Sunday, November 9, 2014

Physics 9702 Doubts | Help Page 10

  • Physics 9702 Doubts | Help Page 10

Question 55: [Momentum > Projectile motion]
A projectile of mass m is fired with velocity v from a point P, as shown below. 


Neglecting air resistance, the magnitude of the change in momentum between leaving P and arriving at Q is
A zero             B ½ mv           C mv√2           D mv               E 2mv

Reference: Past Exam Paper – N80 / II / 6



Solution 55:
Velocity, v is a vector has can be resolved into 2 perpendicular components as follows:
Define the upwards direction as positive for the vertical and the direction from P to Q as positive for the horizontal.

At P,
Horizontal component of velocity = (+) v cos45o
Vertical component of velocity = (+) v sin45o  

Similarly, momentum, p is also a vector and can be resolved into 2 perpendicular components.
At P,
Horizontal component of momentum = (+) mv cos45o
Vertical component of momentum = (+) mv sin45o  
Resultant momentum = [(mv cos45o)2 + (mv sin45o)2]0.5 = mv

Now, arriving at Q, velocity is NOT zero (it may be zero at the instant just after hitting the ground but not when arriving at Q).
Since air resistance is negligible, the horizontal component of velocity at Q is the same as at P. However, the vertical component of the velocity is now downwards due to gravity (its magnitude is the same as at P since the (maximum) vertical distance travelled and the acceleration (due to gravity) are the same).

At Q,
Horizontal component of velocity = (+) v cos45o
Vertical component of velocity = (-) v sin45o  
Horizontal component of momentum = (+) mv cos45o
Vertical component of momentum = (-) mv sin45o  

Change in (resultant) momentum = mΔv = m (vf – vi)
where vf is the final velocity and vi is the initial velocity

The horizontal velocity (and hence, horizontal momentum) is unchanged (change = 0).
Change in vertical component of momentum = (-)mv sin45o – mv sin45o  = - 2mv sin45o  

Change in (resultant) momentum = [02 + (- 2mv sin45o)2]0.5 = [4 x (mv)2 x 0.5]0.5
Change in (resultant) momentum = [2 x (mv)2]0.5 = mv √2

Answer: C








Question 56: [Kinematics > Graphs]
The graph shows the variation with time t of the velocity v of a bouncing ball, released from rest. Downward velocities are taken as positive.
At which time does the ball reach its maximum height after bouncing?




Reference: Past Exam Paper – J99 / I / 3



Solution 56:
The graph is that of a velocity-time graph. The area enclosed under the graph (= v x t) gives the displacement.
The ball is released from rest (from a height) and downward velocities are taken as positive. So, the upper area in the graph shows the distance fallen (not height) while the area enclosed by the graph under the x-axis gives the height.

From the graph, it can be seen that speed increases from t = 0 to t = A (the ball falls under the acceleration due to gravity).
From t = A to t = B, there is a steep decrease in speed and from t = B to t = C, the speed increases in the opposite direction (upward). So, t = A to t = C represents the time of collision of the ball with the ground.
At t = C, the (upward) speed is maximum and it decreases (due to the acceleration due to gravity acting downwards) until t = D (speed is zero, so maximum height has been reached).
Answer: D







Question 57: [Momentum]
A group of students investigating the principle of conservation of momentum use a small truck travelling over a frictionless surface.
Sand is dropped into the truck as it passes X. At Y, a trapdoor in the bottom of the truck opens and the sand falls out.

How does the velocity of the truck change when the sand is added to the truck at X and then leaves the truck at Y?

Reference: Past Exam Paper – November 2011 Paper 12 Q10



Solution 57:
Answer: B.
In this problem, you should consider conservation of the total momentum of the truck and sand, rather than considering the truck in isolation. 

From conservation of momentum (p = mv), when mass (sand) is added to the truck, the total mass (of the truck and sand) increases, resulting in speed v (of the moving truck containing the sand) to decrease so that momentum p is unchanged (conserved).  So, at X, the speed of the truck decreases.
Then the truck (containing the sand) will move with this new speed.

Consider at Y where the sand falls. So, the mass of truck has returned to normal since it no longer contains the sand. But, as the sand falls [it’s obviously falling at some speed which is equal to the speed of the truck], it has some momentum such that the sum of momentum of the (separate) truck and sand together is unchanged and equal to the total momentum at X. So, at Y, the speed of the truck remains the same.

So, what is important to note from this question is that at X, the sand is in the truck which is moving. So, they can be considered as a single body. But at Y, the sand falls from the truck. Now, they are 2 separate bodies. But since the sand was previously in the truck which was moving, it falls at the same speed of motion of the truck.  So, the momentum of the sand should also be considered, in addition to the momentum of the truck.








