__P____RACTICE EXAM 2.__

**Examples of
problems from the past exams, probably 3 or 4 problems will be given at the
test.**

A. A mass of 4 kg is
moving from left to right with a velocity of 3 m/s. It strikes a mass of 5 kg,
which is initially at rest at the table and the two masses stick together.

a) determine the
final velocity of the two masses

b) determine the
linear momentum of the two masses after they join in the table reference frame

c) determine the
linear momentum of the two masses with the velocity measured relative to the
velocity of the center of mass

d) determine the
kinetic energy of the two masses when they stick together in the table
reference frame

e) determine the
kinetic energy of the two masses when they stick together in the center of mass
reference frame.

B. Two blocks joined
by a massless string are sliding down the inclined plane at the angle of 40
degrees with respect to the horizontal. The upper block has mass M=5 kg, and
the lower has mass m=1 kg. The coefficient of friction between the M block and
the plane is m_M=0.4 , and that between
the m block and the plane is m_m=0.10.

a) Find the
tension in the string.

b) Find the
acceleration of the masses

c) Find the
velocity of the masses after they slide by 1 m along the inclined plane (use
the energy conservation methods)

C. An 800 kg car
coasts down a hill 40 m high with its engine off and the driver's foot pressing
on the brake pedal. At the top of a hill the car's speed is 6 m/s, and at the
bottom it is 20 m/s. How much energy was converted to the heat on the way down
?

D. A 1200 kg car
travelling North at 30 km/h collides with a 1800 kg car travelling at 20 km/h.
The cars stick together after the collision. What is the velocity vector of the
wreckage ?

E. Two blocks are
arranged at the ends of a massless string. One block (m=1.5 kg) is resting on
the table, the other (M=2 kg) is hanging freely off the pulley. The system
starts from rest, and when M has fallen through s=40 cm, its downward speed is
1.25 m/s.

Assume that the
string and the pulley are massless.

a) What is the
frictional force between the block and the table ?

b) What is the
coefficient of friction between the block m and the table ?

c) How much work
was done to overcome friction by the time the block M has fallen through L= 1 m
?

__P____RACTICE EXAM 2 (WITH
SOLUTIONS)__

**Examples of
problems from the past exams, probably 3 or 4 problems will be given at the
test.**

A. A mass of 4 kg is
moving from left to right with a velocity of 3 m/s. It strikes a mass of 5 kg,
which is initially at rest at the table and the two masses stick together.

a) determine the
final velocity of the two masses

(m_1)(v_1)=(m_1+m_2)V_f

V_f=(m_1)(v_1)/(m_1+m_2)=(3 m/s)(4
kg)/(9 kg)=4/3 m/s.

b) determine the
linear momentum of the two masses after they join in the table reference frame

The linear momentum is conserved,
p_i=p_f.

p_i=(m_1)(v_1)=(3 m/s)(4 kg)=12
kgm/s.

p_f=(m_1+m_2)(v_f)=(9 kg)(4/3
m/s)=12 kgm/s.

c) determine the
linear momentum of the two masses with the velocity measured relative to the
velocity of the center of mass

The center of mass reference
frame is tied to the center of mass of the two masses. After the collision when
the two masses stick together, their center of mass position is where the two
masses are. Since the two masses move with V_f, the center of mass velocity is
V_cm=V_f. If the velocities in the cms are denoted by U, they are connected with
the velocities in the original (table) frame (in which the two masses were
moving with V_f):

U=V-V_cm=V-V_f

Specifically,

U_f=V_f-V_cm=V_f-V_f=0

The center of mass is AT REST in the center of mass reference frame.

Thus the linear momentum of the two mass system in the cms reference frame is
zero:

P=(m_1+m_2)U_f=0

d) determine the
kinetic energy of the two masses when they stick together in the table
reference frame

E_k(table)=(1/2)(m_1+m_2)(V_f)^2=(1/2)(9 kg)(4/3 m/s)^2=8 J

e) determine the
kinetic energy of the two masses when they stick together in the center of mass
reference frame.

