Consider the system shown in the figure below with m 1 = 28.0 kg, m 2 = 11.1 kg,
ID: 1260070 • Letter: C
Question
Consider the system shown in the figure below with m1 = 28.0 kg, m2 = 11.1 kg, R = 0.280 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.20 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley.
(a) Calculate the time interval required for m1 to hit the floor.
?t1 = s
(b) How would your answer change if the pulley were massless?
?t2 = s
A 1 170.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 900.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.
(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.)
(b) What is the change in mechanical energy of the cartruck system in the collision?
J
Explanation / Answer
1.
Use the energy equation
Initial Energy = Final Energy
Initial Energy is only the potential energy of mass one (PE1).
Final Energy is the final kinetic energy of mass one (KE1), the final kinetic energy of mass two (KE2), the kinetic energy of the pulley (KEp) and the potential energy of mass two (PE2)
PE1 = KE1 + KE2 + KEp + PE2
PE1 = m1*g*h
PE2 = m2*g*h
KE1 = (1/2) m1 * v^2
KE2 = (1/2) m2 * v^2
KEp = (1/2) I w^2
I = (1/2) M r^2
http://en.wikipedia.org/wiki/List_of_mom...
and we know that w = (v/r)
PE1 = KE1 + KE2 + KEp + PE2
m1*g*h = (1/2) m1 * v^2 + (1/2) m2 * v^2 + (1/2) I w^2 + m2*g*h
Now substitute in I = (1/2) M r^2 and w = (v/r)
m1*g*h = (1/2) m1 * v^2 + (1/2) m2 * v^2 + (1/2) (1/2) M r^2 (v/r)^2 + m2*g*h
m1*g*h = (1/2) m1 * v^2 + (1/2) m2 * v^2 + (1/4) M v^2 + m2*g*h
m1*g*h - m2*g*h = v^2 [(1/2) m1 + (1/2) m2 + (1/4) M]
v^2 = (m1*g*h - m2*g*h)/[(1/2) m1 + (1/2) m2 + (1/4) M]
v^2 = g*h*(m1 - m2)/[(1/2) m1 + (1/2) m2 + (1/4) M]
v = sqrt[g*h*(m1 - m2)/[(1/2) m1 + (1/2) m2 + (1/4) M]]
v = sqrt[9.81 m/s^2 *4.20 m *(28.00 kg - 11.1 kg)/[(1/2) 28.00 kg + (1/2) 11.1 kg + (1/4) 5.00 kg]]
v = 5.78 m/s
There is one motion equation without acceleration
h = h0 + (1/2) * (v - v0) * t
h0 = 0 m
v0 = 0 m/s
h = v*t/2
t = 2 * h / v
t = 2 * 4.20 m / 5.8 m/s
t = 1.45 sec
(b) This is the same up to here
PE1 = KE1 + KE2 + KEp + PE2
There is no KEp, everything else is the same.
So v = sqrt[g*h*(m1 - m2)/[(1/2) m1 + (1/2) m2]]
v = sqrt[9.81 m/s^2 *4.20 m *(28.00 kg - 11.1 kg)/[(1/2) 28.00 kg + (1/2) 11.1 kg]]
v = 5.97 m/s
t = 2 * h / v
t = 2 * 4.20 m / 5.97 m/s
t = 1.41 sec
2.
m1 = 1,170 kg
u1 = 25.0 m/s
v1 = 18 m/s
m2 = 9,900 kg
u2 = 20.0 m/s
Here's the equation... so enter values and solve
v2 = u1 * { [2 * m1 ] / [ m1 + m2 ] } - { [ m1 -m2 ] / [ m1 + m2 ] } * u2
v2 = (25.0 m/s) * { [ 2 * (1170 kg) ] / [ (1170 kg) + (9,900 kg) ] } - { [ (1170 kg) - (9,900 kg) ] / [ (1170 kg) + (9,900 kg) ] } * (20.0 m/s)
v2 = (25.0 m/s) * { [ 2,340 kg ] / [ 11070 kg ] } - { [ -8730 kg ] / [ 11070 kg ] } * (20.0 m/s)
v2 = (25.0 m/s) * { 0.21138 } - { -0.78861 } * (20.0 m/s)
v2 = (5.2845 m/s) - (-15.7722 m/s)
v2 = 21.0567 m/s
Calculate total kinetic energy pre-collision, KE = 0.5 * m * v^2
Car
KE = 0.5 * (1,170 kg) * (25.0 m/s)^2
KE = (585 kg) * (625 m^2/s^2)
KE = 365625 J
Truck
KE = 0.5 * (9,900 kg) * (20.0 m/s)^2
KE = (4,950 kg) * (400 m^2/s^2)
KE = 1,980000 J
Total pre-collision energy = 2,345625 J
Post-collision energy
Car
KE = 0.5 * (1,170 kg) * (18.0 m/s)^2
KE = (585 kg) * (324 m^2/s^2)
KE = 189540 J
Truck
KE = 0.5 * (9,900 kg) * (21.0567 m/s)^2
KE = (4,950 kg) * (443.3846m^2/s^2)
KE = 2194753.844 J
Total post-impact energy = 2384293.844 J
A loss of 38668.8437 Joules.