A box of mass \"m1\" slides from rest a distance \"L\" down on an incline at an
ID: 2005734 • Letter: A
Question
A box of mass "m1" slides from rest a distance "L" down on an incline at an angle "theta", and on the flat section collides with another box of mass "m2". After the collision, the second box travels some distance "d" into a rough flat section (kinetic friction coefficient "u") before stopping.a) Find "d" if the boxes are made of elastic rubber
b) find "d" if the box sides are lined with double-sided sticky tape
c) Compute "d" from parts a) and b) if "theta"=30 degrees, "L"=2 m, "u"=0.5, "m1"=0.1kg and "m2"=0.3 kg
Explanation / Answer
Given Mass of the first box, is m1 Mass of the second box is m2 Length of the incline is L Distance travelled by the second box is d Coefficient of kinetic friction is µ a) If the boxes are made of rubber : Acceleration of m1 is a = gsin? The velocity of m1 at the end of the incline is v^2 = 2 as = 2 *gsin? *L v = v2gL sin? Assume that during collision , the total energy of m1 is transformed to m2, then the initial velocity of m2 isv2gL sin? The acceleration of m2 is a = -µg The distance travelled by m2 is (0)^2 - (v2gL sin?)^2 = 2( -µg ) d d = L sin? / µ ----1 b) If the box sides are lined with double-sided sticky tape: The velocity of m1 at the end of the incline is v2gL sin? At the collision , the two boxes stick to gether and move .Then the velocity of the two boxes using law of conservation of momentum is V = m1v2gL sin? / (m1 + m2) The distance travelled by m1 + m2 is (0)^2 - [(m1v2gL sin? / (m1 + m2)]^2 = 2( -µg ) d d = m1^2 L sin? / (m1+m2)^2*µ ----2 c) Taking eq 1 d = (2m) sin30 / 0.5 = 2m Taking eq 2 d = (0.1 kg)^2 (2m) sin30 / (0.1 kg +0.3 kg) ^2 *0.5 d = 0.125 m
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