SOLUTION. will destroy the integrity of the work and Anis not permitted

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1 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. lso neglect friction. Take k = 300 N>m. 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 : and is provided solely for the he use of instructors in teaching sale of any part of this work (includingng on the World Wide Web) will destroy the integrity of the work and nis not permitted.

2 * The free-rolling ramp has a mass of 40 kg. 10-kg crate is released from rest at and slides down 3.5 m to point.if the surface of the ramp is smooth, determine the ramp s speed when the crate reaches.lso, what is the velocity of the crate? 3.5 m 30 Conservation of Energy: The datum is set at lowest point. When the crate is at point, it is 3.5 sin 30 = 1.75 m above the datum. Its gravitational potential energy is = N # m. pplying Eq , we have T 1 + V 1 = T 2 + V = v2 C v2 R = 5v 2 C + 20 v 2 R (1) Relative Velocity: The velocity of the crate is given by v C = v R +v C>R = -v R i + 1v C>R cos 30 i - v C>R sin 30 j2 = v C>R - v R 2i v C>R j The magnitude of v C is v C = 2( v C>R - v R v C>R 2 2 = 2v 2 C>R + v 2 R v R v C>R Conservation of Linear Momentum: If we consider the crate and the ramp as a system, from the FD, one realizes that the normal reaction N C (impulsive force) is internal to the system and will cancel each other. s the result, the linear momentum is conserved along the x axis. 0 = m C 1v C 2 x + m R v R ( : + ) 0 = v C>R - v R v R 2 v R C> R2 2 (2) (3) work is protected ed by United States copyright laws and norprovided is solely for the use of instructors in teaching their courses nd assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) 0 = v C>R - 50v R (4) Solving Eqs. (1), (3), and (4) yields v R = m>s = 1.10 m>s v C = 5.43 m>s v C>R = m>s From Eq. (2) v C = i j = i j6 m>s Thus, the directional angle f of is v C f = tan = 35.8 cf 4.403

3 lock has a mass of 3 kg and is sliding on a rough horizontal surface with a velocity 1v 2 1 = 2m>s when it makes a direct collision with block, which has a mass of 2 kg and is originally at rest. If the collision is perfectly elastic 1e = 12, determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is m k = 0.3. ( : + ) a mv 1 = a mv 2 (v ) = 31v v 2 2 ( : + ) e = 1v 2 2-1v v 2 1-1v = 1v 2 2-1v Solving lock : lock : T 1 + a U 1-2 = T d = 0 d = m T 1 + a U 1-2 = T d210. d = 0 d = m 1v 2 2 = m>s : 1v 2 2 = 2.40 m>s : and is provided solely for the use of instructors in teaching sale 32dof any part of this work (including on the World Wide Web) d = d - d = m

4 * The girl throws the ball with a horizontal velocity of v 1 = 8ft>s. If the coefficient of restitution between the ball and the ground is e = 0.8, determine (a) the velocity of the ball just after it rebounds from the ground and (b) the maximum height to which the ball rises after the first bounce. v 1 3ft 8ft/s h Kinematics:y considering the vertical motion of the falling ball, we have (+T) (v 1 ) 2 y = (v 0 ) 2 y + 2a c Cs y - (s 0 ) y D (v 1 ) y = ft>s Coefficient of Restitution (y): (v 1 ) 2 y = (32.2)(3-0) e = v g 2 - (v 2 ) y (v 1 ) y - v g 1 (+ c) 0.8 = (v 2) y (v 2 ) y = ft>s Conservation of x Momentum: The momentum is conserved along the x axis. World : + m(v x ) 1 = m(v x ) 2 ; (v x ) 2 = 8ft>s : The magnitude and the direction of the rebounding velocity for rthe ball l is v 2 = 2(v x ) v y 2 2 = = 13.7 ft>s u = tan - 1 a b = Kinematics: y considering the vertical lmotion on any of tfie ball after it rebounds from the ground, we have (+ c) (v) 2 y = (v 2 ) 2 y + 2a c Cs y - (s 2 ) y D 0 = (-32.2)(h - 0) and is provided solely o the use of nstructors in teaching their courses and assessing tudent learning. Dissemination or sale of part of this work (including on the Wide Web) h = 1.92 ft

5 * Two smooth disks and each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision travels along a line, 30 counterclockwise from the y axis y (v ) 1 6m/s x (v ) 1 4m/s mv 1 = mv 2 ( : + ) 0.5(4)( 3 5 ) - 0.5(6) = -0.5(v ) 2x + 0.5(v ) 2x = -(v ) 2x + (v ) 2x (+ c) 0.5(4)( 4 5 ) = 0.5(v ) 2y (v ) 2y = 3.20 m>s c (v ) 2x = 3.20 tan 30 = m>s ; (v ) 2x = m>s = m>s ; ( : + ) e = (v ) 2 - (v ) 2 (v ) 1 - (v ) 1 e = ( ) 4( 3 5 )-(-6) = and is provided solely for the use of instructors in teaching sale of any part of this work (including on the World Wide Web) will destroy the integrity of the workork and nis not permitted.