Download Phys 2A Sample Mid-Term 2 Dr. Ray Kwok 2. 3. 4. rad t 10)5.2( 2 08

Phys 2A
1.
Dr. Ray Kwok
The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off.
The turntable comes to rest 2.5 s after being turned off. Through how many radians does
the turntable rotate after being turned off? Assume constant angular acceleration.
a.
b.
c.
d.
e.
2.
Sample Mid-Term 2
12 rad
8.0 rad
10 rad
16 rad
6.8 rad
8+0
∆θ = ωaverage ∆t = 
(2.5) = 10rad
 2 
A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s2. The
diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel
at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s2?
a.
b.
c.
d.
e.
39 cm/s
42 cm/s
45 cm/s
35 cm/s
53 cm/s
aT = Rα = (0.2)(4) = 0.8
a = aT2 + a R2
1.2 2 = 0.82 + a R2
a R = 0.8 = Rω 2 = (0.2)ω 2
ω = 2.11
vT = Rω = (0.2)(2.11) = 0.422m / s
3.
The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-s time interval.
During this same time interval, the velocity of the object changes its direction by 90°.
What is the magnitude of the average total force acting on the object during this time
interval?
a.
b.
c.
d.
e.
4.
r
r
F∆ t = ∆ ( m v )
r
F (5) = (2.0)(40 yˆ − 30 xˆ )
r
F = 16 yˆ − 12 xˆ
30 N
20 N
40 N
50 N
6.0 N
F = 16 2 + 12 2 = 20 N
A 3.0-kg mass sliding on a frictionless surface has a velocity of 5.0 m/s east when it
undergoes a one-dimensional inelastic collision with a 2.0-kg mass that has an initial
velocity of 2.0 m/s west. After the collision the 3.0-kg mass has a velocity of 1.0 m/s east.
How much kinetic energy does the two-mass system lose during the collision?
a.
b.
c.
d.
e.
22 J
24 J
26 J
20 J
28 J
m1v1i + m2 v2 i = m1v1 f + m2 v2 f
(3)(5) + ( 2)( −2) = (3)(1) + (2)v2 f
v2 f = 4
1
1
1
 1

∆KE =  m1v12f + m1v22 f  −  m1v12i + m1v22i 
2
2
2
2

 

