Boxes are placed on a long assembly line as shown above. All of a sudden the ass

Question

Boxes are placed on a long assembly line as shown above. All of a sudden the assembly line malfunctions and stops. One of the boxes is on the inclined position of the assembly line which stops down. Each of the boxes in the above picture have a mass M. The coefficient of kinematic friction between the boxes and the ground and the inclined plane is mu. The inclined plane makes an angle of theta with the ground. The coefficient of restitution for the collisions between the boxes is e. How many impacts occur before the train of boxes comes to a rest. Note - There are many boxes, say hundreds all of equal mass M and equally spaced by d on the assembly line.

Explanation / Answer

The energy of the sliding box just before 1st impact is E = mgLsin - mgLcos, velocity is v

after 1st impact, the velocity of the 2nd box is (1 + e)v/2, the energy becomes (1 + e)2/4 * E

after moving distance d, the energy becomes

E1 = (1 + e)2/4 * E - mgd = aE - b, where a = (1 + e)2/4, b = mgd

Similarly, after 2nd impact and movind d, the energy of the box becomes

E2 = aE1 - b = a2E - b(a + 1)

after 3rd impact and movind d, the energy of the box becomes

E3 = aE2 - b = a3E - b(a2 + a + 1)

after 4th impact and movind d, the energy of the box becomes

E4 = aE3 - b = a4E - b(a3 + a2 + a + 1)

after nth impact and movind d, the energy of the box becomes

En = aEn-1 - b = anE - b(an-1 + an-1 + ... a + 1) = anE - b(an - 1)/(a - 1)

To find n, solve anE - b(an - 1)/(a - 1) = 0, or anE = b(1 - an)/(1 - a)

anL(sin - cos) = d(1 - an)/(1 - a)

an = 1/[1 + (1 - a)L(sin - cos)/d]

n = ln[1 + (1 - a)L(sin - cos)/d]/(-lna)

n = ln[1 + (3 + e)(1 - e)L(sin - cos)/(4d)]/[ln4 - 2ln(1 + e)]

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