On a horizontal, frictionless table, an open-topped 5.20-kg box is attached to an ideal horizontal spring having force constant 375 N/m. Inside the box is a 3.44-kg stone. The system is oscillating with an amplitude of 7.50 cm. When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box. Find (a) the period and (b) the amplitude of the resulting motion of the box. (c) Without doing any calculations, is the new period greater or smaller than the original period? How do you know?

Solution 32E Introduction Using the conservation of momentum we can calculate the velocity of the box at that time. Then using the conservation of energy we can calculate the amplitude of the motion. Step 1 After picking up the stone the time period will be m 5.20 kg T = 2 kox= 2 375 N/m= 0.740 s Step 2 Suppose v i,max was the maximum velocity of the box and stone before picking up the stone. Now after picking up, suppose the maximum velocity of the box is v f,max. Hence from the conservation of momentum we can write that v i,maxmbox+ m stone = vf,max box v (m +m ) v f,max= i,maxbox stone mbox With stone the amplitude of the box was A =i7.50 cm = 0.0750 m Hence equating the maximum kinetic energy and the maximum potential energy we can write that 1 2 1 2 2kA i (m2 box+ m stonevi,max 2 v = kAi i,max (mboxmstone Therefore we have kAi2 (mboxmstone v f,max= (m +m ) m box stone box (375 N/m)(0.0750 m) (5.20 kg+3.44 kg) = (5.20 kg+3.44 kg) 5.20 kg = 3.00 m/s Now if the amplitude of the motion after picking up the stone is A f then we can write that Hence the amplitude of the motion will be 0.204 m or 20.4 cm.