1
1
1

 
 35 75
∆KE =  (3)(12 ) + ( 2)(4 2 )  −  (3)(5 2 ) + 0 =
−
= −20 J
2
2
2
 2
 2
Phys 2A
5.
Sample Mid-Term 2
Dr. Ray Kwok
A 3.0-kg mass is sliding on a horizontal frictionless surface with a speed of 3.0 m/s when
it collides with a 1.0-kg mass initially at rest as shown in the figure. The masses stick
together and slide up a frictionless circular track of radius 0.40 m. To what maximum
height, h, above the horizontal surface will the masses slide?
m1v1i + m2 v2 i = ( m1 + m2 )v f
(3)(3) + 0 = ( 4)v f
0.40 m
vf = 9/ 4
1 2
mv = mgh
2
h
v
a.
b.
c.
d.
e.
6.
0.18 m
0.15 m
0.21 m
0.26 m
0.40 m
A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. Immediately after
the collision, the 2.0-kg object has a velocity of 1.5 m/s directed 30° from its initial
direction of motion. What is the x-component of the velocity of the 1.0-kg object just after
the collision?
a.
b.
c.
d.
e.
7.
2
1 9
m  = m(10)h
2 4
h = 0.253
3.7 m/s
3.4 m/s
1.5 m/s
2.4 m/s
4.1 m/s
m1v1ix + m2 v2ix = m1v1 fx + m2 v2 fx
( 2)(3) + 0 = ( 2)(1.5 cos 30 o ) + (1)v2 fx
v2 fx = 3.4m / s
Two bodies of equal mass m collide and stick together. The quantities that always have
equal magnitude for both masses during the collision are
a.
b.
c.
d.
e.
their changes in momentum.
the force each exerts on the other.
their changes in kinetic energy.
all of the above.
only (a) and (b) above.
Phys 2A
8.
1 2
mv = mgh + KE
2
1
2
(2)(6 ) = ( 2)(9.8)(3 − 3 cos 50 o ) + KE
2
KE = 15 J
0.18 kJ
0.97 kJ
0.89 kJ
0.26 kJ
0.40 kJ
1 2
mv + mgh = KE
2
1
2
(2)(24 ) + (2)(10)(20) = KE = 976 J
2
A spring (k = 600 N/m) is placed in a vertical position with its lower end supported by a
horizontal surface. A 2.0-kg block that is initially 0.40 m above the upper end of the spring
is dropped from rest onto the spring. What is the kinetic energy of the block at the instant
it has fallen 0.50 m (compressing the spring 0.10 m)?
a.
b.
c.
d.
e.
11.
21 J
15 J
28 J
36 J
23 J
A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity
of 24 m/s at some unknown angle above the horizontal. Disregard air resistance and
assume the ground is level. What is the kinetic energy of the mass just before it strikes the
ground?
a.
b.
c.
d.
e.
10.
Dr. Ray Kwok
A 2.0-kg mass swings at the end of a light string (length = 3.0 m). Its speed at the lowest
point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the
string makes an angle of 50° with the vertical?
a.
b.
c.
d.
e.
9.
Sample Mid-Term 2
5.3 J
6.8 J
6.3 J
5.8 J
6.5 J
mgh = KE +
1 2
kx
2
( 2)(9.8)(0.5) = KE +
1
(600)(0.1) 2
2
KE = 6.8 J
A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a
maximum height of 18 m above the launch point. How much work is done by the
dissipative (air) resistive force on the projectile during this ascent?
a.
b.
c.
d.
e.
–0.64 kJ
–0.40 kJ
–0.52 kJ
–0.28 kJ
–0.76 kJ
W = ∆E = E final − Einitial
W = mgh −
1 2
1
mv = (12)(9.8)(18) − (12)( 20) 2 = −283 J
2
2
Phys 2A
12.
Sample Mid-Term 2
A large spring is used to stop the cars after they come down the last hill of a roller coaster.
The cars start at rest at the top of the hill and are caught by a mechanism at the instant
their velocities at the bottom are zero. Compare the compression of the spring, xA, for a
fully loaded car with that, xB, for a lightly loaded car when mA= 2mB.
A
b.
1
xB.
2
xA = xB.
c.
d.
e.
xA = 2 xB.
xA = 2 xB.
xA = 4 xB.
a.
13.
B
mgh =
xA =
1 2
kx
2
Double the mass double x2.
i.e. (√2)x
A champion athlete can produce one horsepower (746 W) for a short period of time. The
number of 16 cm high steps a 70 kg athlete could ascend in one minute while expending
one horsepower is
a.
b.
c.
d.
e.
14.
Dr. Ray Kwok
4.
7.
65.
408.
4567.
In one sec,
W = mgh
746 = (70)(9.8)(0.16 n)
n = 6.80 steps/sec
In one minute, n = 6.80 * 60 = 408 steps/minute
A 2.5-kg object falls vertically downward in a viscous medium at a constant speed of 2.5
m/s. How much work is done by the force the viscous medium exerts on the object as it
falls 80 cm?
a.
b.
c.
d.
e.
+2.0 J
+20 J
–2.0 J
–20 J
+40 J
r r
W = F ⋅ ∆x = −( mg )( h) = −( 2.5)(10)(0.80) = −20 J
Phys 2A
15.
Sample Mid-Term 2
Dr. Ray Kwok
A 4.0-kg block is lowered down a 37° incline a distance of 5.0 m from point A to point B.
