Download Exam1-PC

Partial Credit - Choose 3 out of 4 problems. (If you do all four problems
I take the best 3 scores)
1. In the overhead view of the Figure, five forces pull on a box of
mass m = 4.0 kg. The force magnitudes are F1 = 11 N, F2 = 17 N, F3
= 3.0 N, F4 = 14 N, and F5 = 5.0 N, and angle q4 is 30°. Find the
box's acceleration (a) in unit-vector notation and as (b) a magnitude
and (c) an angle relative to the positive direction of the x axis.
(2points each)
With SI units understood, the net force on the box is
r
Fnet = ( 3.0 + 14 cos 30° − 11) ˆi + (14 sin30° +5.0 − 17 ) ˆj
r
which yields Fnet = (4.1 N) ˆi − ( 5.0 N) ˆj .
(a) Newton’s second law applied to the m = 4.0 kg box leads to
r
Fnet
r
a =
= (1.0 m/s 2 )iˆ − (1.3 m/s 2 )ˆj .
m
r
2
2
2
(b) The magnitude of a is a = 1.0 + ( −1.3) = 1.6 m s .
(c) Its angle is tan–1 (–1.3/1.0) = –50° (that is, 50° measured clockwise from the rightward
axis).
2. The figure shows a kimchi container of mass m1 = 3.0 kg
connected to a block of mass m2 by a cord looped around a
frictionless pulley. The cord and pulley have negligible mass. When
the container is released from rest, it accelerates at 1.0 m/s2 across the
horizontal frictionless surface. What are (a) the tension in the cord (4
Points) and (b) mass m2? (2 Points)
Using the usual coordinate system (right = +x and up = +y) for
both blocks has the important consequence that for the m1 = 3.0 kg block to have a positive
acceleration (a > 0), block m2 must have a negative acceleration of the same magnitude (–a).
Thus, applying Newton’s second law to the two blocks, we have
T = m1a = (3.0 kg) (1.0 m/s 2 )
T − m2 g = m2 ( − 1.0 m/s 2 )
along x − axis
along y − axis.
(a) The first equation yields the tension T = 3.0 N.
(b) The second equation yields the mass m2 = 3.0/8.8 = 0.34 kg.
3. During volcanic eruptions, chunks of solid rock
can be blasted out of the volcano; these projectiles
are called volcanic bombs. The Figure shows a
cross section of Mt. Fuji, in Japan.
(a) At what initial speed would a bomb have to
be ejected, at angle q0 = 35° to the
horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical
distance h = 3.30 km and horizontal distance d = 9.40 km? Ignore the effects of air on the
bomb's travel. (4 Points)
(b) What would be the time of flight (2 Points)
(a) By rearranging Eq. 4-25, we obtain the initial speed:
v0 =
x
cosθ 0
g
2( x tan θ 0 − y )
which yields v0 = 255.5 ≈ 2.6 × 102 m/s for x = 9400 m, y = –3300 m, and θ0 = 35°.
(b) From Eq. 4-21, we obtain the time of flight:
t=
9400
x
=
= 45 s.
. cos 35°
v0 cosθ 0 2555
4. A proton moves along the x axis according to the equation x = 50t + 10t2, where x is in meters
and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its
motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous
acceleration of the proton at t = 3.0 s. (2 points each)
In this solution, we make use of the notation x(t) for the value of x at a particular t. Thus,
x(t) = 50t + 10t2 with SI units (meters and seconds) understood.
(a) The average velocity during the first 3 s is given by
v avg =
x (3) − x (0) (50)(3) + (10)(3) 2 − 0
=
= 80 m / s.
Δt
3
(b) The instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in SI units. At t = 3.0
s, v = 50 + (20)(3.0) = 110 m/s.
(c) The instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2. It is constant, so
the acceleration at any time is 20 m/s2.