A mass of 0.50 kg is attached to a massless spring with a spring constant k = 60

Question

A mass of 0.50 kg is attached to a massless spring with a spring constant k = 600 N/m (see figure). The system rests on a level, friction-free surface and is initially at rest. A second mass of 0.20 kg makes an elastic head-on collision with the mass attached to the spring; thereafter, the oscillating system vibrates with an amplitude of 0.25 m.

A. What was the incident speed of the second mass (Ans - 16m/s need help with work)

B.What is the maximum value of the potential energy stored in the spring during the subsequent oscillations?

Explanation / Answer

A. First we can determine the velocity of mass 1 using the amplitude of the oscillations. When it is hit it will have all kinetic energy, but at the amplitude all the energy will be potential. We can equate these two because of the law of conservation of energy.

PE=KE

(1/2)kx2=(1/2)mv2

v2=kx2/m

v2=600(.25)2/(.5)

v=8.66 m/s (mass one)

Now we can use conservation of momentum and kinetic energy (because elastic) to get the velocity of the second mass.

MOMENTUM CONSERVATION

m2v2=m2v'2+m1v

.2v2=.2v'2+.5(8.66)

v2=v'2+21.65 ((1))

KINETIC ENERGY CONSERVATION

m2v22=m2v'22+m1v2

.2v22=.2v'22+.5(8.66)2

v22=v'22+187.5 ((2))

Now we just have to solve the system of equations. I will substitute the expression for v2, equation 1, into equation 2.

(v'2+21.65)2=v'22+187.5

42.3v'2=187.5-21.652

v'2=-6.495 m/s <---- velocity of mass 2 after the collision

Now use equation 1 to get what we are looking for

v2=-6.495+21.65

v2=15.155 m/s

B. This part is less involved. The maximum potential energy will occur at the amplitude.

PEmax=(1/2)kA2 <------ A=amplitude

PEmax=(1/2)(600)(.25)2

PEmax=18.75 joules

hope that helps

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