2007: Problems 37-38

Topic: Energy
Concepts: 1D elastic collision, 1D inelastic collision, Energy dissipation


Solution:

For the perfectly inelastic collision, we conserve linear momentum to find the velocity of both objects after the collision,

mv0=(m+3m)vmv_0=(m+3m)v’
v=v04v’=\frac{v_0}{4}

Since the combined mass breaks the cord and ends up at rest, all the kinetic energy became spring potential energy (right before it breaks),

U=12(m+3m)v2=12(4m)(v04)2=mv028U=\frac{1}{2}(m+3m)v’^2=\frac{1}{2}(4m)\left(\frac{v_0}{4}\right)^2=\frac{mv_0^2}{8}

In other words, it costs energy UU to break the spring. For the elastic collision, recall the elastic collision equations for a mass mm colliding with MM at rest,

v1=mMm+Mvv_1=\frac{m-M}{m+M}v
v2=2mm+Mvv_2=\frac{2m}{m+M}v

In our case, we have

v2=2mm+3mv0=v02v_2=\frac{2m}{m+3m}v_0=\frac{v_0}{2}

Since the mass 3m3m breaks the cord and ends up at speed vfv_f, we conserve energy to obtain

12(3m)v22U=12(3m)vf2\frac{1}{2}(3m)v_2^2-U=\frac{1}{2}(3m)v_f^2

Substituting our earlier equations,

3m2(v02)2mv028=3m2vf2\frac{3m}{2}\left(\frac{v_0}{2}\right)^2-\frac{mv_0^2}{8}=\frac{3m}{2}v_f^2
32vf2=14v02\frac{3}{2}v_f^2=\frac{1}{4}v_0^2
vfv0=16\frac{v_f}{v_0}=\frac{1}{\sqrt{6}}

so the answer is C.

Topic: Energy
Concepts: Conservation of energy, Energy dissipation


Solution:

The initial kinetic energy of the system is

Ki=12mv02K_i=\frac{1}{2}mv_0^2

The final kinetic energy after the elastic collision and the cord breaking is

Kf=12mv12+12(3m)vf2K_f=\frac{1}{2}mv_1^2+\frac{1}{2}(3m)v_f^2

but since an elastic collision conserves energy, the final kinetic energy is also equal to

Kf=KiUK_f=K_i-U

where UU is the energy cost to break the spring. From the previous problem, we found

U=18mv02U=\frac{1}{8}mv_0^2

Thus,

KfKi=1UKi=11/81/2=34\frac{K_f}{K_i}=1-\frac{U}{K_i}=1-\frac{1/8}{1/2}=\frac{3}{4}

so the answer is D.

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