A.20 kg mass is sliding on a horizontal, frictionless airtrack with speed of 3.0

Question

A.20 kg mass is sliding on a horizontal, frictionless airtrack with speed of 3.0 m/sec It is heading from left toright. It instanteously hits and sticks to a 1.3 kgmass initially at rest on the track. The 1.3 kg mass isconnected to one end of a massless spring. Spring has springconstant of 100 newtons /meter. The other end of the springis fixed. Determine the following for the for the .20kg mass immediatelybefore the impact a. its linear momentum b. its kinetic energy Determine the following for the combined masses immediatelyafter impact a. the linear momentum b. The kinetic energy After the collision the 2 masses undergo simple harmonicmotion about their position at impact Determine amplitude of harmonic motion Determine period of harmonic motion A.20 kg mass is sliding on a horizontal, frictionless airtrack with speed of 3.0 m/sec It is heading from left toright. It instanteously hits and sticks to a 1.3 kgmass initially at rest on the track. The 1.3 kg mass isconnected to one end of a massless spring. Spring has springconstant of 100 newtons /meter. The other end of the springis fixed. Determine the following for the for the .20kg mass immediatelybefore the impact a. its linear momentum b. its kinetic energy Determine the following for the combined masses immediatelyafter impact a. the linear momentum b. The kinetic energy After the collision the 2 masses undergo simple harmonicmotion about their position at impact Determine amplitude of harmonic motion Determine period of harmonic motion

Explanation / Answer

   a.   Linearmomentum   p   =   m *v       For 0.50 kgmass,   p1   =   0.20*3.0   =   0.60   kg-m/s    b.   kineticenergy   k1   =   (1/2)* m * v2   =   0.5 * 0.20*3.02   =   0.90   J    c.   This is case ofinelastic collision where linear momentum remain conserved but notthe k.e..       Total finalmomentum   p2   =   p1   =   0.60   kg-m/s    c.   Finalmomentum   p2   =   M* V          M   =   totalmass   =   1.3 +0.20   =   1.50   kg          =>   velocityof combinedmass   V   =   p2/ M   =   0.60 /1.5   =   0.40   m/s    d.   Kineticenergy   K         =   (1/2)* M * V2   =   0.5 * 1.50*0.402   =   0.120   J    e.   K.E. of combinedmass   =   p.e. stored in spring onfull compression   =   (1/2) * C *A2          C   =   Springconstant   =   100   N/m,   A   =   amplitude          0.120   =   0.5* 100 * A2          amplitude   A   =   (0.120/50)   =   0.049   m   =   4.9   cm    f.   timeperiod   T   =   2* (M/C)   =   2 * 3.14 *(1.5 /100)   =   0.769   s .

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