PHYS211 WS2 SOL

Center for Academic Resources in Engineering ( CARE ) Mid-Semester Review Session Phys211 − University Physics: Mechanics Midterm 2 Worksheet Solutions The problems in this review are designed to help prepare you for your upcoming exam. Questions pertain to material covered in the course and are intended to reflect the topics likely to appear in the exam. Keep in mind that this worksheet was created by CARE tutors, and while it is thorough, it is not comprehensive. In addition to exam review sessions, CARE also hosts regularly scheduled tutoring hours. Tutors are available to answer questions, review problems, and help you feel prepared for your exam during these times: Session 1: Sun. October 15th, from 4-6pm Josh, Lucy, Matthew Session 2: Mon. October 16th, from 4-6pm Jon, Wyatt, Josh Can’t make it to a session? Here’s our schedule by course: https://care.grainger.illinois.edu/tutoring/schedule-by-subject Solutions will be available on our website after the last review session that we host. Step-by-step login for exam review session: 1. Log into Queue @ Illinois: https://queue.illinois.edu/q/queue/847 2. Click “New Question” 3. Add your NetID and Name 4. Press “Add to Queue” Please be sure to follow the above steps to add yourself to the Queue. Good luck with your exam!

Phys211 − University Physics: Mechanics Midterm 2 Exam Review 1. A putty ball of mass M = 4 kg is traveling horizontally at v = 7 m/s. (Ignore the effects of gravity). It strikes a block of the same mass, which is adjacent to a relaxed ideal spring attached to an infinitely massive wall with a spring constant of k = 8 N/m. The putty ball sticks to the block. After the collision, the spring is compressed. v M M M M M M ∆ x k What is the maximum compression of the spring? Conservation of Momentum: P i = P f m p v p = m b + p v f v f = 3.5 m/s to the left One thing to note is that conservation of momentum is only valid in cases where there’s not an external force (otherwise momentum changes). So we use momentum during the collision, but once the spring starts compressing, we must resort to energy conservation. Energy conservation here is justified because the spring force is conservative. KE = PE 1 2 ( m p + m b ) v 2 f = 1 2 kx 2 x = 3 . 5 m 2 of 12

Phys211 − University Physics: Mechanics Midterm 2 Exam Review 3. A block of mass m 1 = 15 kg hangs from the ceiling on an ideal, massless spring with spring constant k = 60 N/m. With the block hang- ing on the spring, the total length of the spring is L = 4.5 m. When a second block with an identical mass of m 2 = 15 kg is tied to the first with a massless string, the spring stretches an additional h 0 = 2.45 m. The string is cut so that mass m 2 falls away. What is the maximum velocity of mass m 1 ? M 1 M 1 M 2 L h 0 Since all the forces acting on the blocks are conservative (spring and gravity), we can use energy conservation. The total potential energy stored in the spring initially is given by PE = 1 2 kx 2 This potential energy, when completely converted to kinetic, will be the point where the mass is moving with the maximum speed. PE i = KE f 1 2 kx 2 = 1 2 mv 2 60 ∗ (2 . 45) 2 = 15 ∗ v 2 v = 4 . 9 m/s 4 of 12

Phys211 − University Physics: Mechanics Midterm 2 Exam Review 4. A cart of mass M = 8 kg rolls without friction on a horizontal surface. It is attached at the cart’s center of mass to a freely pivoting initially-horizontal massless rod of length L to a ball of mass m = 4 kg. The system is initially at rest when the ball is released. The pendulum swings down and to the left, and at the bottom of its swing the ball is observed to have a v b = 3 m/s to the left. M m v c v b (i) Which one of the following remains constant as the pendulum swings down? A) horizontal component of the momentum of the ball B) horizontal component of the momentum of the cart C) horizontal component of the momentum of the ball + cart (ii) What is the speed of the cart when the ball is at the bottom? (iii) What is the length L of the pendulum? (iv) How far to the right has the cart moved, when the ball is at the bottom (in terms of L)? (i) Momentum separately is not conserved because they both start at rest but end with some non-zero velocity. As a system, however, momentum is conserved because the momentum of the ball will be the same magnitude but opposite direction of the momentum of the cart. The answer is (C) . (ii) P i = P f = 0 = P ball + P cart ⃗ P ball = − ⃗ P cart m b ⃗v b = − M c ⃗v c v c = 1 . 5 m/s (iii) The only force acting on this cart-ball system is gravity, and that acts vertically. So we can use energy conservation here since gravity is conservative. PE ball = KE ball + KE cart m b gL = 1 2 m b v 2 b + 1 2 m c v 2 c (4)9 . 81 L = 1 2 (4)( − 3) 2 + 1 2 (8)(1 . 5) 2 5 of 12

