Question

Two identical steel balls, each of mass 70.8 g, are moving in opposite directions at 4.20 m/s.They collide head-on and bounce apart elastically. By squeezing one of the balls in a vise while precise measurements are made of the resulting amount of compression, you find that Hooke's law is a good model of the ball's elastic behavior. A force of 16.5 kN exerted by each jaw of the vise reduces the diameter by 0.120 mm. Model the motion of each ball, while the balls are in contact, as one-half of a cycle of simple harmonic motion. Compute the time interval for which the balls are in contact.

Answer #1

From Hooke's law, we get

F = k x k = F / x

where, F = force exerted by each jaw to the ball = 16.5 x
10^{3} N

x = displacement due to force = 0.120 x 10^{-3} m

then, we get

k = [(16.5 x 10^{3} N) / (0.120 x 10^{-3}
m)]

k = 1.375 x 10^{8} N/m

Compute the time interval for which the balls are in contact.

using a formula, we have

T = 2_{}m
/ k

where, m = mass of each steel ball = 70.8 g = 0.0708 kg

then, we get

T = (6.28 rad) _{
}[(0.0708 kg) / (1.375 x 10^{8} N/m)]

T = (6.28 rad) _{
}5.1490 x 10^{-10} s^{2}

T = [(6.28 rad) (2.27 x 10^{-5} s)]

**T = 1.42 x 10 ^{-4} sec**

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