A 2.2 kg disk traveling at 2.9 m/s strikes a 1.3 kg stick of length 4.0 m that i
ID: 1731485 • Letter: A
Question
A 2.2 kg disk traveling at 2.9 m/s strikes a 1.3 kg stick of length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of figure (a). The disk strikes the stick at a distance r = 1.30 m from the stick's center. Assume the collision is elastic and the disk does not deviate from its original line of motion. Find the translational speed of the disk, the translational speed of the stick, and the angular speed of the stick after the collision. The moment of inertia of the stick about its center of mass is 1.73 kg · m2.
Comparison of Values Before and After the Collision v(m/s) ?(rad/s) p(kg · m/s) L(kg · m2/s) Ktrans (J) Krot (J) Before Disk - - Stick 0 0 0 0 0 0 Total for system - - 0 After Disk - 0 Stick Total for system - - Note: Linear momentum, angular momentum, and total kinetic energy of the systems are all conserved.Explanation / Answer
From conservation of linear momentum,
m1u1 = m1v1 + m2v2
2.2*2.9 = 2.2*vdf + 1.3*vs
vs = 4.907 - 1.692*vdf ......(1)
From conservation of angular moementum,
m1u1*r = m1v1*r + lw
2.2*2.9*1.3 = 2.2*vdf*1.3 + 1.73*w
w = 4.794 - 1.653*vdf ......(2)
From conservation of kinetic energy,
(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2 + (1/2)lw^2
2.2*2.9^2 = 2.2*vdf^2 + 1.3*vs^2 + 1.73*w^2
14.23 = 1.692*vdf^2 + vs^2 + 1.33*w^2
Put the values of w from eq(2),
14.23 = 1.692*vdf^2 + vs^2 + 1.33*(2.732*vdf^2 - 15.84*vdf + 22.98)
14.23 = 1.692*vdf^2 + vs^2 + 3.63*vdf^2 - 21.06*vdf + 30.56
Put the value of vs from eq(1),
14.23 = 1.692*vdf^2 + (2.86*vdf^2 - 16.6*vdf + 24.078) + 3.63*vdf^2 - 21.06*vdf + 30.56
8.182*vdf^2 - 37.66*vdf + 40.4
By solving the quadratic equation,
vdf = 2.89 m/s
Put the value of vdf in eq(1),
vs = 4.9 - 1.692*2.89
vs = 0
Put the value of vdf in eq(2),
w = 4.794 - 1.653*2.89
w = 0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.