Physics 131
Exam (Ch 6 to 9)
Name
Chapter 06, Problem 051
An airplane is flying in a horizontal circle at a speed of 500 km/h.
If its wings are tilted at angle θ = 30° to the horizontal, what is
the radius of the circle in which the plane is flying?
Fc = tan43 mg
Fc = m v 2 / r
tan θ = Fc / mg
22/22
11/11
470 km/hr (1hr/3600s) (1000m/1km)
= 130.56 m/s
tan30° g = 130.562 / r
r = 1865 m
Chapter 09, Problem 068
Block 1 of mass m1 slides from rest along a frictionless ramp from
height h = 3.0 m and then collides with stationary block 2, m 2 =
4*m1. After the collision, block 2 slides into a region where the
coefficient of kinetic friction μk is 0.40 and comes to a stop in
distance d within that region.
m1ghtop + ½mvtop2 = m1ghbot + ½mvbot2
vbot2 = 2ghtop
vbot = 7.75 m/s
5/5
Inelastic
Elastic
m1 v + m2v
= m1 v1 + m2 v2
m(7.75) + 0
= m v1 + 4m v2
7.75 = v1 + 4 v2
Now it collides
m1v + m2v = (m1+m2) v1-2
mv + 0 = (m + 4m) v1-2
v1-2 = 1.55 m/s
What is the value of distance d if the collision
is (a) elastic and (b) completely inelastic?
½m1 v2 + ½m2v2 = ½ m1 v12 + ½ m2 v22
m(7.75)2 + 0
= m v12 + 4m v22
60 = v12 + 4 v22
v2 = 3.1 m/s (check; v1 = -4.65)
12/12
3/3
2/2
Energy, post collision, is reduced by friction through a distance
4/4
4/4
4/4
Post collision energy is reduced by friction through a distance
½ m1&2 v2 –
Ff
d = ½ mvf2
½ 5m 1.552 – 0.4 (5m)g d = 0
6
– 20d
= 0
dInelas = 0.30 m
½ m2 v2 –
Ff
d = ½ mvf2
½ 4m 3.12 – 0.4 (4m)g d = 0
20 – 16d
delas = 1.2 m
Block 2 (mass 2.00 kg) is at rest on a frictionless
surface and touching the end of an unstretched spring of spring constant 4
N/cm. The other end of the spring is fixed to a wall. Block 1 (mass 3.00 kg), with
an initial speed of speed v1 = 8.00 m/s travels 4 meters on a dirty surface with
an effective coefficient of friction of 0.2, then collides with block 2 (clean and
frictionless at that this point), and the two blocks stick together.
Chapter 09, Problem 058
Initial energy is reduced by friction through a dist.
2/2
3/3
3/3
½m v2 – Ff
d = ½mvf2
2
½m 8 – 0.2(m“g”) 4 = ½mvf2
vf = 6.92 m/s
pinitial
7/7
=
pfinal
7/7
m1 v + m2v = m1 v1 + m2 v2
3(6.92) + 0 = (3 + 2) vf
vf = 4.16 m/s
When the blocks momentarily
stop, by what distance is the
spring compressed?
Init energy transfers and does work on the spring
TETop 6/6
=
TEfinal 7/7
½ m v2 + ½kx2 = ½ m vf2 + ½ k
x2
2
½(5)4.16 + 0 = 0 + ½(400N/m) x2
x = 0.465 m
Chapter 10, Problem 057
REPLACEMENT PROBLEM, only answer
to replace another problem; clearly mark out other problem
A pulley, with a rotational inertia of 2.00 × 10-3 kg·m2 about its axle and
a radius of 10.0 cm, is acted on by a force applied tangentially at its
rim. The force magnitude varies in time as F = t + 0.90t2, SI units. The
pulley is initially at rest. At t = 4.00 s what is its angular speed?
τ
= I
α
rF
= I dω/dt
F dt
= ( I / r) ∫dω
2
∫tdt+ .9∫t dt = ( I / r) Δω
½t2 + .9/3 t3 = (2×10-3/.1) Δω
Δω = 1360 rad/sec