Question 58: [Projectile motion]
When a rifle is fired horizontally at a target P on a screen at a range of 25m, the bullet strikes the screen at a point 5.0mm below P. The screen is now moved to a distance of 50m and the rifle again fire horizontally at P in its new position.
Assuming that air resistance may be neglected, what is the new distance below P at which the screen would now be struck?




A 5√2mm        B 10mm          C 15mm          D 20mm          E 25mm

Reference: Past Exam Paper – November 1982 / II / 4



Solution 58:
Answer: D.
The equations for uniformly accelerated motion involving displacement are
s = ut + ½at2    and      v2 = u2 +2as
We need to find a vertical displacement. For the first equation, t is not known. For the second equation, v is not known. So, the calculation cannot be carried out directly. Since there are 2 cases, we can write separate equations and solve them simultaneously.

For case 1,
Let (horizontal) velocity with which the bullet leaves the rifle = v
Let time taken for bullet to travel the horizontal distance of 25m = t1
Speed = distance / time
Time, t1 = 25 / v

Now, consider the vertical motion of the bullet,
Initial (vertical) velocity of bullet, u = 0
Acceleration, a = g, the acceleration due to gravity
Displacement, s = 5.0mm = 5.0x10-3m
Time to be displaced by vertical displacement s, t1 = 25 / v
s = ut + ½ t2
5.0x10-3 = 0 + 0.5 g (25 / v)2  ……………….(1)

For case 2,
Time to travel (horizontal) distance of 50m, t2 = 50 / v
Let the new displacement below P = x
s = ut + ½at2
 x = 0 + 0.5 g (50 / v)2 ………………………(2)

To find x, divide equation (2) by equation (1)
x / (5x10-3) = (50 / 25)2 = 4
New displacement, x = 20.0x10-3 = 20mm







Question 59: [Projectile motion]
A ball, dropped on a 45o inclined plane, makes an elastic collision with the surface. Stroboscopic photographs (a series of exposures on the same film at equal time intervals) are taken of the path of the ball. Which one of the following diagrams best represents the photographs?



Reference: Past Exam Paper – June 1984 / II / 3



Solution 59:
Answer: A.
Under the influence of acceleration due to gravity, the (downward) speed of the ball increases every second. So, in a specific time interval, the vertical separation of the exposures should increase (since the distance travelled every second increases).

Just after the collision, the vertical speed is zero and the horizontal speed is equal to the speed just before collision [for elastic collision, relative speed of approach is equal to the relative speed of separation]. So, the separation of exposures just after collision should be equal to that just before collision. [B is eliminated]

The horizontal speed is not affected by any forces (like air resistance), the horizontal separation should be equal. [C and E are eliminated]

Let’s assume that each exposure was taken after 1 second each. Since there are 8 exposures, the final speed before collision would be v = u + at = 0 + g8 = 8g. After the collision, the horizontal sped is constant and equal to 8g (as explained before). However, the vertical speed of collision is zero and will only be equal to the horizontal speed at the 8th exposure. In such a situation, if a line is drawn between 2 exposures there, it should be parallel to the inclined plane [since horizontal and vertical components are equal]. It also means that for the earlier exposures, the horizontal component is more dominant. 

But looking at diagram A and D, it is seen that for D, a parallel line between 2 exposures is obtained before the 8th exposure is reached. [So, D is also incorrect]. And if we look at diagram A, it is seen clearly that the horizontal is more dominant than the vertical for the earlier exposures.

 

 

9 comments:

  1. Assalamualaikum!
    Are relative speeds of approach and separtion always equal, even when objects of different masses collide, or when the directions are along different axis(as in q59, where the horizontal velocity has been equated to the vertical velocity)?
    should we take into account the directions of the velocities when usinf the relative spped of separation and approach formula?
    why can q48 be nt solved this way?
    ---rel.speed of app=relspeed of sep
    v+0=- 6v+speed of rocket
    speed of rocket= +7v

    What does this mean(q59):
    "In such a situation, if a line is drawn between 2 exposures there, it should be parallel to the inclined plane [since horizontal and vertical components are equal]. It also means that for the earlier exposures, the horizontal component is more dominant."
    Kindly explain this in detail.
    I would really appreciate it if i could help with this.

    ReplyDelete
    Replies
    1. Wslm.
      Yes, for elastic collision, relative speed of approach is equal to the relative speed of separation, even when the masses are different.

      Usually, 2d problems like the above are no longer in the syllabus. Currently, momentum calculations are only for linear cases. Normally, you'll have to equate the velocity itself, not its components but the above is a special case since before collision, the speed is vertical (there is no horizontal component) and just after the collision, the speed is horizontal (there is no vertical component).

