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2 5 2 C H APTE R 1 5 K I N ET I C S O F A PART I C L E : I M P U LS E A N D M O M E NTU M

EXA M P L E 1 5 .9

\

CD 6 lb

(a)

@ 1D,,"m

3 ft

j (b)

(VA) l = 13.90 [tis Just before impact

Just after impact ( c)

Fig. 15-16

The bag A, having a weight of 6 Ib, is released from rest at the position () = 0°, as shown in Fig. 15-16a. After falling to () = 90°, it strikes an 18-lb box B. If the coefficient of restitution between the bag and box is e = 0.5, determine the velocities of the bag and box just after impact. What is the loss of energy during collision? SOLUTION This problem involves central impact. Why? Before analyzing the mechanics of the impact, however, it is first necessary to obtain the velocity of the bag just before it strikes the box. Conservation of Energy. With the datum at () = 0°, Fig. 15-16b, we have To + Vo = TJ + Vj

0 + 0 = !( 6 1b

2) (VA )t - 6 1b(3 ft) ; (vA h = 13.90 ft/s 2 32.2 ft/s

Conservation of Momentum. After impact we will assume A and B travel to the left. Applying the conservation of momentum to the system, Fig. 15-16c, we have ( � ) mB(vBh + mA(vAh = mB(vBh + mA(vAh

( 6 Ib

) ( 18 Ib

) ( 6 Ib

) 0 + 32.2 ft/s2

(13 .90 ft/s) = 32.2 ft/s2

(vBh + 32.2 ft/s2

(vAh

(vAh = 13.90 - 3 (VBh (1) Coefficient of Restitution. Realizing that for separation to occur after collision (vBh > (vAb Fig. 15-16c, we have

( � ) e = (vBh - (vAh . 0 5 =

(vBh - (vAh (vAh - (vsh ' .

13 .90 ft/s - 0 (vAh = (vBh - 6.950

Solving Eqs. 1 and 2 simultaneously yields (2)

(vAh = -1 .74 ft/s = 1 .74 ft/s � and (vBh = 5.21 ft/s +- Ans. Loss of Energy. Applying the principle of work and energy to the bag and box just before and just after collision, we have "2.UJ -2 = T2 - TJ ;

"2.Ul-2 = [! ( 18 1b

2) (5.21 ft/s)2 + !( 6 1b

2) ( 1 .74 ft/S)2 ] 2 32.2 ft/s 2 32.2 ft/s

- [! ( 6 1b

2) ( 13.9 ft/s? ] 2 32.2 ft/s

"2.Ul-2 = -10.1 ft · lb Ans. NOTE: The energy loss occurs due to inelastic deformation during the collision.