| AS Mechanics: Exam Pack 3: Connected ParticlesName:______________________ Date:________________Year 12Q1.A car of mass 800 kg pulls a trailer of mass 200 kg along a straight horizontal road using a light towbarwhich is parallel to the road. The horizontal resistances to motion of the car and the trailer havemagnitudes 400 N and 200 N respectively. The engine of the car produces a constant horizontal drivingforce on the car of magnitude 1200 N. Find(a) the acceleration of the car and trailer,(3)(b) the magnitude of the tension in the towbar.(3)The car is moving along the road when the driver sees a hazard ahead. He reduces the force producedby the engine to zero and applies the brakes. The brakes produce a force on the car of magnitudenewtons and the car and trailer decelerate. Given that the resistances to motion are unchanged and themagnitude of the thrust in the towbar is 100 N,(c) find the value ofFF.(7)(Total 13 marks)
(a) Let's first find the force that provides an engine of the car: "P=Fv,""F=\\dfrac{P}{v}=\\dfrac{14.28\\cdot10^3\\ W}{6\\ \\dfrac{m}{s}}=2380\\ N."Then, we can find the acceleration of the car from the Newton's Second Law of Motion: "F-R_1-R_2=(m_1+m_2)a,""a=\\dfrac{F-R_1-R_2}{m_1+m_2},""a=\\dfrac{2380\\ N-630\\ N-280\\ N}{1400\\ kg+700\\ kg}=0.7\\ \\dfrac{m}{s^2}."(b) Applying the Newton's Second Law of Motion, we get: "F-R_1-T=m_1a,""T=F-R_1-m_1a,""T=2380\\ N-630\\ N-1400\\ kg\\cdot0.7\\ \\dfrac{m}{s^2}=770\\ N."Find more answers or ask new questions now. Find More Answers 
I was wondering if anyone could check my answers and help with some questions... 1.) A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope. The mass of the car is 1400kg. The mass of the trailer is 700kg. The car and the trailer are modelled as particles and the tow-rope as a light inextensible string. The resistances to motion of the car and the trailer are assumed to be constant and of magnitude 630N and 280N respectively. The driving force on the car, due to its engine, is 2380N. Find: a) The acceleration of the car a=F/m 2380-(630+280)/1400+700=0.7 a=0.7ms^-2 b) the tension in the tow-rope T-muR=ma T-0=1400(0.7) T=980N When the car and trailer are moving at 12ms^-1, the tow-rope breaks. Assuming that the driving force on the car and the resistances to motion are unchanged, c) find the distance moved by the car in the first 4s after the tow-rope breaks. s= ? , u=12, a=0.7, t=4 s=ut+1/2at^2 =12(4) +0.5(0.7) * 4^2 =48+10 = 53.6m d) State how you have used the modelling assumption that the tow-rope is inextensible. The tension is the same throughout the string & the acceleration of the masses is the same.
2.) A railway truck P of mass 2000kg is moving along a straight horizontal track with speed 10ms^-1. The truck P collides with a truck Q of mass 3000kg, which is at rest on the same track. Immediately after the collision, Q moves with speed 5ms^-1. Calculate a) the speed of P immediately after the collision m1u1+m2u2=m1v1+m2v2 2000(10)+3000(0)=2000v+3000(5) 20000=2000v+15000 2000v=5000 Pv=2.5ms^-1 b) the magnitude of the impulse exerted by P on Q during the collision. mv-mu=I 2000(2.5)-2000(10)=I 5000-20000=-15000 I=15000Ns
3.) A fixed wedge has two plane faces, each inclined at 30 degrees to the horizontal. Two particles A and B, of mass 3m and m respectively, are attached to the ends of a light inextensible string. Each particle moves on one of the plane faces of the wedge. The string passes over a small smooth light pulley fixed at the top of the wedge. The face on which A moves is smooth. The face on which B moves is rough. The coefficient of friction between B and this face is mu. Particle A is held at rest with the string taut. The string lies in the same vertical plane as lines of greatest slope on each plane face of the wedge, as shown in the figure above. The particles are released from rest and start to move. Particle A moves downwards and B moves upwards. The accelerations of A and B each have magnitude 1/10g. a) By considering the motion of A, find, in terms of m and g, the tension in the string. T-mg=ma T-3mg=3m(1/10g) T=3m(1/10g)+3mg =0.3mg+3mg =32.34mN b) By considering the motion of B, find the value of mu. R-mgcos60=ma R-mgcos60=1/10gm R=1/10gm+mgcos60 =5.88m sin60mg-mu(5.88m)=1/10gm 0.866g-5.88mu=0.98 8.4868-5.88mu=0.98 8.4868=0.98+5.88mu 5.88mu=7.5068 mu=1.28 c) Find the resultant force exerted by the string on the pulley, giving its magnitude and direction. R=F/mu =9.8sin90m/1.28 =6.63m
(Original post by najinaji) I was wondering if anyone could check my answers and help with some questions... 1.) A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope. The mass of the car is 1400kg. The mass of the trailer is 700kg. The car and the trailer are modelled as particles and the tow-rope as a light inextensible string. The resistances to motion of the car and the trailer are assumed to be constant and of magnitude 630N and 280N respectively. The driving force on the car, due to its engine, is 2380N. Find: a) The acceleration of the car a=F/m 2380-630/1400=1.25ms^-1 You might want to reconsider your answers to parts b) and c), as these are in error too.
