Question

Two ice skaters, Daniel (mass 65.0kg ) and Rebecca (mass 45.0kg ), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00m/s at an angle of 53.1? from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

Part A

What is the magnitude of Daniel's velocity after the collision?

Part B

What is the direction of Daniel's velocity after the collision?

Part C

What is the change in total kinetic energy of the two skaters as a result of the collision?

Answer 1

Mass of Daniel, m_D= 65.0 kg ; Mass of Rebecca, m_R= 45.0 kg

Initial velocity of Daniel, u_D= 0 ; Initial velocity of Rebecca, u_R= 13.0 m/s

(Part A). Let the velocity of Daniel, after collision, be v_D as shown (seee figure at the bottom). According to coservation of momentum,

m_Rv_Rsin 53.1 = m_Dv_Dsin    ....(i) [Total momentum along perpendicular direction is zero]

And m_Rv_R cos 53.1 + m_Dv_D cos = m_Ru_R   [Total momentum along horizontal]

     m_Ru_R - m_Rv_R cos 53.1 = m_Dv_D cos ....(ii)

Squaring and adding equations (i) and (ii), we get

(m_Rv_R)² sin² 53.1 + (m_Rv_R?)² cos² 53.1 + (m_Ru_R)² - 2( m_Ru_R)m_Rv_R cos 53.1 = (m_Dv_D)²  

(m_Rv_R)² + (m_Ru_R)² - 2( m_Ru_R)m_Rv_R cos 53.1 = (m_Dv_D)²  

m_Dv_{D =} sqrt [ (45.08.00)² + (45.013.0)² -245.0²13.08.000.6 ]

= sqrt [ 219105]

v_D = 468.086/65.0 = 7.2 m/s (magnitude of Daniel's velocity)

(Part B). Direction of Daniel's velocity: Using equation (i),

m_Rv_Rsin 53.1 = m_Dv_Dsin   

= sin ^-1(m_Rv_Rsin 53.1 / m_Dv_D) = 37.96 degree with Rebecca'a initial direction of motion.

(Part C). Change in kinetic energy = [ m_Rv_R² +m_Dv_D²- m_Ru_R² ] /2

= [ 45.0(64.00 - 169.00 ) + 65.0 (7.27.2)]/2 = - 677.7 J

Thus total kinetic energy is reduced by 677.7 J.