A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf. Find (a) the angle and (b) the speed vf, assuming that friction can be ignored.

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lublana lublana · Answer 1 · 2019-11-05T09:19:32+01:00

Answer:

a.[tex]\thta=71^{\circ}[/tex]

b.[tex]v_f=3.78 m/s[/tex]

Explanation:

We are given that

[tex]m_1=53 kg[/tex]

[tex]v_1=2.9 m/s[/tex]

[tex]m_2= 72 kg[/tex]

[tex]v_2=6.2 m/s[/tex]

a.We have to find the angle

[tex]\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}[/tex]

[tex]\theta=71^{\circ}[/tex]

b. We have to find the speed [tex]v_f[/tex]

According to law of conservation of momentum

[tex]m_1v_1=(m_1+m_2)v_fcos\theta[/tex]

[tex]53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f[/tex]

[tex]v_f=\frac{153.7}{40.7}=3.78 m/s[/tex]