479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 506
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*R1–40. The assembly consists of two blocks A and B,
which have masses of 20 kg and 30 kg, respectively.
Determine the distance B must descend in order for A to
achieve a speed of 3 m>s starting from rest.
3 s A + s B - l
3 ¢s A = - ¢s B
A
B
3 v A = - v B
v B = -9 m>s
T 1 + V 1 = T 2 + V 2
(0 + 0) + (0 + 0) = 1 2 (20)(3)2 + 1 2 (30)(-9)2 + 20(9.81)a s B
3 b - 30(9.81)(s B)
s B = 5.70 m
Ans.
R1–41. Block A, having a mass m, is released from rest,
falls a distance h and strikes the plate B having a mass 2 m.
If the coefficient of restitution between A and B is e,
determine the velocity of the plate just after collision. The
spring has a stiffness k.
Just before impact, the velocity of A is
B
A
h
T 1 + V 1 = T 2 + V 2
k
0 + 0 = 1 2 mv A 2 - mgh
v A = 22gh
A +TB e = (v B) 2 - (v A ) 2
22gh
e22gh = (v B ) 2 - (v A ) 2
(1)
A +TB ©mv 1 =©mv 2
m(v A ) + 0 = m(v A ) 2 + 2m(v B ) 2
(2)
Solving Eqs. (1) and (2) for (v B
) 2
yields
(v B ) 2 = 1 22gh (1 + e)
3
Ans.
506