Two gliders are set in motion on an air track. Glider one has mass m1 = 0.220 kg

Question

Two gliders are set in motion on an air track. Glider one has mass m1 = 0.220 kg and velocity 0.730ihatbold m/s. It will have a rear-end collision with glider number two, of mass m2 = 0.380 kg, which has original velocity 0.150ihatbold m/s. A light spring of force constant 40.0 N/m is attached to the back end of glider two as shown in the figure below. When glider one touches the spring, superglue instantly and permanently makes it stick to its end of the spring.

(a) Find the common velocity the two gliders have when the spring compression is a maximum.


(b) Find the maximum spring compression distance.


(d) Find the energy of the center-of-mass motion.


(e) Find the energy of the oscillation.

Explanation / Answer

a )   the conservation of momentum for the system oftwo gliders gives
   Pi = Pf
   m1 v1 + m2v2 = (m1 + m2) v

   v = (m1 v1 + m2v2) / (m1 + m2)

   v = ( 0.220 kg * 0.730 m/s + 0.380 kg * 0.150 m/s )

     = 0.2176 m/s

b )     conservative forces act we get that
   E = 0
   (1 / 2) m1 v12 +(1 / 2) m2 v22 = (1 / 2)(m1 + m2) v2 + (1 / 2) kxm2
   on substitutiong v from a we get that and solving for xm we get
   xm = (v1 -v2) [(m1 m2) / k(m1 + m2)]

         = ( 0.730-0.150) (0.220*0.380)/40.0 (0.220 + 0.380)

         = 0.034 m

d ) E = 1/2 ( m _ 1+ m_2 ) v^2

            = 1/2 ( 0.220 kg + 0.380 kg ) ( 0.2176)^2

           = 14.2 *10^-3 J

e )   E = 1/2 k x_m ^2

             = 1/2 * 40 * ( 0.034 m )^2

            = 23.1 *10^-3 J

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