(Original post by ghostwalker) DRAW a diagram, if you haven't already. It should be clear that if you treat the car by itself, then the forces acting on it include the tension in the tow rope. For this first part of the question, the car and trailer will be accelerating at the same rate, so it's easier to treat the two masses together, as if they were one mass, and this eliminates the necessity to consider the tension in the tow rope. You might want to reconsider your answers to parts b) and c), as these are in error too. Ah, I thought so. So is it: 2380-(630+280)/1400+700=0.7a=0.7ms^-2?
(Original post by najinaji) Ah, I thought so. So is it: 2380-(630+280)/1400+700=0.7 a=0.7ms^-2? Yep.
(Original post by ghostwalker)
(Original post by najinaji) Edit: I assume you're still correcting 1b. The methodology you're using is incorrect since there is no friction, but there is a resistive force.
4.) a) Two particles A and B, of mass 3kg and 2kg respectively, are moving in the same direction on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of A is 4ms^-1 and the speed of B is 1.5ms^-1. In the collision, the particles join to form a single particle C. Find the speed of C immediately before the collision. m1u1+m1v1=m2u2+m2v2 3(4)+2(1.5)=3v+2v 15=5v v=3ms^-1 b) Two particles P and Q have mass 3kg and m kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table. Each particle has speed 4ms^-1 when they collide directly. In this collision, the direction of motion of each particle is reversed. The speed of P immediately after the collision is 2ms^-1 and the speed of Q is 1ms^-1. Find i) the value of m 3(4)+3(2)=4m+m 12+6=5m 5m=18 m=3.6kg ii) the magnitude of the impulse exerted on Q in the collision mv-mu=Ft 3.6(4)-3.6=Ft 14.4-3.6=Ft I=10.8Ns
(Original post by ghostwalker) T-630=1400(0.7)?
(Original post by najinaji) Ah, I see. so should it be: T-630=1400(0.7)? There is also the driving force. And take care what's positive and what's negative.You need to assign one direction as positive, and then assign signs appropriately.
(Original post by ghostwalker) There is also the driving force. And take care what's positive and what's negative. You need to assign one direction as positive, and then assign signs appropriately. Is it:2380-T-630=1400(0.7)?
5.) A stone S is sliding on ice. The stone is moving along a straight horizontal line ABC, where AB=24m and AC=30m. The stone is subject to a constant resistance to motion of magnitude 0.3N. At A the speed of S is 20ms^-1, and at B the speed of S is 16ms^-1. a) the deceleration of S s=24, u=20, v=16, a=? v^2=u^2+2as 16^2=20^2+2a(24) 256=400+48a 48a=-144 a=-3ms^-2 b) the speed of S at C s=6, u=16, v=?, a=-3 v^2=u^2+2as v^2=16^2+2(-3*6) v^2=256-36=220 v=14.8ms^-1 c) Show that the mass of S is 0.1kg At C, the stone S hits a vertical wall, rebounds from the wall and then slides back along the line CA. The magnitude of the impulse of the wall on S is 2.4Ns and the stone continues to move against a constant resistance of 0.3N. d) Calculate the time between the instant that S rebounds from the wall and the instant that S comes to rest.
(Original post by najinaji) Is it: 2380-T-630=1400(0.7)? Yep.Now try part c), after the tow rope breaks.
(Original post by ghostwalker) Yep. Now try part c), after the tow rope breaks. s= ? , u=12, a=0.7, t=4 s=ut+1/2at^2 =12(4) +0.5(0.7) * 4^2=48+10 = 53.6m ?
(Original post by najinaji) s= ? , u=12, a=0.7, t=4 s=ut+1/2at^2 =12(4) +0.5(0.7) * 4^2 =48+10 = 53.6m ? Your acceleration of 0.7 comes from calculations where the tow rope is intact. You have a new situation where there is no tow rope, so you need to calculate the new acceleration, and take it from there.
(Original post by ghostwalker) =48+10=58 ?
(Original post by najinaji) a=F/m 2380-630/1400=1.25 s=?, u=12, a=1.25, t=4 s=ut+1/2at^2 =12(4) +0.5(1.25) * 4^2 =48+10=58 ? Yep
(Original post by ghostwalker)
Sorry to bump, but I still need someone to check the rest of my answers. |
A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope
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