A bumper car with mass m1 = 109 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2-83 kg is moving

Question

A bumper car with mass m1 = 109 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2-83 kg is moving to the left with a velocity of v2 =-3.6 m/s. The two cars have an elastic collision. Assume the surface is frictionless.

1) What is the velocity of the center of mass of the system?
2) What is the initial velocity of car 1 in the center-of-mass reference frame?
3) What is the final velocity of car 1 in the center-of-mass reference frame?
4) What is the final velocity of car 1 in the ground (original) reference frame? –
5) What is the final velocity of car 2 in the ground (original) reference frame?

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Sigridomena 3 years 2021-08-22T01:26:08+00:00 1 Answers 12 views 0

Answers ( )

  1. binhan
    0
    2021-08-22T01:27:47+00:00
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    Answer:

    (1)

    The velocity of the center of mas of the system is= 0.801 m/s

    (2)

    Initial velocity of car 1 in the center of mass reference frame is =4.099 m/s

    (3)

    The final velocity of car 1 in the center of mass reference frame is

     – 4.099 m/s

    (4)

    The final velocity of car 1 in the ground (original ) reference frame is = -3.298 m/s

    (5)

    The final velocity of car 2 in the ground (original) reference frame is = 7.166  m/s

    Explanation:

    Given

    m₁ = 109 kg

    v₁= 4.9 m/s

    m₂= 83 kg

    v₂= -3.6 m/s

    The two cars have an elastic collision.

    (1)

    The velocity of the center of mas of the system is

    V_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}

           = \frac{109.4.9+83(-3.4)}{109+83} m/s

          = 0.801 m/s

    (2)

    Initial velocity of car 1 in the center of mass reference frame is

    V_{1,i} = initial velocity – V_{cm}

        = (4.9 – 0.801) m/s

        =4.099 m/s

    (3)

    Since the collision is elastic, the car 1 will bounce of opposite direction.

    The final velocity of car 1 in the center of mass reference frame is

     V_{1,f} = – 4.099 m/s

    (4)

    The final velocity of car 1 in the ground (original ) reference frame

    V'_{1,f}  =  V_{cm}+V_{1,f}

          =(0.801- 4.099) m/s

          = – 3.298 m/s

    (5)

    The momentum is conserved,

    m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

    \Rightarrow v'_2=\frac{m_1}{m_2}(v_1-v'_1) +v_2  

    Here v'_1= V'_{1,f} =  – 3.298 m/s

    \Rightarrow v'_2=\frac{109}{83}[4.9-(-3.298)]+(-3.6)

          =7.166 m/s

    The final velocity of car 2 in the ground (original) reference frame is = 7.166  m/s

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