Answer:
speed of each marble after collision will be 1.728 m/sec
Explanation:
We have given mass of the marble [tex]m_1=41gram=0.041kg[/tex]
Velocity of marble [tex]v_1=2.30m/sec[/tex]
Its collides with other marble of mass 25 gram
So mass of other marble [tex]m_2=25gram=0.025kg[/tex]
Second marble is at so [tex]v_2=0m/sec[/tex]
We have to find the velocity of second marble
From momentum conservation we know that
[tex]m_1v_1+m_2v_2=(m_!+m_2)v[/tex], here v is common velocity of both marble after collision
So [tex]0.041\times 2.30+0.025\times 0=(0.041+0.025)v[/tex]
v = 1.428 m /sec
So speed of each marble after collision will be 1.728 m/sec
The speed of each marble immediately after the collision = 1.43mls
The formula used to solve for perfect collision of particles = m1v1 +m2v2 = (m1 + m2) VC
Where m1 = 41.0g
m2 = 25g
v1 = 2.3 m/s
V2= 0 m/s
Therefore, VC = ?
Substitute the following variables into the equation above.
41 × 2.3 + 25 ×0 = ( 41 + 25) VC
94.3. + 0 = 66 VC
make VC the subject of formula,
VC = 94.3/66
VC = 1.43mls
Therefore, the speed of each marble immediately after the collision = 1.43mls
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