Introductory Physics Questions (19th June 2015) These questions are taken fro end-of-ter and ake-up exas in 2013 and 2014. After today s class, you should e ale to answer Q1, Q2, Q3 and Q7. Q4, Q5, and Q6 will e covered in Week 11. Q1 Fiure shows a lock A with ass M and a lock B with ass are connected y a liht strin that passes over a frictionless pulley. Masses of the pullery and the strin are nelected. (1) Write the equations of the otion for oth locks (2) Find the tension in the strin. (3) Find the anitude of the acceleration of each ass. M A1 We note that the strin is taut - this eans the velocity and acceleration of oth locks have sae anitude with opposite direction. The equations of otion are Ma = M T (1) (for lock A) ( a) = T (2) (for lock B) (1)+M (2) leads to 0 = 2M ( + M)T T = 2M M + a = T M = 2 M + = M M + 1
Q2 O N P A fixed hollow sphere has center O and a sooth inner surface of radius. A particle P, which is inside the sphere, is projected horizontally with speed v 0 fro the lowest interior point. In the susequent otion, the particle kept in contact with the sphere. (1) Show that v 2 = v 2 0 2(1 cos θ). (2) Find the reaction force N as a function of θ. (3) How lare ust the initial speed v 0 to e for the particle to coplete circle? A2 Because the surface is sooth, the noral reaction force N is perpendicular to the velocity of the particle, thus does no work. The echanical enery is conserved. 1 2 v2 0 = 1 2 v2 + (1 cos θ) Thus v 2 = v 2 0 2(1 cos θ) The equations of otion are (usin polar coordinate syste) Sustitutin v = θ, θ 2 = cos θ N θ = sin θ (r-direction) (θ-direction) N = cos θ + θ 2 = cos θ + v2 = cos θ + v2 0 = v2 0 To coplete the spere, v > 0 and N > 0 for all θ. + (3 cos θ 2) v 2 > 0 v 2 = v 2 0 2(1 cosθ) > 0 v 2 0 > 4 2(1 cos θ) N > 0 Thus v 2 0 + (3 cos θ 2) > 0 v0 2 > 5 v 0 > 5 2
Q3 A particle is initially at the top of a sooth heisphere of radius a. It eins to slide down on the surface, with a neliile initial speed. Then at a certain point, the particle leaves the surface. Let θ e the anle shown in the fiure. v (1) Find the speed of the particle on the surface. (2) Find the anitude of noral reaction force R exerted y the surface. a θ θ C (3) Find cos θ c, where R ecoes zero. This is where the particle leaves the surface of the heisphere. A3 Because the surface is sooth, the noral reaction force R does no work. So the echanical enery of the particle conserves. 1 2 v2 + a cos θ = a, which leads to v 2 = 2a(1 cos θ) The equation of otion for r direction is v = 2a(1 cos θ) ( a θ 2 ) = R cos θ, 2 (a θ) which leads to R = cos θ a The particle leaves the surface when R = 0. 3 cos θ c 2 = 0 cos θ c = 2 3 = cos θ 2(1 cos θ) = (3 cos θ 2) (θ c = tan 1 2 3 48 ) = cos θ v2 a 3
Q4 A ullet of ass strikes a pendulu o (weiht) of ass M horizontally with speed v, and then ecoes eedded in the o. The o is initially at rest, and is suspended y a stiff rod of lenth l and neliile ass. The o is free to rotate in the vertical direction. (1) Find the velocity of the o (with eedded ullet) riht after hit y the ullet. (2) Find the iniu value of v which causes the o to execute a coplete vertical circle. M A4 The linear oentu conservation: v = (M + )V 0 V 0 = M + v The tension is perpendicular to the velocity of the particle, thus the echanical enery is conserved. 1 2 (M + )V0 = 1 2 2 (M + )V 2 + (M + )l(1 cos θ) V 2 = V0 2 2l(1 cos θ), where θ is the anle etween the rod and the vertical line. This ust e positive for any θ. V 2 0 > 4l v > M + 4l Note that the rod is solid, so that we don t have to worry aout the sin of tension force the rod exerts on the o. In case the o is suspended y a flexile strin, the condition would e v > M + 5l. See Q2. 4
