HCI Assignment 04b Answer Newton s laws of motion

Dynamics 1 Name: Index No: Date: Class: Assignment 04b: Newton's laws of motion (IP) 1. (a) The metal head of a hammer is loose. To tighten it, it is being dropped down onto a table. To do so most efficiently, should we • drop the hammer with handle end down, • drop the hammer with the head down, or • either way will give us the same results? Explain your answer. [2] Drop hammer with handle down. [1] [1]: Either one explanation below When the handle hits table, it suddenly stops moving but the inertia of the hammer head causes the head to continue moving . Since the hammer head has a greater mass , it moves more downward and tightens the head to handle more efficiently. Or When the head hits the table, it suddenly stops moving but the inertia of the handle also causes the handle to continue moving. Since the hammer handle has a smaller mass , it moves downward to a smaller extent and tightens the head to the handle less efficiently. (b) An inelastic cord C is suspended from the ceiling on one end and tied to a block of mass M on the other end. Another cord D is attached to the bottom of the block. ceiling C M D Explain, using Newton's 1st law, why "If you give a sudden jerk to D, it will break first , but if you pull D with a gradually increasing force, C will break first". [2] If a sudden jerk is given to D, inertia of M prevents M from moving . Tension in string at D (or pull force acts on D only) will cause D to break. If D is pulled with a gradual increasing force, M tends to move. Tension in string at C is equal to applied force and weight of M will cause C to break instead. HWA CHONG INSTITUTION Sec 3 Physics

Dynamics 2 2. A 110 kg motorbike carrying a 50 .0 kg rider coasts to a stop in 51 m. It was originally travelling at 15 m s-1. Calculate the braking force exerted by the road on the motorbike and rider. [3] v2 = u2 + 2a(s) 0 = (15 m s-1)2 + 2a(51 m) [1] a = - 2.2 m s-2 Fnet = ma = (110 kg + 50.0 kg)(- 2.206 m s-2) [1] = - 350 N (to 2 s.f.) braking force = 350 N [1] 3. Fig. 3.1 shows the horizontal forces acting on a moving car of mass 9.0 x 102 kg. The forward driving force remains constant throughout its journey. Fig. 3.2 shows the total resistive force that acts on the car as time passes. Fig. 3.3 shows the speed-time graph for the first 24 s of the motion of the car along a straight road. Fig. 3.3: Fig. 3.2: Total resistive force / N time / s F 2000 Fig. 3.1 forward driving force total resistive force

Dynamics 3 (a) State the terminal velocity (speed) of the car. [1] 16 m s-1 [1] (b) Hence, state the (magnitude of) forward driving force of the car. [1] When the speed is constant, forward driving force = total resistive force = 2000 N [1] (c) For the first 8.0 s of its journey, calculate (i) the acceleration of the car, [1] a = (v – u)/(t) = (12.8 – 0)/8.0 = 1.6 m s-2 (to 2 s.f.) [1] (ii) the total resistive force, F, on the car. [2] forward driving force – total resistive force = ma 2000 N – total resistive force = (9.0 x 102 kg)(1.6 m s-2) [1] total resistive force = 560 N (to 2 s.f.) [1] (d) Complete the following statements and underline the correct answer to describe the acceleration and forces experienced by the car for the whole journey. [6] Assume rightward direction to be positive velocity. From 0 to 8 s, its acceleration is …constant and positive………………………………………. [1] Since the constant total resistive force is ( smaller than /greater than/equal to ) the constant forward driving force, the net force acting on the car is …constant and positive…………….. [1] From 8 to 14 s, its acceleration is ...decreasing and positive…...………………………..……. [1] Since the total resistive force ( decreases/increases/remains constant ) as the speed increases while forward driving force remains constant, thus net force acting on the car (decreases/increases/remains constant). [1] From 14 to 24 s, its acceleration is …zero………………………………………………………….. [1] Since the total resistive force is ( smaller than/greater than/ equal to) the forward driving force, the net force acting on the car is …zero…………………………………………………………….. [1]

Dynamics 4 4. An elevator (lift) starts from rest on the ground floor and comes to rest at a higher floor. Its motion is controlled by an electric motor. A simplified graph of the variation of the elevator’s velocity with time is shown below. 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 time / s velocity / m s –1 The elevator is supported by a cable. The diagram below is a free -body force diagram for when the elevator is moving upwards during the first 0.5 s. tension weight (a) In the space below, draw the tension acting on the elevator during the following time intervals.[2] (i) 0.5 s to 11.5 s tension tension weight weight [1]: Tension must be equal to weight. (ii) 11.5 s to 12.0 s tension tension weight weight [1]: Tension must be less than weight. Note: Point of application (POA) from top of elevator as given in the diagram. tension

Dynamics 5 (b) A person is standing on weighing scales in the elevator. Before the elevator rises, the reading on the scales is W. On the axes below, sketch a graph to show how the reading on the scales varies during the whole 12.0 s upward journey of the elevator. ( Note that this is a sketch graph – you do not need to add any values.) [3] [1]: a value greater than W before 0.5 s and smaller than W after 11.5 s. [1]: a constant value equal to W from 0.5 to 11.5 s [1]: constant value of W from 0 s to 0.5 s and from 11.5 s to 12.0 s. 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 time / s Reading on scales W 5. An inextensible string , connecting the objects m1 and m2 over a frictionless pulley , is shown in the diagram below. Let T be tension in string. The 3.00 kg mass, m1, moves on a frictionless surface while the 5.00 kg mass, m2, moves vertically when it is released from rest. Take the acceleration due to gravity to be 10 m s-2. Assume the string is taut throughout the time when the two masses are moving. (a) Calculate (i) the acceleration a of the whole system, [2] By considering both masses as one system, since Fnet = ma, 5.00 X 10 N = (3.00 kg + 5.00 kg) a [1] a = 6.25 m s-2 (to 3 s.f.) [1] (Note: Net force, which pulls both masses, is weight of mass m2.) lab bench m1 m2 T 50.0 N T T T Can we experience weightlessness on Earth? https://www.youtube.com/watch?v=yd 8_FNSAEgM https://www.gozerog.com/reservations / (Cost of zero-g flight) https://www.youtube.com/watch?v=44 gekCy_yhQ

Dynamics 6 (ii) the tension T in the string. [2] To find tension in the string, consider only one free body, i.e. either m1 or m2. Consider only 3.00 kg mass. T = Fnet = ma = (3.00 kg)(6.25 m s-2) [1] = 18.75 N = 18.8 N (to 3 s.f.) [1] Consider only 5.00 kg mass. Fnet = ma 50.0 N - T = (5.00 kg)(6.25 m s-2) [1] T = 18.75 N = 18.8 N (to 3 s.f.) [1] (b) A rough wooden surface was placed on the lab bench and the whole experiment was repeated with the mass m1 moving across the wooden surface when the ma