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Dominik [7]
3 years ago
8

(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it

at 4 m/s. What is the speed of the big fish immediately after swallowing the small fish? (Ignore any affects due to the drag of the water).
Physics
1 answer:
faust18 [17]3 years ago
3 0

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

Now, plug in the given values and solve for 'v'. This gives,

v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

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A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo
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Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

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\Delta L = 21.5 - 11.9 = 9.6 cm

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

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Now the period of oscillation of spring is given as

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Now plug in all values in it

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hope this answer will help you

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follow me

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