A grinding wheel of 0.08 m radius operates under power at a steady angular velocity of 120 radians per second.  Power is shut off at time t =  0 and the wheel decelerates uniformly at 1.5 rad /s2.
A   B
1D. 1D  The initial rate of rotation of the wheel, in RPM, is closest to:
120rad/sec = 120* (rad/2p) / (1/60 min) = 120 * 60 / 2p = 1145
2A. 2D  The linear speed of a point on the rim of the wheel when time t =  0 s, in SI units, is closest to:
v = w*r = 120 * 0.08 = 9.6
3C. 3B  The time t, in seconds, at which the wheel comes to rest is closest to:
w = w₀ - a Dt  so  Dt = w₀ / a = 80
4C. 4C  The angular displacement, in radians, when time t = 10 sec  is closest to:
Q = Q₀ + w₀t - 1/2 a t² = 120*10 - 1/2*1.5*10² = 1125
5C. 5C   The number of revolutions made by the wheel in the time interval during which the angular velocity drops from 90 rad/s to 80 rad / sec is closest to:
w² = w₀² - 2 a DQ     so  DQ = (90² -80²) / (2 *1.5) = 566 rad = 90

A pulley wheel of inner radius, R₁ = 0.4 m and outer radius R₂ = 0.6 m rotates about a frictionless axis through its center.  A mass M is hung from the outer pulley hub and accelerates downward at the rate of 2 m/s².  A cord is wound around the inner pulley and a horizontal force F of 50 N is applied as shown.

6C. 6B  If the tension in the outer cord connecting the mass M to the pulley is 39 N the mass M in kilograms is close to
T - Mg = M a   so     39 - M*9.8 = M * (-2)    and   M = 5
7B.  7B The net torque on the pulley wheel in Nm, assuming the conter-clockwise direction to be positive is close to:
 t = +39*R₂ - 50*R₁ = 39*0.6 - 50*0.4 = 23.4 - 20 = 3.4
8C. 8B  The angular acceleration of the pulley wheel in rad/s² is closest to :
a = a*R₂   so a = a/R₂ = 2/0.6 = 3.33
9B.9B   The moment of inertia of the pulley wheel, in kgm², is closest to:
t = I*a      so    I =t /a = 3.4 / 3.33 = 1
10C.10D   The moment of inertia of the wheel in the figure is 0.50 kgm2, and the bearing is frictionless.  The acceleration of the 6-kg mass is approximately:
t = I*a    so  -T*R =  I*(-a) = I*(-a/R)    and   for mass M:     T - Mg = M(-a)
 T = M(-a) + Mg = M (g-a)    then  M (g-a) *R =  I*a/R   Mg = a [ I/R2 + M ]  a = 60 / [0.5/(0.5)² + 6] = 60/8 = 7.5

Block  having a mass of 6 kg and initially at rest is struck by a baseball bat which imparts an impulse to it.  The time the bat and block A are in contact is 0.04 seconds.  As a result of being struck block A acquires a velocity, v₁ = 10 m/s.  Block A then slides along a frictionless surface and collides with block B which is initially; at rest.  The two blocks stick together and after the collision move with a velocity of v₂ = 4 m/s.

11D. 11D   The average force acting on block A, during the time interval when the block and bat are in contact, in SI units, is closest to:
F_avg*Dt = m*Dv     then   F_avg = 6*10 / 0.04 = 1500
12D. 12A  The mass of block B, in SI units, is closest to:
conservation of momentum:   before collision = 6*10 = after collision = (6+m_B)*4    and find m_B
m_B = (60 - 24)/4 = 9
13B.  13C The magnitude of the impulse acting on block A due to contact with the bat, in SI units is closest to
J = Dp   so J = 6*10 - 0 = 60
14C. 14C  The magnitude of the impulse acting on block A due to the collision with block B, in SI units, is closest to:
J = Dp  so J = final p- initial p = 6*4 - 6*10 = - 36  so magnitude is 36
15C. 15C  The energy dissipated in the collision between blocks A and B, in SI units, is closest to
K_final = 1/2 (6+ 9) 4² = 120 J   K_initial = 1/2 (6) 10² = 300 J   then energy dissipated during the collision is 300 - 120 = 180
16B. 16C   A particle of mass m moving at 5.0 m/s in the positive x direction makes a glancing elastic collision with a particle of mass 2m that is at rest before the collision.  After the collision, m moves off at an angle of 60^o above the  x-axis and 2m moves off at 45^o below the x-axis .  The speed of m after the collision is about:
Linear momentum is conserved during collision in both x and y direction:
In x direction linear momentum is: before = m *5 = after = m v_m cos60 + 2m v_2m cos45^o
In y direction initial linear momentum as well as final is zero.    m v_m sin60 - 2m v_2m sin45^o = 0
from y direction:   v_2m =  (m v_m sin60)/(2m sin45^o) = 0.612 v_m
and use it in x direction:  m *5 = m v_m cos60 + 2m (0.612 v_m) cos45^o
and factoring:   v_m = 5 / [cos60 + 2*0.612*cos45] =  5 / 0.865 = 5 / 0.5 + 1.224 * 0.707 = 5 / 1.365 = 3.661
17B. 17B  Two identical masses are hung on strings of the same length.  One mass is released from a height h above its free hanging position and strikes the second mass.  The two stick together and move off. They rise to a height H given by:
inelastic collision:  to calculate velocity  for free falling mass just before the collision use conservation of energy:
mgh = 1/2 mv²   so:     v = m*sqrt(2gh)
then conservation of momentum during collision;   m*v_initial = m*sqrt(2gh) = 2m v_final    v_final = 1/2 sqrt(2gh)
again use the conservation of energy to raise mass 2m:     2m g H = 0.5*2m*[1/2 sqrt(2gh)]² =  1/4  m*2*g*h
finally  H = 1/4 h
18B. 18C  Two identical cars approach an intersection.  One is traveling east at 24 m/s.  The second is traveling north at 18 m/s.  They collide violently, sticking together.  Immediately after the crash they are moving:
conservation of momentum:   initial = m(24,0) + m(0,18) = final = 2m(vcosQ, vsinQ)
then:  24=2vcosQ     and    18 = 2vsinQ
dividing:   18/24 = tanQ  so Q = +37^o   and v = 9/sinQ = 15
19B. 19B  The graph shows the momentum of a body as a function of time.  The time at which the force acting on the body is greatest is
20D. 20C  A 2-lb ball is dropped from a height of 16 ft and rebounds to a height of 9 ft.  The ball is in contact with the ground for 0.02s.  The impulse acting on the ball in (slug  ft/s) is
1 slug (mass)= 1 lb (weight) / 32 ft/s² (gravity constant)
in coordinate axis pointing up:
J = Dp = (2lb/32 ft/s²) [ + sqrt(2*32*9)  (- -)  sqrt(2*32*16) ] = 2/32 ( 24 + 32) = 2*56/32 = +3.5 you can find J