we can use lagrange\'s Equation if you think it will help. Two masses, m1 and m2
ID: 1700353 • Letter: W
Question
we can use lagrange's Equation if you think it will help.Two masses, m1 and m2, are attached to the two ends of a spring of
force constant k. The system lies horizontally on a perfectly smooth
surface. A third mass, m3, is thrown with a velocity of Vo
horizontally onto the plane to hit mass m2. THe two masses m2 and m3 stick together at the moment of collision. The sticking process occurs
almost immediately, so that the length of the spring does not
change. It is given that m1=2m, m2=m, m3=0.5*m. (see attached diagram) q
a). What is the velocity of the cenre of mass before and after the
collision?
b). How much kinetic energy is lost during the collision?
c). What are the velocities of the masses m1 and m2+m3 immediately after
the collision?
d). What is the maximum that the spring shrinks? (ie change in length from initial)
Explanation / Answer
a). center of mass of system remain the same. vG*(m1+m2+m3)=v0*m3 so vG=v0*0.5m/3.5m=v0/7. b). velocity of m2-3. v23*(m2+m3)=v0*m3. so v23=v0/3. so that energy lost. (m2+m3)*v0^2/2*9-m3*v0^2/2=1.5m*v0^2/18-0.5*v0^2/2=-mv0^2/6. c)after collision. v23=v0/3 and v1=0. d)in the center of mass (vG=v0/7). velocity of m23 and m1 is. v23'=v0/3-v0/7=v0*4/21. v1'=v0/7 -- so conservation of energy. 1.5m*(4v0/21)^2/2+2*(v0/7)^2/2=kx^2/2. so mv0^2*2/21=kx^2. so x=sqrt(2mv0^2/k21)
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