Two gliders are set in motion on an air track. Glider 1 has mass m 1 = 0.200 kg

Question

Two gliders are set in motion on an air track. Glider 1 has mass m1 = 0.200 kg and moves to the right with a speed 0.660 m/s. It will have a rear-end collision with glider 2, of mass m2 = 0.540 kg, which initially moves to the right with a speed 0.200 m/s. A light spring of force constant 59.0 N/m is attached to the back end of glider two as shown in the following figure. When glider 1 touches the spring, superglue instantly and permanently makes it stick to its end of the spring.

a) Find the common speed the two gliders have when the spring is at maximum compression.

(b) Find the maximum spring compression distance.

The motion after the gliders become attached consists of a combination of (1) the constant-velocity motion of the center of mass of the two-glider system found in part (a) and (2) simple harmonic motion of the gliders relative to the center of mass.

(c) Find the energy of the center-of-mass motion.

(d) Find the energy of the oscillation.

Explanation / Answer

(A) Applying momentum conservation,

(0.200 x 0.660) + (0.540 x 0.200) = (0.200 + 0.540) v

v = 0.324 m/s .........Ans

(B) Applying energy conservation,

Pei + KEi = PEf + KEf

0 + (0.200 x 0.660^2 / 2) + (0.540 x 0.2^2 / 2) = 59 d^2 /2 + (0.200 + 0.540)(0.324^2) / 2

0.05436 = (59/2) d^2 + 0.03884

d = 0.023 m

(C) energy of center of mass motion = (0.200 + 0.540) (0.324^2)/2

= 0.0388 J  

(D) Energy of oscillations = (59/2)(0.023^2) = 0.0156 J

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