PRACTICE 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 ?

PRACTICE 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.