A mass M = .454kg moves with an intial speed v = 2.88m/s on a level frictionless

Question

A mass M = .454kg moves with an intial speed v = 2.88m/s on a level frictionless air track. The mass is intially a distance D = .250m away from a spring with k = 876N/m, which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance d, before reversing direction. After bouncing off the spring, the mass travels with the same speed v, but in the opposite direction.

A) Determine the maximum distance that the spring is compressed.

B) Find the total elasped time until the mass returns to its starting point.

Explanation / Answer

When the spring is compressed to the maximum distance, speed of mass is zero

So, change in kinetic energy of mass = 1/2*0.454*2.882 = 1.88 J

All this energy goes into the elastic potential energy of spring = 1/2*k*x2,

where k is spring constant and 'x' is spring compression

So, 1/2*876*x2 = 1.88

Solving, we get : x = 0.065 m

Now, total time elapsed = time to reach the spring from initial point + time for compression of spring + time for re-elongation of spring + time to reach initial point from spring

Since, velocity of mass is same after returning from spring, so,

time to reach the spring from initial point = time to reach initial point from spring = distance/speed = 0.250/2.88 = 0.086 s

Also, time for compression = time for re-elongation

Now, the compression of spring by the mass would have been a case of Simple harmonic Motion if the mass got attached to spring on contact

Assuming SHM :

The time period of SHM is given by : T = 2*pi*(m/k)1/2 = 2*pi*(0.454/876)1/2 = 0.143 s

Thus, time for compression + time for elongation = Time peroid of SHM of spring/2 = 0.143/2 = 0.0715 s

Thus, total time = 0.086+0.0715+0.086 = 0.2435 s

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