dynamics hw problem I need help in. A 2100-g collar A released from rest, slides

Question

dynamics hw problem I need help in.

A 2100-g collar A released from rest, slides down a frictionless rod, and strikes a 3150-g collar Bwhich is at rest and supported by a spring of constant 500 N/m. The coefficient of restitution between the two collars is 0.9 and h-2.1 m. mA k = 500 N/m mB 30° value: 10.00 points Required information Determine the maximum distance collar A moves up the rod after impact. (Round the final answer to two decimal places.) The maximum distance collar A moves up the rod after impact is

Explanation / Answer

given, mass of collar A, m = 2.1 kg
mass of collar B, M = 3.15 kg
coefficient of restitution = e = 0.9
h = 2.1 m
initial speed of collar A when it strikes collar B = u
from conservationof eenrgy, 0.5mu^2 = mghcos(30)
u^2 = 2ghcos(30) = 2*9.81*2.1*cos(30)
u = 5.973 m/s

after it strikes the collar B, let its speed be v and that of collar B be V
then from conservation of momentum
mu = mv + MV

also, e = u/(V - v) = 0.9
u = 0.9(V - v) = 5.973
6.63715 = V - v

2.1*5.973 = 2.1*v + 3.15(6.63715 + v)
v = -1.59309 m/s ( along the rod, upwards)

hence after collision let the distance collar A moves up the rod be d
then from conservation of energy
0.5v^2 = gdcos(30)
0.5*(-1.59309)^2 = 9.81*d*cos(30)
hence
d = 0.14936 m

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