Kinematics, Dynamics, Mass Weight and Density Question

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1. It takes 10 minutes for a MRT train to travel 7.0 km from the Tanah Merah MRT Station to the Changi airport.

Which statement is not correct?
(A) The displacement is 7.0 km.
(B) The average speed is about 42.0 km / h.
(C) The average velocity is smaller than 42.0 km / h.
(D) The instantaneous speed can be greater than 42.0 km / h.

Analysis:
This question is actually testing you on the difference on displacement and distance, speed and velocity, and the concept in instantaneous speed.
Option A is wrong as 7.0km is the distance travelled from MRT station to the Airport. Displacement is unlikely to have 7.0km unless the MRT track is a straight line, which is not possible! So, displacement should be less than 7.0km.
Option B is true as average speed = distance / time = 7 / (10/60) = 42.0km/h
Option C is true as average velocity = displacement / time. With displacement less than 7.0km, average velocity should be smaller than 42.0km/h
Option D is true as average speed is 42.0km/h. Therefore, instantaneous speed can be greater than, smaller than or equal to 42.0km/h

2. Darren has recently taken up snowboarding as a hobby. He is practising making smooth turns while travelling up sloped inclines. The velocity-time graph in Figure shows his motion travelling up an embankment and part-way down.
                  

(a) Determine
      (i) his acceleration at t = 8.0 s.
      (ii) the distance he travelled from 0.0 to 5.0 seconds.
      (iii) at what time he began to travel back down the embankment.
(b) Sketch the acceleration-time graph until t = 20.0 s. Label all appropriate numerical values on the graph.

(a)(i) Comment: A very common mistake is to simply read the value of speed at t = 8.0s as 6.0m/s, and then divide it by 8.0 to get the answer as acceleration of 6/8 = 0.75ms-2

Analysis: Recall acceleration is the gradient of the v-t graph. At t = 8.0s, v-t graph is a downslope straight line. So we can simply choose any two points on the line to find the gradient.
Two points chosen: (5, 12) and (14, -6), so a = (-6 – 12) / (14 – 5) = -2.00 ms-2
(ii) Recall distance is the area under v-t graph
      d = area under v-t graph
         = 12 * 5 = 60.0 m
(iii) As Darren travels up the embankment first and then travels down, the time he began to travel back down is when he has a speed of 0m/s. Recall the speed at highest point is 0m/s. Therefore, t = 11.0 s
(b) Recall acceleration is the gradient of v-t graph. Therefore, we find the respective gradient for each region from 0s to 5s, from 5s to 14s and from 14s to 20s. We should be able to draw the a-t graph as shown below.

                  

3. A boy applies a force of F to push an object which is placed on a surface. The force R is the reaction force according to Newton’s Third Law of Motion.

                                    

Which of the following statements is correct ?
(A) The object will not move because F and R are equal and opposite.
(B) The object will not move because F and the friction are equal and opposite
(C) The object will move because F is greater than R.
(D) The object will move because F is greater than the friction.

Analysis:
To determine whether an object will move or not, we have to look at the forces acting on the object.
F: a force acting on the object by the boy to the right
R: a force acting on the boy by the object to the left (by Newton's Third Law)
friction: a force acting on the object by the rough surface to the left

Based on the analysis, the forces acting on the object are only F and friction. Thus, answer is D as F is greater than friction as shown in the Figure.

4. Ryan is pulling a block with a rope across a floor. The block has a mass of 10.0 kg. Ryan pulls with a horizontal force of 60.0 N and the constant force of friction is 20.0N.
(a) Calculate the net force acting on the block.
(b) Calculate the acceleration of the block.
(c) If the rope is broken, what happen to the motion of the block.

(a) Fnet = Fapplied - frictional force
              = 60 - 20 = 40.0 N
(b) Fnet = ma
     40.0 = 10.0 a
          a = 4.0ms-2
(c) If the rope is broken, 60.0N of force is no longer acting on it. The only force acting on it now is friction.
      Fnet = ma
     20.0 = 10.0 a
          a = 2.0ms-2
So, it will decelerates with a value of 2.0ms-2

5. A spaceship descends vertically toward the surface of the Moon with a constant speed of 3.0 m/s. The total mass of the spaceship is 15.0 kg. Assume the gravitational strength on the Moon to be 1.6 N/kg. During the first part of the descent, the upward force acting on the space ship is 24.0 N.
(a) Show that the spaceship is moving with a constant speed.
(b) Now, the upward force is increased to 30.0 N for the last 5 s. Calculate
      (i) the deceleration of the spaceship during this time.
      (ii) the velocity of the spaceship just before it lands.

