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Anna35 [415]
3 years ago
7

After a hold up, Robin Banks flees in his 1575 kg getaway car at 20 m/s. He crashes into a 45 kg highway barrel which is at rest

. If Robin Bank’s car moves at 18.9 m/s after the collision, how fast does the barrel move after being hit?
Physics
2 answers:
fiasKO [112]3 years ago
7 0

Answer:

v = 38.9 m/s

Explanation:

Dimas [21]3 years ago
6 0

v = 38.5 m/s

Explanation:

P1 = P2 (conservation law of linear momentum)

(1575 kg)(20 m/s) + (45 m/s)(0) = (1575 kg)(18.9 m/s) + 45v

Solving for v

v = 38.5 m/s

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Explanation:

The initial kinetic energy, KE_i is given by

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The final kinetic energy, KE_f is given by

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Change in kinetic energy, \triangle KE is given by

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Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

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