to the moment when it compresses the spring and comes to rest. There are no externalforces acting on the spring-block-floor system in the horizontal direction, so from conservation ofenergy, we find

    Emec Eth = K U Eth = 0 K = U Eth

    0 1

    2m v

    2 = 1

    2k d

    2 0 0.46 J

    1

    2m v

    2 = 1

    2k d

    2 0.46 J = 1

    2320 N/m 0.075 m2 0.46 J = 1.36 J

    v = 2 1.36 J

    2.5kg v = 1.04 m/s

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    Physics 40A - Final Exam Solutions - Winter 2013

    Problem 4[25 points]A steel ball of mass 0.5 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. Theball is then released when the cord is horizontal (see Figure). At the bottom of its path the ballstrikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find

    (a) the velocity of the ball and (b) the velocity of the block both just after the collision.

    Solution:

    (a) First we find the speed of the ball (of mass m1) right before it hits the block. When the cord is

    horizontal the ball is at height h and has zero velocity. Juts before the ball hits the block it haszero height and has speed v. From conservation of mechanical energy we find

    m1g h = l

    2m1 v

    2

    v = 2gh = 29.8 m/s20.7m = 3.7 m/s

    We now treat the elastic collision and find the velocity of the ball just after the collision using

    v 1f = m

    1 m

    2

    m1 m2v =

    0.5 kg 2.5 kg

    0.5 kg 2.5 kg3.7 m/s

    v 1f = 2.47 m/si.e. the ball moves to the left with speed 2.47 m/s.

    (b) The velocity of the block just after the collision is given by

    v 2f = 2m

    1

    m1 m2v =

    20.5 kg

    0.5 kg 2.5 kg3.7 m/s

    v2f

    = 1.23 m/s

    i.e. the block moves to the right with a speed of 1.23 m/s.

    4