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2 years ago

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a 2.00 kg object is moving east at 4.00 m/s when it collides with a 6 kg object that is initially at rest. after the collision t

he larger object moves east at 1 m/s. what is the final velocity of the smaller object after the collision

1 answer:

baherus [9]2 years ago

6 0

Answer:

v = 1.00 m/s east

Explanation:

Conservation of momentum

Let east be the positive direction

2.00(4.00) + 6.00(0.00) = 2.00(v) + 6.00(1.00)

v = 1.00 m/s east

The two items have stuck together.

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Answer:

$\Delta f=f_{Lr}-f_{Se}$

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The difference in frequency is given by

$\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}$

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$f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz$

$f_{Se}=22\ kHz$

$\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz$

The difference in wavelength is 147.45 Hz

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