The diagram below shows the initial positions of two disks at rest on a very smo

Question

The diagram below shows the initial positions of two disks at rest on a very smooth ice rink. Disk 2 is four times as massive as disk 1. Starting at the same time, a constant force F is exerted on each disk.

Part 1. When asked which disk has more momentum when it crosses the finish line one student says “The less massive disk is moving faster when it crosses the line. The faster speed compensates for the lower mass, so, each disk has the same momentum as it crosses the line.” Do you agree? Explain briefly why you do or do not.

Part 2. Apply the Momentum Principle to this situation in ways that allow you to determine the speed of each disk as it crosses the finish line and the time it took each disk to get there.

Part 3. When asked which disk has more kinetic energy when it crosses the finish line one student says “The less massive disk is moving faster when it crosses the line. Since kinetic energy depends on the speed squared that more than compensates for that disk’s lower mass, so, it has more kinetic energy than the more massive one.” Do you agree? Explain briefly why you do or do not.

Part 4. Apply the Energy Principle to this situation in ways that allow you to determine the speed of each disk as it crosses the finish line. They should agree with the speeds you predicted in part 2. Compare and contrast the work involved in solving parts 2 and 4.

I'm more interested in the theory and mechanics than just answers.

Explanation / Answer

momentum of a body is defned as the product of mass and velocity

it is clear from this to create accleration easily in a body, its mass must be as small as possible

try imaging moving a hill and a stone. it is obvious that momentum can be created in less mass body with ease

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according to the law of conservation of momentum

m1 u1 + m2u2 = 0

m1 u1 = -m2u2

as m2 = 4 m1

m1 u1 = -4 m1u2

u2/u1 = 1/4

u2 = 0.25 times of u1

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KE of disk 1 = 0.5 m1 u1^2

KE of disk 2 = 0.5 m2 u2^2 = 0.5 * 4 m1 * 1/14 u1^2

KE2 = 1/4 times of KE 1
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use KE1 + KE2 = 0

KE 1 = KE2

0.5 m1 u1^2 = 0.5 m2u2^2

u2^2/u1^2 = 4

u2/u1 = sqrt4 = 2

so u2 = 2 u1

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