E_k(cms)=(1/2)(m_1+m_2)(U_f)^2=0

B. Two blocks joined
by a massless string are sliding down the inclined plane at the angle of 40 degrees
with respect to the horizontal. The upper block has mass M=5 kg, and the lower
has mass m=1 kg. The coefficient of friction between the M block and the plane
is m_M=0.4 , and that between
the m block and the plane is m_m=0.10.

a) Find the
tension in the string.

Let's write the second Newton's law for each of the masses. x-axis points
down the inclined plane, parallel to it; the y axis is perpendicular to the
inclined plane, pointing up.

M)

T+Mgsina-Mgcosa(m_M)=Ma

m)

mgsina-mgcosa(m_m)-T=ma

We have two equations for two unknowns, a and T. Solving for a:

a=(gsina(M+m)-gcosa(m_m+m_M))/(M+m)

T can be then found from one of the equations (second Newton's law for m or M)

T=mgsina-mgcosa(m_m)-ma

b) Find the
acceleration of the masses

as above

c) Find the
velocity of the masses after they slide by 1 m along the inclined plane (use
the energy conservation methods)

When the masses slide by L along the inclined plane, they move
vertically down by H=Lsina.

Considering the two masses, the gravitational field and the frictional
interactions as one system we

write (as the total energy is conserved):

E_k(i)+U_g(i)=E_diss+E_k(f)+U_g(f)

E_k(i)=0

E_k(f)=U_g(i)-U_g(f)-E_diss=(m+M)gLsina-Lgcosa(M(m_M)+m(m_m)=(1/2)(M+m)V^2

From which one easily find V.

C. An 800 kg car
coasts down a hill 40 m high with its engine off and the driver's foot pressing
on the brake pedal. At the top of a hill the car's speed is 6 m/s, and at the
bottom it is 20 m/s. How much energy was converted to the heat on the way down
?

As above:

E_k(i)+U_g(i)=E_diss+E_k(f)+U_g(f)

(1/2)m(v_i)^2+mgH=E_diss+(1/2)m)(v_f)^2

Find E_diss.

D. A 1200 kg car
travelling North at 30 km/h collides with a 1800 kg car travelling West at 20
km/h. The cars stick together after the collision. What is the velocity vector
of the wreckage ?

The linear momentum is conserved, and since the collision is in two
dimensions, the y-component (North) and x-component (East) are conserved
independently:

x: p_ix=p_fx

(1200
kg)(0)-(1800 kg)(20 km/h)=-36000 kgm/h=-(3000 kg)(v_fx) =>
v_fx=-12 km/h.

y: p_iy=p_fy

(1200 kg)(30
kg/h)+(1800 kg)(0)=36000 kgm/h=(3000 kg)(v_fy) =>
v_fy=12 km/h

Thus, v_f=(v_fx,v_fy)=(-12 km/h,12 km/h); or NW at 45 degrees with velocity (12
km/h)sqrt(2).

E. Two blocks are
arranged at the ends of a massless string. One block (m=1.5 kg) is resting on
the table, the other (M=2 kg) is hanging freely off the pulley. The system
starts from rest, and when M has fallen through s=40 cm, its downward speed is
1.25 m/s.

Assume that the
string and the pulley are massless.

Let's consider the two masses, frictional forces and the gravitational field
as one system. The total energy is conserved as there are no external forces
acting on the system.

E_k(i)+U_g(i)=E_diss+E_k(f)+U_g(f)

Mass m remains on the same level, its potential energy does not change,
while mass M moves down by H.Mass m moves horizontally by H, and energy is
dissipated due to friction along this distance. Initial kinetic energy is zero.

U_g(i)-U_g(f)=E_diss+E_k(f)

MgH=E_diss+E_k(f)

a) What is the
frictional force between the block and the table ?

f=mgm

b) What is the coefficient
of friction between the block m and the table ?

MgH=Hmgm+E_k(f)

MgH=Hmgm+(1/2)(M+m)(V)^2

From this we find m.

c) How much work
was done to overcome friction by the time the block M has fallen through L= 1 m
?

W_to overcome friction=E_diss=(1 m)mgm.