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

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.

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Tài Đức 5 months 2021-08-10T06:08:15+00:00 1 Answers 159 views 0

Answers ( )

    0
    2021-08-10T06:09:38+00:00

    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:

     K.E=70.23J

    Explanation:

    From the question we are told that:

    Mass of ball m_b=10kg

    Speed V_{b1}=15 km/hr ( m/s)

                V_{b1} = 4.1667 m/s

                V_{b1} = 4.1667 m/s

    Mass of Skater m_s=50kg

                       

    Generally the equation for conservation of momentum is mathematically given by

      m_sV_{s1}+m_bV_{b1}=(m_s+m_b)V

      V=\frac{m_sV_{s1}+m_bV_{b1}}{(m_s+m_b)}

      V=\frac {50+10*4.1667}{(50+10)}

      V=1.53m/s

    Generally the equation for  Kinetic energy is mathematically given by

     K.E=\frac{1}{2}(m_s+m_b)V^2

     K.E=\frac{1}{2}(50+10)(1.53)^2

     K.E=70.23J

    Therefore kinetic energy K.E after collision is given as

     K.E=70.23J

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