Two acrobats flying through the air grab and hold onto each other in midair as part of a circus act. One acrobat has a mass of 60 kg and has a horizontal velocity of 5 m/s just before the grab. Another acrobat has a mass of 50 kg and has a horizontal velocity of -3 m/s just before the grab. Their horizontal velocity immediately after they grab onto each other is:
A. 1.4 m/s
B. 3.0 m/s
C. 0.6 m/s
D. 2.0 m/s
E. 4.1 m/s

Question

contestada

Respuesta :

aristeus aristeus · Answer 1 · 2019-11-14T07:17:11+01:00

Answer:

Final velocity will be 1.4 m/sec

So option (a) is correct option

Explanation:

We have given there are two acrobats

Mass of first acrobats [tex]m_1=60kg[/tex] and its velocity [tex]v_1=5m/sec[/tex]

Mass of second acrobats [tex]m_2=50kg[/tex] and its velocity [tex]v_2=-3m/sec[/tex]

Now from conservation of momentum we know that

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]60\times 5+50\times (-3)=(50+60)v[/tex]

[tex]v=1.36=1.4m/sec[/tex]

So option (a) is correct option

nuuk nuuk · Answer 2 · 2019-11-14T08:03:19+01:00

Answer:1.4 m/s

Explanation:

Given

mass of first acrobat [tex]m_1=60 kg[/tex]

velocity [tex]v_1=5 m/s[/tex]

mass of second acrobat [tex]m_2=50 kg[/tex]

velocity [tex]v_2=-3 m/s[/tex]

let us take v be the final velocity of two acrobats

[tex]m_1\cdot v_1+m_2\cdot v_2=(m_1+m_2)v[/tex]

[tex]60\times 5-50\times 3=(60+50)v[/tex]

[tex]v=\frac{300-150}{110}[/tex]

[tex]v=\frac{15}{11}=1.36 a\pprox 1.4 m/s[/tex]