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    A box on a sledge is moving down a snow slope at a uniform speed.

    Question
    SLPaper 2

    A box on a sledge is moving down a snow slope at a uniform speed.

    Image

    1.

    Draw the free-body diagram for the sledge at the position shown on the snow slope.

    [2]
    Verified
    Solution

    arrow vertically downwards labelled weight «of sledge and/or girl»/W/mg/gravitational force/ FgF_gFg​ / FgravitationalF_{gravitational}Fgravitational​ AND arrow perpendicular to the snow slope labelled reaction force/R/normal contact force/N/F 1 mark

    friction force/F/f acting up slope «perpendicular to reaction force» 1 mark

    2.

    After leaving the snow slope, the box on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the box must be in equilibrium as it moves over the horizontal region.

    [3]
    Verified
    Solution

    gravitational force/weight from the Earth «downwards» 1 mark

    reaction force from the sledge/snow/ground «upwards» 1 mark

    no vertical acceleration/remains in contact with the ground/does not move vertically as there is no resultant vertical force 1 mark

    3.

    When the sledge is moving on the horizontal region of the snow, the box falls off the sledge. The box has no horizontal velocity after the fall. The velocity of the sledge immediately after the box falls off is 3.2 m s−1^{-1}−1. The mass of the box is 15 kg and the mass of the sledge is 4.5 kg. Calculate the speed of the sledge immediately before the box falls from it.

    [2]
    Verified
    Solution

    mention of conservation of momentum

    OR

    4.5×3.2=(15+4.5)«v»4.5 × 3.2=(15+4.5) « v » 4.5×3.2=(15+4.5)«v» 1 mark

    0.74«ms−1»0.74 «m s^{-1}» 0.74«ms−1» 1 mark

    4.

    The box can fall in a way that it lands on loosely-packed snow rather than frozen ice. Outline why it would be safer to land on the snow to protect the box from the damage.

    [2]
    Verified
    Solution

    same change in momentum/impulse

    the time taken «to stop» would be greater «with the snow» 1 mark

    F=ΔpΔtF=\frac{\Delta p}{\Delta t}F=ΔtΔp​ therefore FFF is smaller «with the snow»

    OR

    force is proportional to rate of change of momentum therefore FFF is smaller «with the snow» 1 mark

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