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Oduvanchick [21]

3 years ago

6

A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli

sion? Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble $m_1=41gram=0.041kg$

Velocity of marble $v_1=2.30m/sec$

Its collides with other marble of mass 25 gram

So mass of other marble $m_2=25gram=0.025kg$

Second marble is at so $v_2=0m/sec$

We have to find the velocity of second marble

From momentum conservation we know that

$m_1v_1+m_2v_2=(m_!+m_2)v$ , here v is common velocity of both marble after collision

So $0.041\times 2.30+0.025\times 0=(0.041+0.025)v$

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

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Answer:

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Explanation:

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Therefore, v can be isolated from equation 4:

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