Complete Question
A 10 kg medicine ball is thrown at a velocity of 15 km/hr ( m/s) to a 50 kg skater who is at rest on the ice. The skater catches the ball and subsequently slides with the ball across the ice.
Calculate the kinetic energy after collision(in joules).
Answer:
[tex]K.E=70.23J[/tex]
Explanation:
From the question we are told that:
Mass of ball [tex]m_b=10kg[/tex]
Speed [tex]V_{b1}=15 km/hr ( m/s)[/tex]
[tex]V_{b1} = 4.1667 m/s[/tex]
[tex]V_{b1} = 4.1667 m/s[/tex]
Mass of Skater [tex]m_s=50kg[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_sV_{s1}+m_bV_{b1}=(m_s+m_b)V[/tex]
[tex]V=\frac{m_sV_{s1}+m_bV_{b1}}{(m_s+m_b)}[/tex]
[tex]V=\frac {50+10*4.1667}{(50+10)}[/tex]
[tex]V=1.53m/s[/tex]
Generally the equation for Kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2}(m_s+m_b)V^2[/tex]
[tex]K.E=\frac{1}{2}(50+10)(1.53)^2[/tex]
[tex]K.E=70.23J[/tex]
Therefore kinetic energy K.E after collision is given as
[tex]K.E=70.23J[/tex]