A horizontal force (F = 10 N) is applied to the block between A and B as shown in the
figure. The kinetic energy of the block at A is 10 J and at B it is 20 J. How much work is
done on the block by the force of friction between A and B?
A
F
37˚
37
37˚
37
a.
b.
c.
d.
e.
16.
W fr + WFx = ∆Etotal = E B − E A = KE B − KE A − mgh
(
)
(
W fr − 10 cos 37 o (5) = (20) − (10) − ( 4)(9.8) 5 sin 37 o
)
W fr = −68 J
Starting from rest at t = 0, a 5.0-kg block is pulled across a horizontal surface by a constant
horizontal force having a magnitude of 12 N. If the coefficient of friction between the
block and the surface is 0.20, at what rate is the 12-N force doing work at t = 5.0 s?
a.
b.
c.
d.
e.
17.
–58 J
–53 J
–68 J
–63 J
–47 J
B
0.13 kW
0.14 kW
0.12 kW
26 W
12 W
∑ F = ma
12 − fr = 12 − µ k N = 12 − µ k mg = 12 − (0.2)(5)(10) = 5a
a = 0. 4
v = vo + at = 0 + 0.4(5) = 2
r r
W F ⋅ ∆x r r
P=
=
= F ⋅ v = (12)(2) = 24W
∆t
∆t
A 10-kg block on a rough horizontal surface is attached to a light spring
(force constant = 1.4 kN/m). The block is pulled 8.0 cm to the right from its equilibrium
position and released from rest. The frictional force between the block and surface has a
magnitude of 30 N. What is the kinetic energy of the block as it passes through its
equilibrium position?
a.
b.
c.
d.
e.
4.5 J
2.1 J
6.9 J
6.6 J
4.9 J
1 2
kx
2
1
− (30)(0.08) = KE − (1400 )(0.08) 2
2
KE = 2.08 J
W fr = ∆E = KE −
Phys 2A
18.
0.
− mgh .
+mgh.
–2mgh.
+2mgh.
A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20
km) track. What is the magnitude of the resultant force on the 80-kg driver of this car?
a.
b.
c.
d.
e.
20.
Dr. Ray Kwok
When a ball rises vertically to a height h and returns to its original point of projection, the
work done by the gravitational force is
a.
b.
c.
d.
e.
19.
Sample Mid-Term 2
0.68 kN
0.64 kN
0.72 kN
0.76 kN
0.52 kN
Fnet = mac = m
v2
40 2
= (80)
= 640 N
R
200
A roller-coaster car has a mass of 500 kg when fully loaded with passengers. The car
passes over a hill of radius 15 m, as shown. At the top of the hill, the car has a speed of 8.0
m/s. What is the force of the track on the car at the top of the hill?
8.0 m/s
mg − N = m
15 m
a.
b.
c.
d.
e.
21.
v2
R
N = (500)(10) − (500)
82
= 2866 N
15
7.0 kN up
7.0 kN down
2.8 kN down
2.8 kN up
5.6 kN down
A split highway has a number of lanes for traffic. For traffic going in one direction, the
radius for the inside of the curve is half the radius for the outside. One car, car A, travels
on the inside while another car of equal mass, car B, travels at equal speed on the outside
of the curve. Which statement about resultant forces on the cars is correct?
a.
b.
c.
d.
e.
The force on A is half the force on B.
The force on B is half the force on A.
The force on A is four times the force on B.
The force on B is four times the force on A.
There is no net resultant force on either as long as they stay on the road while
turning.
Phys 2A
22.
Sample Mid-Term 2
Dr. Ray Kwok
A rock attached to a string swings in a vertical circle. Which free body diagram could
correctly describe the force(s) on the rock when it is at the lowest point?
(b)
23.
A car enters a level, unbanked semi-circular hairpin turn of 300 m radius at a speed of 40
m/s. The coefficient of friction between the tires and the road is µ = 0.25 . If the car
maintains a constant speed of 40 m/s, it will
a.
b.
c.
d.
e.
24.
attempt to dig into the road surface.
v2
40 2
F
=
m
=
m
= 5.33m
tend to veer toward the center of the
c
R
300
semicircle.
frmax = µ s N = (0.25)( m)(10) = 2.5m < Fc
arrive safely at the end of the semicircle.
tend to veer toward the outside of the circle.
veer toward the center for the first quarter-circle, then veer toward the outside for the
second quarter-circle.
A 500 g firework explodes into two pieces of equal mass at an instant when it is traveling
straight up at 10 m/s. If one half shoots off horizontally to the left at 20 m/s, what is the
velocity, in m/s, of the other half immediately after the explosion? (The x-axis is directed
right; the y-axis up.)
a.
b.
c.
d.
e.
−20 ˆi − 20ˆj
−20ˆi + 20ˆj
+20 ˆi − 20ˆj
+20ˆi + 20ˆj
−20 ˆi + 20kˆ
r  m r  m r
mvi =  v1 +  v2
2
2
1
1r
10 yˆ = ( −20 xˆ ) + v2
2
2
r
v2 = 20 xˆ + 20 yˆ