A) Bill and Phil are playing the age old game of throwing rocks at a wall of bub

Question

A) Bill and Phil are playing the age old game of throwing rocks at a wall of bubble gum. All the rocks stop moving when they stick to the wall. They are trying to see who can heat up the bubble gum wall more. For simplicity let's say the rocks don't change temperature and the wall doesn't move. Bill can throw his rocks 10 times as fast as Phil. Phil tries to compensate by throwing 10 times as many rocks as Bill (assume all rocks are the same mass). Who will heat up the bubble gum wall more? Ignore air resistance. Eth = mCmT, |Eb| = |mH|, KEtrans = 1/2 mv2 , PEg = mgh, Cm,water = 4.2J/g0C, Cm,dry ice= 1.9 J/g0C, Cm, solid water = 2.0 J/g0C, Hsub,C02 = 930kJ/kg, Hmelt,H20 = 330kJ/kg

B) In the distant future, people only have springs, iron balls and physics to entertain themselves. You shoot a 0.5 kg ball at 3 m/s horizontally at an initially uncompressed/unstretched spring. The spring's spring constant is 5N/m. The ball hits the spring and compresses it. After the spring compresses 0.5m, how fast is the ball moving? Assume no friction or air-resistance. If you don't show your work, you will be thrown out the airlock.

Explanation / Answer

Using consevation of energy:

All the kinetic energy of rocks will go in heating the wall.

Between Bill and Phil who generates greatest total kinetic energy heats the wall the most.

Heat generated by Bill.. throwing 1 rock

KE1= 1/2* m*v'^2

here v' is 10 time the normal speed v.

KE1= 50 m*v^2.

Heat generated by Phill..by throwing 10 rocks at same time taken by Bill for throwing one rock:

KE2= 10*(1/2)*m*v^2

=5*m*v^2

Hence, Bill heat the wall the most.

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