physics problem?

A student is shopping for study snacks at a local grocery store. Somebody broke a very large bottle of olive oil on the floor. The student slips on the oil, sliding along the aisle. A helpful clerk rushes to the rescue, but also slips and crashes into the student. Entangled, they slide along the floor together. Assume no friction or air drag.

If before the collision the student (weighing 41.2 kg) (indicated orange in the figure above) was moving at 1.68 m/s, and the clerk (weighing 45.3 kg) (indicated green) was moving at 3.26 m/s, how fast will they be moving together after the collision?

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3 Answers

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  • Ash
    Lv 7
    2 months ago
    Favorite Answer

    Conservation of momentum in x direction

    m₁ux₁ + m₂ux₂ = (m₁+m₂)vx

    vx = (m₁ux₁ + m₂ux₂)/(m₁+m₂)

    vx = [(41.2 kg)(1.68 m/s) + (45.3 kg)(0)] / (41.2 kg + 45.3 kg)

    vx = 0.800 m/s

    Conservation of momentum in y direction

    m₁uy₁ + m₂uy₂ = (m₁+m₂)vy

    vy = (m₁uy₁ + m₂uy₂)/(m₁+m₂)

    vy = [(41.2 kg)(0) + (45.3 kg)(3.26 m/s)] / (41.2 kg + 45.3 kg)

    vy = 1.71 m/s

    Resultant velocity = √[(0.800)² + (1.71)²] = 1.89 m/s  ← They are moving at this velocity

  • oubaas
    Lv 7
    2 months ago

    Conservation of momentum shall apply :

    Final momentum M :

    M = √ (41.2*1.68)^2+(45.3*3.26)^2 = 163 kg*m/sec 

    Final velocity V = M/(WF1+WF2) = 163/(41.2+45.3) = 1.88 m/sec 

  • Jim
    Lv 7
    2 months ago

    A picture is worth...

    Remember to do the vector addition.

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