A 15 g marble moves to the right at 3.5 m/s and makes an elastic head-on collision with a 22 g marble. The final velocity of the 15 g marble is 5.4 m/s to the left, and the final velocity of the 22 g marble is 2.0 m/s to the right.
What is the initial velocity of the 22 g marble?

a) 5.3 m/s to the left
b) 5.3 m/s to the right
c) 4.1 m/s to the right
d) 4.1 m/s to the left

Question

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Respuesta :

zainsubhani zainsubhani · Answer 1 · 2020-01-25T05:47:54+01:00

Answer:

[tex]v_2[/tex]≅-4.1 m/s (-ve for left)

[tex]v_2\\[/tex] ≅4.1 m/s (To the left)

Explanation:

According to the conservation of momentum:

Momentum before collision=Momentum After Collision

[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]

Where:

[tex]m_1[/tex] is the 15g marble

[tex]m_2[/tex] is the 22 g marble

[tex]v_1[/tex] is the velocity of 15g marble before collision

[tex]v_2[/tex] is the velocity of 22g marble before collision

[tex]v'_1[/tex] is the velocity of 15g marble after collision

[tex]v'_2[/tex] is the velocity of 22g marble after collision

Note: -ve sign for left, +ve sign for right

We have to calculate v_2:

Above equation on rearranging will become:

[tex]v_2=\frac{m_1v'_1+m_2v'_2-m_1v_1}{m_2} \\v_2=\frac{(0.015kg)(-5.4m/s)+(0.022kg)(2m/s)-(0.015kg)(3.5)}{0.022kg} \\v_2=-4.06 m/s\\[/tex]

[tex]v_2[/tex]≅-4.1 m/s (-ve for left)

[tex]v_2\\[/tex] ≅4.1 m/s (To the left)