Phys211 − University Physics: Mechanics Midterm 2 Exam Review 5. A block of mass M = 2.0 kg is released from rest h = 1.5 meters above the ground and slides down a frictionless ramp. It slides across a floor that is frictionless, except for a small section with width d =0.5 meters that has a coefficient of kinetic friction µ k = 0.5. At the left end, is a spring with spring constant k = 300 N/m. The box compresses the spring, and is accelerated back to the right. M h µ k k d (a) What is the speed of the box at the bottom of the ramp? (b) What is the maximum distance the spring is compressed by the box? (c) What is the maximum height to which the box returns on the ramp? (a) PE = KE mgh = 1 2 mv 2 v = p 2 gh v = 5 . 42 m/s (b) Here we need to use the Work-Kinetic Energy Theorem: W = ∆ KE 1 2 mv 2 i − W f = 1 2 mv 2 f 1 2 mv 2 i − mgµ k d = 1 2 mv 2 f v 2 f = v 2 i − 2 gµ k d v 2 f = 24 . 5 m/s This gives us the velocity of the box after it has lost some energy due to friction. We can now use energy conservation again with the spring. KE = PE 1 2 mv 2 f = 1 2 k ∆ x 2 ∆ x = s mv 2 f k ∆ x = 0 . 404 m 7 of 12

Phys211 − University Physics: Mechanics Midterm 2 Exam Review (c) Note: v i in this part is v f in the previous part W = ∆ KE 1 2 mv 2 i − mgµ k d = 1 2 mv 2 f v 2 f = v 2 i − 2 gµ k d v 2 f = 19 . 6 m/s This is the velocity of the box right before it begins climbing back up the ramp (taking into account the lost energy from friction again). We can use energy conservation to find that its maximum height is at the point where all of its initial kinetic energy is converted into potential. KE = PE 1 2 mv 2 f = mgh h = v 2 f 2 g h = v 2 i − 2 gµ k d 2 g h = 1 m 8 of 12

Phys211 − University Physics: Mechanics Midterm 2 Exam Review (ii) p i = p f p 10 . 4 ,i + p 10 . 6 ,i = p 10 . 4 ,f + p 10 . 6 ,f p cm,i = p 10 . 4 ,f + p 10 . 6 ,f p 10 . 6 ,f = 10 . 6 ∗ (1 . 4 , 1 . 2) p cm,i = 21 ∗ (1 . 48 , 1 . 01) p 10 . 4 ,f = p cm,i + p 10 . 6 ,f = (16 . 64 , 8 . 32) v 10 . 4 ,f = p 10 . 4 ,f 10 . 4 v 10 . 4 ,f = (1 . 6 , 0 . 8) m/s 7. Show that for a system of particles with total mass M and total momentum ⃗ P total , the center of mass follows ⃗ P total = M ⃗ V CM The velocity of the center of mass for N particles is, by definition, ⃗ V CM = ∑ N n =1 m n ⃗v n M If we now multiply each side by the total mass M , we get M ⃗ V CM = N X n =1 m n ⃗v n = ⃗ P total Where the middle term is the total momentum of the system. 8. Two blocks A and B are on a path to collide with each other. M A = 10 kg and v A = 30 m/s to the right while M B = 30 kg and v B = 10 m/s to the right. What is the velocity of each in the center of mass frame? M A M B v A v B Let’s establish that rightward is the +ˆ x direction since both blocks are moving that way. We must calculate the velocity of the center of mass in the lab frame. We can do that by 10 of 12

Phys211 − University Physics: Mechanics Midterm 2 Exam Review ⃗ V CM,lab = ∑ N n =1 m n ⃗v n M = 10 · 30 + 30 · 10 10 + 30 = 15 m/s Now we use the relative velocity relation: v A,B = v A,C + v C,B For Block A: v A,CM = v A,lab + v lab,CM v A,CM = 30 − 15 = 15 m/s For Block B: v B,CM = v B,lab + v lab,CM v B,CM = 10 − 15 = − 5 m/s 9. A ball of mass M = 2 kg is moving left at constant velocity v = -3 m/s when it hits a wall. The force applied by the wall onto the ball is plotted against time as shown below. What is velocity of the ball after hitting the wall? 50 100 0.1 0.2 0.3 t F Impulse is given by ∆ P = F ave ∆ t . We find F ave by taking the integral of the Force function and dividing by the change in time. F ave = Area ∆ t = 1 2 ∗ ( . 3 − . 1) ∗ 100 0 . 2 = 50 N Plugging into the impulse formula, ∆ P = F ave ∆ t ∆ P = 50 ∗ . 2 = 10 Finally, ∆ P = mv f − mv i 10 = (2) v f − (2)( − 3) v f = 2 m/s 11 of 12