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A 5 g bullet moving at 210 m/s hits a fixed wooden target. Half of its kinetic energy heats the bullet and the other half heats the wood. If the bullet's specific heat is 0.030 cal/(g-deg C) (1 cal = 4.2 J), find the approximate rise in the bullet's temperature.
- 83.3 deg C
- 87.5 deg C
- 119.2 deg C
- 38.4 deg C
Correct answer: 87.5 deg C
Solution
Half the kinetic energy heats the bullet; equating (1/4) m v² to m c deltaT gives deltaT = v²/(4 c), about 87.5 deg C.
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