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Two ice skaters, Daniel (mass 65 kg) and Rebecca (mass 45 kg), are practicing. Daniel is intially at rest when struck by Rebecca moving at 13 m/s. After the collision, Rebecca has a velocity of 8 m/s at an angle of 53.1º from her initial direction. a) What are the magnitude and direction of Daniel’s velocity after the collision? b) What is the change in kinetic energy of the two skaters as a result of the collision?

User Philkark
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Answer:

Step-by-step explanation:

Given

mass of Daniel
m_d=65 kg

mass of Rebecca
m_r=45 kg

Initial velocity of daniel
u_d=0

Initial velocity of rebecca
u_r=13 m/s

Let us suppose velocity of Daniel is v m/s at an angle of \theta w.r.t horizontal

conserving Momentum in Horizontal direction


m_d* u_d+m_r* u_r=m_d* v_d\cos (\theta )+m_r* u_r\cos (53.1)


45* 8=65\v_d\cos \theta +45* 8\cos (53.1)


65v_d\cos \theta =368.848------1

Conserving momentum in y direction


45* 8* \sin (53.1)=65* v_d\sin \theta


65* v_d\sin \theta =45* 8* \sin (53.1)---------2

divide 1 & 2


(\sin \theta )/(\cos \theta )=(287.886)/(368.848)


\tan \theta =0.7805


\theta =37.97^(\circ)

substitute the value of \theta in equation 2


v_d=(45* 8* \sin (53.1))/(65* \sin (53.1))


v_d=7.193\approx 7.2 m/s

(b)Change in kinetic Energy of Rebecca


\Delta K.E._(rebecca)=(45)/(2)(13^2-8^2)=2362.5 J


\Delta K.E._(Daniel)=(65)/(2)(7.2^2)=1684.8 J

Two ice skaters, Daniel (mass 65 kg) and Rebecca (mass 45 kg), are practicing. Daniel-example-1
User Sakin
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