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2 years ago

Help please!

Physics

1 answer:

JulsSmile [24]2 years ago

7 0

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

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Answer:

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