Q5 A shell of ass is shot with an initial velocity v 0 at an anle of θ 0 with the horizontal. At the top of the trajectory, the shell explodes into two fraents (fraent 1 and fraent 2) of equal ass. Fraent 1, whose speed iediately after the explosion is zero, falls vertically. (1) Which fraent does reach the round earlier? v 0 θ 0 Explosion Fraent 1 Fraent 2 (2) How far fro the un does fraent 2 land, asuin that the terrain is level and the air dra is neliile? A5 (1) Both fraents reaches the round at the sae tie. The vertical coponent of the shell s velocity was zero when the shell was exploded at the top of the trajectory. The fact the speed of fraent 1 iediately after the explosion was zero eans the vertical coponent of the fraent 2 was also zero riht after the explosion (ecause of the linear oentu conservation). (2) First, we consider the shell s otion until it reaches to the top. The differential equations are with initial conditions These yield z =, ẍ = 0 ż(0) = v 0 sin θ 0, ẋ(0) = v 0 cos θ 0 z(0) = 0, x(0) = 0 ż = v 0 sin θ 0 t ẋ = v 0 cos θ 0 z = (v 0 sin θ 0 )t 1 2 t2 x = (v 0 cos θ 0 )t The shell s vertical velocity is zero when it reaches the top. By solvin v 0 sin θ 0 t = 0, we understand the shell reaches the top at t = v 0 sin θ 0. The shell has travelled horizontally x t = (v 0 cos θ 0 ) v 0 sin θ 0 = v2 0 sin θ 0 cos θ 0. The heiht of the top of the trajectory is z t = (v 0 sin θ 0 ) v 0 sin θ 0 1 v0 2 sin 2 θ 0 2 1 2 v2 0 sin 2 θ 0 2 = Second, we consider the otion of fraent 2 after the explosion. Let the velocity of the shell just efore the explosion v, the velocity of the fraent 1 and 2 iediately after the explosion v 1 and v 2 respectively. The asses of the fraent 1 and 2 are 1 and 2. Linear oentu conservation says v = 1 v 1 + 2 v 2 5
The speed of fraent 1 iediately after the explosion is zero : v 1 = 0. The asses of fraent 1 and 2 are 1 2. The velocity of the shell at the top was v = (v 0 cos θ) e x. By sustitutin these conditions, we otain that (v 0 cos θ 0 ) e x = 1 2 v 2 v 2 = (2v 0 cos θ 0 ) e x Thus fraent 2 is projected horizontally with initial speed of 2v 0 cos θ 0 fro the heiht of h = 1 v0 2 sin 2 θ 0. The tie it takes for fraent 2 to reach the round is t = v 0 sin θ 0 2 (which is sae as the tie it takes for the shell to reach the top). Durin this period, fraent 2 travels horizontally y (2v 0 cos θ 0 ) v 0 sin θ 0 = 2v2 0 sin θ 0 cos θ 0 Addin toether, the fraent 2 will land at x = v2 0 sin θ 0 cos θ 0 + 2v2 0 sin θ 0 cos θ 0 = 3v2 0 sin θ 0 cos θ 0 6
Q6 Two skaters, one with ass 65k and the other with ass 40k, stand on an ice rink holdin a pole of lenth 10 and neliile ass. Startin fro the ends of the pole, the skaters pull theselves alon the pole until they eet. How far does the 40k skater ove? A6 Let the positions of the skaters to e x 1 and x 2, and asses to e 1 and 2. Initially, two skaters are separated y 10. x 2i x 1i = l After pullin theselves alon the pole, they eet each other. Fro these two equations, x 2f x 1f = 0 (x 2i x 2f ) (x 1i x 1f ) = l Durin this process, no horizontal external forces are actin on the skaters. So the liniar oentu of the syste is conserved. The skaters are at rest initially. Thus, The 40k skater oves y 1 ẋ 1 + 2 ẋ 2 = 0 1 x 1 + 2 x 2 = const. 1 x 1i + 2 x 2i = 1 x 1f + 2 x 2f 1 (x 1i x 1f ) + 2 (x 2i x 2f ) = 0 ( 1 + 2 )(x 2i x 2f ) = 1 l 1 (x 2i x 2f ) = l 1 + 2 ( 1 + 2 )(x 1i x 1f ) = 2 l (x 1i x 1f ) = 2 1 + 2 l 65 k 65 10 = 10 6.2. 65 k + 40 k 105 7
Q7 A particle P of ass is situated on the axis of a circular loop of ass M and radius a. Find the ravitational force that the loop exerts on the particle. z P(0, 0, ) O θ a y A7 We consider the eleent of lenth dl on the loop. dl has ass Mdl, and attracts P with 2πa the force of anitude GMdl 2πaR. 2 By syetry, we need to add the zcoponent of the force to otain the resultant force actin on P. The z coponent is GMdl 2πaR 2 R The distance fro dl to P is R = a 2 + 2. Finally dl = adθ. By interatin over the loop, we otain F = ( l = GM 2πa GMdl 2πaR 2 = GM (a 2 + 2 ) 3/2 2π R dl = 0 2πa (a 2 + 2 ) 3/2 = GM 2 1 (1 + a )3/2 ) x GM 2πa adθ (a 2 + 2 ) 3/2 8