      For Q48, it is not a collision, and even if it was, it has not been mentioned that it's an elastic collision. For that case, the law of conservation of momentum still applies (actually this applies to all systems): the sum of momentum before collision is equal to the sum of momentum after collision.

      As I said in the qoute, after 8 sec, the speed would become 8g. Acceleration is the rate of change of velocity. That is, if the acceleration is g and the initial speed is zero,

      after 1sec, speed = g
      after 2 sec, speed = 2g
      This is given from the equation v = u + at = 0 + at

      after 8 sec, speed = 8g

      The horizontal speed is not affected by gravity and is, after the collision 8g. After 8sec, the vertical speed becomes 8g and the horizontal speed is still 8g. So, the overall speed is at an angle of 45 degrees to the horizontal since both the horizontal and vertical components are the same.

      Since the plane is inclined at 45degrees, a vertical line at the 8th exposure would be parallel to the plane.

      Delete
    2. JazakAllah
      should we take into account the directions of the speeds when calculating the relative spped of approach/separation?
      for e.g., if a truck, moving at 5m/s towards the right, is about to have a head-on collision with another truck moving to the left at 10 m/s, would the relative speed of approach be 5+10=15m/s or 5+(-10)= -5m/s
      what do you mean by "vertical line" when you say "a vertical line at the 8th exposure "?

      Delete
    3. The relative speed of approach is 10 + 5 = 15m/s

      Here, look how a single word could cause confusion to students. Any student understanding english would know what the word 'approach' or 'separation' mean, but they fail to implement it in their calculations.

      If you think about it yourself, you know that 'approach' means coming together, .. but you failed to fail to implement it. you were confused with whether it's 10+5 or 10-5 when it's obviously the former from the definition of approach.

      So, I would advise you and other students to think carefully and avoid 'silly' mistakes like these. And these types of questions are actually considered to be tricky or difficult, but you can see how obvious the answer can be.

      This is the same case in different situations.

      For example, another common mistake is as follows:
      It is known that gradient of a graph = change in y / change in x

      Now, consider the current-voltage (I-V) graph of a resistor, filament lamp, ...
      Ohm's law is as follows: V = IR

      It is clear from Ohm's law that we are not considering changes in V or changes in R but we consider only the actual values of V or R.

      But from the graphs, students usually use the gradient to find current. Gradient considers the CHANGE in values, so it's obviously not appropriate for finding currents ....

      I hope you understood what I tried to say.



      'vertical line': actually it should be straight line, sorry

      Delete
    4. are you saying that the answer is 10+5 because it is "speed" and not "velocity"?if this is not what you meant in the first 4 paragraphs of your reply, i dd not understand anything unfortunately.
      as for the current voltage graph, the gradient should give resistance, right?are you tryin to say that for a voltage-resitance graph with y-intercept as 0,0, will the not be the current?

      Delete
  2. The actual equation relating the speeds are:
    RELATIVE speed of APPROACH = RELATIVE speed of SEPARATION

    since it's RELATIVE, it cannot be velocity. Also, it's APPROACH.

    consider the following case:
    Person A is moving towards person B and B is not moving. So, A is approaching B. Can A approach B while moving in a direction different from the direction towards B. NO. If it was the case, it would not be called an 'approach'.

    Now consider A moving towards B and B is also moving towards B. So, they are approaching faster. If A is moving towards B and B is moving AWAY from A, then this can only be called an 'approach' if A is moving faster than B. the relative speed of approach is then smaller than before.


    FOr the current-voltage case, you CANNOT use gradient because gradient considers the change (between 2 values) in current and voltage.

    From, Ohm's law, resistance can be obtained by considering single values of each current and voltage, NOT changes in current and voltage.

    To find resistance, you only need to take a value of voltage and the corresponding value of current at that voltage.

    ReplyDelete
    Replies
    1. In IGCSE we learnt that to calculate a value for the fixed resistance of an unknown resistor, connect it in series in a circuit and measure the values of voltage and currents. take atleast 3 sets of readings, and plot a graph, and use the gradient to calculate the resistance. is this method wrong?

      Delete
    2. It's not necessarily wrong if when calculating the resistance, you use only a point, as I said before.

      YOu may measure some values, but you won't measure all the values of current and voltages. But through the graph, you can have values of currents and voltages that you did not actually measure.

      FOr example, if it is a straight line, by using the graph, you can use any point on the line for the currents and voltages even if you did not measure these. BUT you cannot use the gradient to find the resistance.


      Also, I just remembered, collision in 2d has been re-introduced in the new physics syllabus for 2016 exams and onwards.

      Delete

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