Question

In a lab experiment, two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of 0.500 m/s, one student holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at 1.300 m/s. (the string and spring both have negligible mass

Answer 1

Answer:

a) vfinal₂ = - 0.30 m/s  (←)

b)

Explanation:

In a lab experiment, two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of 0.50 m/s, one student holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at 1.30 m/s. (a) What velocity does the other glider have? (b) Is the total kinetic energy of the two gliders after the collision greater than, less than, or equal to the total kinetic energy before the collision? If greater, where did the extra energy come from? If less, where did the "lost" energy go?

a)

Known values:

vinitial₁ = vinitial₂ = v = 0.5 m/s

vfinal₁ = 1.30 m/s

m₁ = m₂ = m

Unknown values:

vfinal₂ = ?

According to conservative Law of Moment we have:

pfinal = pinitial

then

m*vinitial₁ + m*vinitial₂ = m*v + m*v = 2m*v = 2*m*0.5 = m   (I)

m*vfinal₁ + m*vfinal₂ = m*1.30 + m*vfinal₂ = m*(1.3+vfinal₂)   (II)

If (I) = (II) we get

m = m*(1.30+vfinal₂)   ⇒  vfinal₂ = (1 - 1.30) = - 0.30 m/s

vfinal₂ = - 0.30 m/s  (←)

b) In order to evaluate the kinetic energy we use the equation

Kinitial = Kinitial₁ + Kinitial₂ = (0.5*m*v²) + (0.5*m*v²) = m*v² = m*(0.5)²  

⇒   Kinitial = 0.25m

Kfinal = Kfinal₁ + Kfinal₂ = (0.5*m*1.30²) + (0.5*m*(-0.30)²)

⇒   Kfinal = 0.89m

we evaluate the ratio of the two kinetic energies, as follows:

Kfinal / Kinitial = 0.89m / 0.25m = 3.56

Finally, the final kinetic energy is greater than the initial kinetic energy.