Answered

A body of mass 4.5 kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/5 of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 4.5 kg body was 8.9 m/s?

Answer :

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Answer with Explanation:

We are given that

Mass of one body=4.5 kg

Initial velocity of one body=8.9 m/s

Initial velocity of another body=0

Final velocity of one body=[tex]\frac{1}{5}\times 8.9=1.78 m/s[/tex]

a.We have to find the mass of the other body.

Energy and momentum are conserved in this system

[tex]p_i=P_f[/tex]

[tex]m_1v_{1i}+0=m_1v_{1f}+m_2v_{2f}[/tex]

[tex]v_{2f}=\frac{m_1v_{1i}-m_1v_{1f}}{m_2}[/tex]  (equation I)

From conservation of kinetic energy

[tex]\frac{1}{2}m_1v^2_{1i}=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v^2_{2f}[/tex]

[tex]m_2=\frac{m_1v^2_{1i}-m_1v^2_{1f}}{v^2_{2f}}[/tex] (equation II)

Substitute equation I in equation II

[tex]m_2=\frac{m_1v^2_{1f}-m_1v^2_{1f}}{(\frac{m_1v_{1i}-m_1v_{1f}}{m_2})^2}[/tex]

[tex]m_2=\frac{(m_1v_{1i}-m_1v_{1f})^2}{m_1v^2_{1i}-m_1v^2_{1f}}[/tex]

Substitute the given values then we get

[tex]m_2=\frac{(4.5\times 8.9-4.5\times 1.78)^2}{4.5\times (8.9)^2-4.5\times (1.78)^2}[/tex]

[tex]m_2=3 kg[/tex]

Hence, the mass of other body=3 kg

b.We have to find the speed of two body center of mass .

The center of mass speed  is given by

[tex]v_{com}=\frac{p_i}{m_1+m_2}[/tex]

Substitute the values then we get

[tex]v_{com}=\frac{m_1v_1+0}{m_1+m_2}[/tex]

[tex]v_{com=\frac{4.5\times 8.9}{4.5+3}=5.34 m/s[/tex]

Hence, the speed of the two body center of mass is given by =5.34 m/s

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