MIDTERM2_FINAL_SOLUTIONS (1)

PCS 211: Mechanics MIDTERM #2 Fall 2023 1 Toronto Metropolitan University – Department of Physics PCS 211 – Physics: Mechanics Midterm Test 2 – November 17, 2023 Check the name of your professor: Read the following attentively: • This test consists of 1 long answer problem and 14 multiple-choice questions. • Each multiple choice question is out of 2 points. In order to score 2 points, you should both select a correct answer and reasonably justify your selection in the space provided with maximum clarity and detail. If you only select a correct answer without meaningful justification, you can only score 1 point. • Select the option that is closest to the correct answer. Circle it in this booklet and fill in the appropriate bubble on the bubble sheet. The bubbles must be filled using a pencil. • This is a closed book test; pens, pencils, erasers, calculators (Sharp EL-546 and the Casio FX-991 only) and your Toronto Metropolitan University ID are the only allowed items on your desk. • Show all your work in this booklet. We will check your work for all questions, and if your choice is not justified by the work done on this booklet, credit will not be given. • All electronic devices INCLUDING WATCHES must be turned off and placed out of reach. • Coats, jackets and bags must be placed out of reach (under the seats). • Your TMU photo ID must be on your desk at all times. • Perform ALL your calculations to an accuracy of 5 significant figures and provide your final answer to an accuracy of 3 significant figures. Sign here to signify you read the instructions Please Print Clearly: Student: Last Name First Name Student Number: _ _ Section: Dr. Da Silva Dr. Falou Dr. Rebello Dr. Singh Dr. Toronov Dr. Yuan

PCS 211: Mechanics MIDTERM #2 Fall 2023 2 PART 1: This is a full answer problem. Ensure that you write out your full solution, with all steps, clearly and completely [6 marks]. The figure below shows three blocks initially at rest on a frictionless table. The blocks have masses, ࠵? ! = 1.00࠵?࠵? , ࠵? " = 2.00࠵?࠵? , ࠵? # = 3.00 ࠵?࠵? . You exert a horizontal force ࠵? ⃗ = 18.0 ࠵? ࠵? on the block on the left. 1) Find the acceleration of the blocks [2 marks]. Write the solution in the space below. (Make sure your equations are understandable, or this part will be marked as zero) Newton’s Second Law - dynamic: ࠵? ⃗ $%& = ࠵? ’( )⃗ !"! ’& = ࠵?࠵? ⃗ +,+ System: All three masses ( ࠵? ! = 1࠵?࠵? , ࠵? " = 2࠵?࠵? , ࠵? # = 3 ࠵?࠵? ) Surroundings: Earth, Surface Ignore verAcal (y) direcAon, because weight of each block balanced by normal reacAon. Focus on x direcAon ࠵? ⃗ $%& = (࠵? ! + ࠵? " + ࠵? # )࠵? ⃗ à ࠵? ⃗ = (࠵? ! + ࠵? " + ࠵? # )࠵? ⃗ à ࠵? 5555⃗ = ࠵? )⃗ (࠵? ࠵? 0࠵? ࠵? 0࠵? ࠵? ) = 3.00 i m/s 2 2) Draw the free body diagram for the middle block in the space below [2 marks]. (Make sure your diagram is understandable, clearly label each force with subscripts like F A on B or this part will be marked as zero) 3) Find the force that the block on the right exerts on the middle block [2 marks]. (Make sure your equations in the space below are understandable, or this part will be marked as zero) Next use Newton’s third law because force of ࠵? # on ࠵? " is equal and opposite to force of ࠵? " o ࠵? # : ࠵? ⃗ # 3$ " = −࠵? ⃗ " 3$ # We do this b/c it is easier to solve for forces on ࠵? # than on ࠵? " , since there is only one force acAng on ࠵? # System: Masse ( ࠵? # = 3 ࠵?࠵? ) n m 2 g F 3 on 2 F 1 on 2

PCS 211: Mechanics MIDTERM #2 Fall 2023 4 PART 2: Multiple choice questions. Enter your response on your bubble sheet starting at question #1. 1. A particle is moving with a kinetic energy of 1 kJ and a momentum of 55.0 kg m/s at a given point in time t . What is the speed of this particle at this point in time? a) 17.6 m/s b) 28.9 m/s c) 36.4 m/s d) 59.8 m/s c) 61.2 m/s ࠵? = 1 2 ࠵?࠵? " ࠵? = ࠵?࠵? → ࠵? = ࠵? ࠵? ࠵? = 2࠵? ࠵? = 2(1000) 55 = 36.4 ࠵?/࠵? 2. A 22.0 g lead bullet is shot at “bullet proof” glass. At the point of impact, the bullet has a velocity of -1250 i m/s. The impact with the wall lasted 0.600 micro-seconds and resulted in the bullet travelling at a velocity of 625 i m/s after the collision. What is the average force exerted on the bullet during this collision? a) 6.88 x 10 7 B̂ N b) 7.98 x 10 7 B̂ N c) 5.99 x 10 6 B̂ N d) (7.21 B̂ + 8.92 D̂ ) x 10 6 N e) (7.25 B̂ + 8.45 D̂ ) x 10 7 N ࠵? = ࠵? ࠵? − ࠵? ࠵? = ࠵?(࠵? ࠵? − ࠵? ࠵? ) = 0.022 kg J625 ࠵? m s − M−1250 ࠵? m s NO = 41.25 ࠵? kg m/s MP ࠵?N ;(< = ࠵? Δ࠵? = 41.25 ࠵? kg m/s 0.600 × 10 =8 s = 6.88 × 10 > ࠵? N

