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For each of these problems, don\'t just write down an answer, even if you can so

ID: 1570250 • Letter: F

Question

For each of these problems, don't just write down an answer, even if you can solve it in your head. Briefly justify your answer by writing down the physics relationship or equation you're using to think about it. e) Starting from rest, a hollow, thin-walled, 33-gram cylinder of radius 4 cm rolls down a ramp and reaches a final speed of 20 cm/s. What would the final speed of a 66-gram cylinder of radius 4 cm be? f) Starting from rest, a hollow, thin-walled, 33-gram cylinder of radius 4 cm rolls down a ramp and reaches a final speed of 20 cm/s. What would the final speed of a 33-gram, 2 cm times 2 cm times 2 cm greased block be, if it slid with negligible friction? g) A 1-ton train car bumps into and couples with a stationary 3-ton train car, and the two then roll off at 20 mph. How fast would they go if the stationary car instead weighed 4 tons? h) A 3 kg piece of space junk moving at 1 km/s collides off-center with an unmoving space station and embeds in it, causing the station to rotate at 10 RPM. How fast would the station rotate (in RPM) if the junk were instead moving at 2 km/s? i) A disk spinning about an axle at 6 rotations per second (RPS) experiences a constant frictional torque, and completes exactly 27 rotations before coming to a stop. How many rotations would it complete if it were initially spinning at 12 RPS?

Explanation / Answer

initial energy Ei = m*g*h

final energy Ef = (1/2)*I*w^2 + (1/2)*m*v^2

Ef = (1/2)*m*r^2*(v/r)^2 + (1/2)*m*v^2

Ef = (1/2)*m*v^2 + (1/2)*m*v^2 = m*v^2


Ef = Ei

m*v^2 = m*g*h

v = sqrt(gh)

speed does not depend on mass and radius


speed = 20 cm/s <<<-------------ANSWER


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f)


for cylinder speed = sqrt(gh)


speed = 20 cm/s = 0.2 m/s


for block

m*g*h = (1/2)*m*v^2

v = sqrt(2*g*h) = sqrt2*20 = 20.3 cm/s

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g)


momentum before collision = momentum after collision


m1*u1 + m2*u2 = (m1+m2)*v


((1*20)) + (3*0) = (1+3)*v


speed v = 5 mph


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h)


initial angular momentum = final angular momentum


m*v*r = I*w

m*v1*r = I*w1


for v2 = 2 km/s


m*v2*r = I*w2

v2/v1 = w2/w1

2/1 = w2/10


w2 = 20 RPM <<<<----------answer

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i)

wf^2 - w1^2 = 2*alpha*theta1

for w2 = 12 RPS

wf^2 - w2^2 = 2*alpha*theta2

theta2/theta1 = (w2/w1)^2

theta2/27 = (12/6)^2

theta2 = 108 revolutions <<<-------answer