SOLUTION. ill Principle of Impulse and Momentum: Referring to Fig. b, 75(0) + 75(9.81)(3) - T(3) = 75vA. vb = T (1) From Fig.

Size: px

Start display at page:

Download "SOLUTION. ill Principle of Impulse and Momentum: Referring to Fig. b, 75(0) + 75(9.81)(3) - T(3) = 75vA. vb = T (1) From Fig."

Neal Shaw
5 years ago
Views:

1 The crate B and cylinder A have a mass of 200 kg and 75 kg, respectively. If the system is released from rest, determine the speed of the crate and cylinder when t = 3 s. Neglect the mass of the pulleys. B SOLUTION Free-Body Diagram: The free-body diagrams of cylinder A and crate B are shown in Figs. b and c. va and vb must be assumed to be directed downward so that they are consistent with the positive sense of sa and sb shown in Fig. a. Principle of Impulse and Momentum: Referring to Fig. b, t2 m(v1)y + (+ T) Fy dt = m(v2)y Lt1 75(0) + 75(9.81)(3) - T(3) = 75vA va = T t2 (+ T) m(v1)y + Lt1 th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th. rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r. b) From Fig. b, (1) Fy dt = m(v2)y 200(0) (9.81)(3) - 4T(3) = 200vB vb = T (2) Kinematics: Expressing the length of the cable in terms of sa and sb and referring to Fig. a, sa + 4sB = l Taking the time derivative, va + 4vB = 0 Solving Eqs. (1), (2), and (4) yields vb = m>s = 2.10 m>s c (3) (4) va = m>s = 8.41 m>s T Ans. T = N 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ A

2 The block has a mass of 50 kg and rests on the surface of the cart having a mass of 75 kg. If the spring which is attached to the cart and not the block is compressed 0.2 m and the system is released from rest, determine the speed of the block with respect to the cart after the spring becomes undeformed. Neglect the mass of the wheels and the spring in the calculation. Also neglect friction. Take k = 300 N>m. SOLUTION B C T 1 + V 1 = T 2 + V 2 (0 + 0) (300)(0.2)2 = 1 2 (50)(v b) (75)(v c) 2 12 = = 50 v 2 b + 75 v 2 c ( : + ) mv 1 = mv = 50 v b - 75 v c v b = 1.5v c v c = m>s ; v b = m>s : v b = v c + v b>c ( : + ) = v b>c v b c = m s : Ans. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ

3 The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the 10-g bullet is traveling at 300 m>s when it becomes embedded in the 10-kg block, determine the distance the block will slide up along the plane before momentarily stopping. 300 m/s A SOLUTION 30 Conservation of Linear Momentum: If we consider the block and the bullet as a system, then from the FBD, the impulsive force F caused by the impact is internal to the system. Therefore, it will cancel out. Also, the weight of the bullet and the block are nonimpulsive forces. As the result, linear momentum is conserved along the axis. x œ m b (v b ) x = (m b + m B ) v x œ 0.01(300 cos 30 ) = ( ) v v = m>s Conservation of Energy: The datum is set at the blocks initial position. When the block and the embedded bullet is at their highest point they are habovethe datum. Their gravitational potential energy is ( )(9.81)h = h. Applying Eq , we have T 1 + V 1 = T 2 + V ( )A B = h h = m = 3.43 mm d = 3.43 > sin 30 = 6.87 mm Ans. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ

4 Two smooth spheres A and B each have a mass m. If A is given a velocity of v0, while sphere B is at rest, determine the velocity of B just after it strikes the wall. The coefficient of restitution for any collision is e. v0 A B SOLUTION Impact: The first impact occurs when sphere A strikes sphere B. When this occurs, the linear momentum of the system is conserved along the x axis (line of impact). Referring to Fig. a, + ) (: mava + mbvb = ma(va)1 + mb(vb)1 mv0 + 0 = m(va)1 + m(vb)1 (va)1 + (vb)1 = v0 e = (vb)1 - (va)1 va - vb e = (vb)1 - (va)1 v0-0 (vb)1 - (va)1 = ev0 Solving Eqs. (1) and (2) yields (vb)1 = a 1 + e b v0 : 2 th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th. rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r. b) + ) (: (1) (2) (va)1 = a 1 - e b v0 : 2 The second impact occurs when sphere B strikes the wall, Fig. b. Since the wall does not move during the impact, the coefficient of restitution can be written as + ) (: e = e = 0 - C - (vb)2 D (vb) (vb)2 c 1 + e d v0-0 2 (vb)2 = e(1 + e) v0 2 Ans Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ

15 93. Disks A and B have a mass of 15 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact.

5 Disks A and B have a mass of 15 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. The coefficient of restitution between them is e = 0.8. y Line of impact A 10 m/s x SOLUTION Conservation of Linear Momentum: By referring to the impulse and momentum of the system of disks shown in Fig. a, notice that the linear momentum of the system is conserved along the n axis (line of impact). Thus, 8m/s B +Q m A Av A B n + m B Av B B n = m A Av AB n + m B Av BB n 15(10)a 3 5 b - 10(8)a 3 5 b = 15v A cos f A + 10v B cos f B 15v A cos f A + 10v B cos f B = 42 (1) Also, we notice that the linear momentum of disks A and B are conserved along the t axis (tangent to? plane of impact). Thus, +a m A Av A B t = m A Av AB t and 15(10)a 4 5 b = 15v A sin f A v A sin f A = 8 +a m B Av B B t = m B Av BB t 10(8)a 4 5 b = 10 v B sin f B v B sin f B = 6.4 Coefficient of Restitution: The coefficient of restitution equation written along the n axis (line of impact) gives +Q e = (v B) n - (v A) n (v A ) n - (v B ) n 0.8 = v B cos f B - v A cos f A 10a 3 5 b - c -8a 3 5 bd (2) (3) This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. v B cos f B - v A cos f A = 8.64 (4) Solving Eqs. (1), (2), (3), and (4), yeilds v A = 8.19 m>s Ans. f A = v B = 9.38 m>s Ans. f B = Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ

= m. 30 m. The angle that the tangent at B makes with the x axis is f = tan-1

= m. 30 m. The angle that the tangent at B makes with the x axis is f = tan-1 1 11. When the roller coaster is at B, it has a speed of 5 m>s, which is increasing at at = 3 m>s. Determine the magnitude of the acceleration of the roller coaster at this instant and the direction angle