PCS 211: Mechanics MIDTERM #2 Fall 2023 5 3. How much work must an external force do to change the length of a spring (with spring constant 50.0 N/m and an unstretched length of 5.00 dm) from 7.00 dm to 6.00 dm? (a) 0.750 J (b) 3.25 J (c) 0.500 J (d) -0.750 J (e) -3.25 J W = kx f 2 /2- kx i 2 /2 = 50/2*[(0.6-0.5) 2 -(0.7-0.5) 2 ] = -0.750 J 4. A constant force of magnitude F = 16.0 N directed at an angle ࠵? = 22.0° below the horizontal (as shown in the figure below) pushes a m = 2.40 kg crate a distance d = 2.00 m along the frictionless floor. Determine the work done by this force on the crate. a. 0.00 J b. 29.7 J c. 60.1 J d. −29.7 J e. −60.1 J The applied force does work given by W = F Δ r cos( ࠵? ) = (16.0 N)(2.00 m)cos(22.0°) = 29.7 J. 5. A 36.5 kg box initially at rest is pushed 5.80 m along a rough, horizontal floor with a constant applied horizontal force of 145 N. If the coefficient of friction between box and floor is 0.300, find the change in kinetic energy of the box. a) 37.1 J b) 62.0 J c) 78.0 J d) 89.4 J e) 219 J ࠵? = ࠵?∆࠵? cos ࠵? = (145)(5.80) = 841 ࠵? ∆࠵? ?$& = ࠵? @ ∆࠵? = ࠵?࠵?࠵? @ Δ࠵? = 622.398 ࠵? ∆࠵? = ࠵? A − ࠵? ? = P ࠵? 3&B%C − ∆࠵? ?$& = 219 ࠵?

PCS 211: Mechanics MIDTERM #2 Fall 2023 7 8. A highway curve has a radius of 0.140 km and is unbanked. A car weighing 12.0 kN is moving around the curve at a speed of 24.0 m/s without slipping. What is the magnitude of the centripetal force on the car? a) 12.0 kN b) 17.0 kN c) 13.0 kN d) 5.00 kN e) 49.0 kN ࠵? = ࠵?࠵? " ࠵? = 12000࠵? 9.8 ࠵?/࠵? " × M24 ࠵? ࠵? N " 0.14 × 10 # ࠵? = 5037.9 ࠵? = 5.0 ࠵?࠵? 9. A crate of mass 50.0 kg rests on a horizontal floor on some planet with g = 10.0 m/s 2 . The coefficient of static friction between the crate and the floor is 0.500, and the coefficient of kinetic friction is 0.400. What is the magnitude of the frictional force on the crate if you push it with horizontal force of 100 N? A) 250 N B) 200 N C) 150 N D) 100 N E) None of the above 1) Find Max force of staAc fricAon = µ s F N = µ s mg. 2) If F applied < µ s F N Then, object is at rest FricAon Force f = F applied . 3) If F applied ³ µ s F N Then, object is moving: FricAon Force f = µ k mg

PCS 211: Mechanics MIDTERM #2 Fall 2023 8 10. Two particles interact by conservative forces. In addition, an external force acts on each particle. They complete round trips, each ending at the points where they started. Which of the following must have the same values at the beginning and end of this trip? a. the total kinetic energy of the two-particle system b. the potential energy of the two-particle system c. the mechanical energy of the two-particle system d. the total linear momentum of the two-particle system e. none of the above 11. The figure shows the view looking down onto a horizontal rink with frictionless ice. A puck, tied with a string to point P, slides on the ice in the circular path shown and has made many revolutions. If the string suddenly breaks with the puck in the position shown, which path best represents the puck’s subsequent motion? a) A b) B c) C d) D e) E 12. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval Δt, resulting in a final speed v for the object. You then repeat the experiment, but with a force that is half as small. What time interval is now required to reach the same final speed v ? a. 4Δt b. 2Δt c. 0.5Δt d. 0.25Δt e. Δt

PCS 211: Mechanics MIDTERM #2 Fall 2023 10 INTENTIONALLY LEFT BLANK

PCS 211: Mechanics MIDTERM #2 Fall 2023 11 PCS 211 Formula Sheet If x = a + b - c If x = ( a x b )/ c x f = x i + v x t v xf = v xi + a x t x f = x i + 1 2 ( v xi + v xf ) t x f = x i + v xi t + 1 2 a x t 2 v 2 xf = v 2 xi + 2 a x ( x f - x i ) ~ v f = ~ v i + ~ a t ~ r f = ~ r i + ~ v i t + 1 2 ~ a t 2 ! f = ! i + ↵ t ✓ f = ✓ i + ! i t + 1 2 ↵ t 2 ! 2 f = ! 2 i + 2 ↵ ( ✓ f - ✓ i ) ✓ f = ✓ i + 1 2 ( ! i + ! f ) t x = - b ± p b 2 - 4 ac 2 a a r = - a c = - v 2 r a c = ( r ! ) 2 r = r ! 2 T = 2 ⇡ r v δ x = p δ a 2 + δ b 2 + δ c 2 δ x x = s ✓ δ a a ◆ 2 + ✓ δ b b ◆ 2 + ✓ δ c c ◆ 2