A mass; m1 = 64 g, sits on a frictionless horizontal surface, and is attached to

Question

A mass; m1 = 64 g, sits on a frictionless horizontal surface, and is attached to a spring of spring constant k = 51 N/m. The other end of the horizontal spring is attached to a wall; the system is in equilibrium. Another mass; m2 = 18 g, strikes the stationary mass m1, and sticks to it. As a result, the spring is compressed by a distance of 24.5 cm before the masses come to a momentary stop.

How fast was m2 moving before the collision?

How long does it take the masses to come to a momentary stop after the collision?

Explanation / Answer

after collision, Applying energy conservation,

(m1 + m2) v^2 / 2 = k x^2 /2

(0.018 + 0.064) (v^2) /2 = (51)(0.245^2)/2

v = 6.11 m/s

Now applying momentum conservation for the collision,

m1 v0 = (m1 + m2)(6.11)

64 v0 = (64 + 18)(6.11)

v0 = 7.83 m/s ...........Ans

For Spring - mass SHM,

time period, T = 2 pi sqrt[ (m1 + m2)/ k]

T = 2pi sqrt[ (0.018 + 0.064) / 51] = 0.25 sec  

from eq. to extreme position,

t = T/4 = 0.063 sec

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