(a) Note that g is 1.6N/kg now!
     W = mg = 15 * 1.6 = 24N
     Fnet = W - Fup
            = 24 - 24 = 0N
Since net force acting on the spaceship is 0N, the acceleration is 0ms-2, which means it is moving at constant speed.
(b)(i) Fnet = ma
     24 - 30 = 15a
               a = -0.4ms-2
(ii) Note that the initial velocity is 3.0m/s.
       a = (v - u)/t
  -0.4 = (v - 3)/5
       v = 1m/s

6. Block A, B and C has a mass of 4.0 kg, 12.0 kg and 13.0 kg respectively as shown in Figure. They are connected by light ropes and the ropes pass over frictionless and massless pulleys. The block B is on the rough horizontal surface.
                                    
(a) State and explain in which direction (left or right) the block B will move after released from rest.
(b) Assume the blocks are moving with an acceleration of 2.0 ms^(-2), calculate
      (i) the tension acting in rope A and C (i.e TA and TC).
      (ii) the frictional force between block B and horizontal surface.
(c) Determine the acceleration of the block C if rope C breaks.

(a) The block B will move to the right because the pulling force from block C is greater than that from block A since block C is heavier.
(b)(i) To do this kind of question, you have to draw the free-body diagram first. When drawing free-body diagram, think about which force is larger. Since block B moves to the right, block C has to move down and block A has to move up. This means WC > TC and TA > WA

         TA - WA = mA a
           TA - 40 = 4 * 2
                  TA = 48 N

         WC - TC = mC a
           30 - TC = 13 * 2
                   TC = 104 N
(ii) TC - TA - f = mB a
    104 - 48 - f = 12 * 2
                     f = 32N
(c) 10 ms-2 undergo free fall

7. A large wrecking ball is held in place by two light steel cables as shown in Figure. Given the weight of the wrecking ball is 400 N, calculate the tension TA and TB in the cables by the means of a scale diagram.
                                    

Analysis:
(a) The only known force is the weight, which means this is the only line you can draw with a value.
(b) TA acts at an angle of 90 deg horizontally to the left;
TB acts at an angle of 40 deg up to the vertical line.

Since the ball is in equilibrium, it is easier to use Tip-to-Tail Method.

For Tip-to-Tail Method,
we first draw the weight followed by drawing TA with an angle of 90 deg to the left, then draw TB with an angle of 40 deg to the tail of the line of action of weight. TA and TB would intersect at a point and from there, we determine the magnitude of TA and TB.

For Parallelogram Method,
Since the ball is in equilibrium, the weight has the same magnitude but in opposite direction with the resultant force of TA and TB. So we start by drawing a vertical line to represent the resultant force with double arrowhead. Then draw a horizontal force, TA, to the left and TB with an angle of 40 deg with the vertical line. After which, we have to construct a parallelogram by shifting the TA line up so that it will intersect with TB. Likewise, shift TB to the left and it will intersect with TA. From there, we can then determine the magnitude of TA and TB

Scale: 1cm to 100N


TA = 340N and TB = 520N

8. Two liquids X and Y were mixed together. The mass of liquid X is 0.40 kg and its density is 2.0 gcm-3. The mass of liquid Y is 0.26 kg and its density is 65.0 gcm-3. Assume that no new substance is produced by the mixing, determine
(a) the volume of the new mixture.
(b) the density of the new mixture.

(a) Recall density = mass / volume
Volume of X = 0.40 * 1000 / 2 = 200 cm3
Volume of Y = 0.26 * 1000 / 65 = 4 cm3
total volume = 200 + 4 = 204 cm3

 Comment: a very common mistake is to forget converting the unit before calculation. You could either convert gcm-3 to kgm-3 or just convert kg to g and leave the volume in cm-3.

(b) Analysis:
We need to first find the total mass of the new mixture, then find the density.

total mass = (0.4 + 0.26) * 1000 = 660 g
density = total mass / total volume = 660 / 204 = 3.24 gcm-3


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