Question

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?

Answer 1

Answer:

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

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

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

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

Conserving momentum in y direction

$45\times 8\times \sin (53.1)=65\times v_d\sin \theta$

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

divide 1 & 2

$\frac{\sin \theta }{\cos \theta }=\frac{287.886}{368.848}$

$\tan \theta =0.7805$

$\theta =37.97^{\circ}$

substitute the value of \theta in equation 2

$v_d=\frac{45\times 8\times \sin (53.1)}{65\times \sin (53.1)}$

$v_d=7.193\approx 7.2 m/s$

(b)Change in kinetic Energy of Rebecca

$\Delta K.E._{rebecca}=\frac{45}{2}(13^2-8^2)=2362.5 J$

$\Delta K.E._{Daniel}=\frac{65}{2}(7.2^2)=1684.8 J$