Tuesday, May 19, 2015

Physics 9702 Doubts | Help Page 146

  • Physics 9702 Doubts | Help Page 146



Question 721: [Matter > Hooke’s law]
Cylindrical samples of steel, glass and rubber are each subjected to gradually increasing tensile force F. Extensions e are measured and graphs are plotted as shown below.

Which row correctly relates the graphs to the materials?
steel     glass    rubber
A         X         Y         Z
B         X         Z          Y
C         Y         X         Z
D         Y         Z          X

Reference: Past Exam Paper – June 2005 Paper 1 Q21 & June 2014 Paper 13 Q24



Solution 721:
Answer: D.
Graph X relates to rubber. It does not obey Hooke’s law.

Graph Y relates to steel. It obeys Hooke’s law until the limit of proportionality. Then, the extension is no longer proportional to the force – plastic deformation occurs. As more force is applied, the material will each a point where it breaks.

Graph Z relates to glass. The graph has a steep gradient and the extension is relatively much less than the other materials. This indicates if a force big enough is applied, the material breaks – so, we cannot extend the graph for large values of extension.










Question 722: [Measurement]
(a) State SI base units of force.

(b) Two wires each of length l are placed parallel to each other a distance x apart, as shown in Fig.1.
Each wire carries a current I. Currents give rise to a force F on each wire given by
F = KI2l / x
where K is a constant.
(i) Determine SI base units of K.
(ii) On Fig.2, sketch variation with x of F. The quantities I and l remain constant.
(iii) Current I in both of the wires is varied.
On Fig.3, sketch variation with I of F. The quantities x and l remain constant.

Reference: Past Exam Paper – June 2013 Paper 23 Q1



Solution 722:
(a) SI Unit of Force: kg m s–2

(b)
(i)
Units: I2: A2               l: m                  x: m
{K = Fx / I2l}
Units of K: kg m s–2 A–2

(ii) The curve of the correct shape (for inverse proportionality) should clearly be shown approaching each axis but never touching the axis
{The graph has the same shape as a graph of y = 1 / x. Plotting the values of x and y that are very close to each other will show that the graph is a curve, not a straight line. When x = 0, y = 1 / 0 = but we can’t plot a graph showing ∞, so the line should approach the axis but never touches the axis. Similarly, when y = 0, x = 1 / 0 = ∞.}



(iii) The graph should be shown curving upwards and through the origin
{The graph has the same shape as a graph of y = x2.}












Question 723: [Dynamics > Collision]
Particle of mass 2m and velocity v strikes a wall.

The particle rebounds along same path after colliding with the wall. The collision is inelastic.
What is a possible change in momentum of the ball during the collision?
A mv                           B 2mv                         C 3mv                         D 4mv

Reference: Past Exam Paper – November 2010 Paper 12 Q9



Solution 723:
Answer: C.
Initial momentum of particle = (2m)v = 2mv
Initial kinetic energy of particle = ½ (2m) v2 = mv2

In an inelastic collision, energy is lost – so the kinetic energy of the particle must be less than it was before the collision. But momentum is still conserved.

The wall is fixed (not moving), so its initial momentum is zero.

After colliding with the wall, the particle rebounds along same path. So, its direction has changed.

Momentum is a vector quantity, so we need to consider its direction too.
Consider the direction towards the wall to be positive.
Initial momentum of particle = (2m)v = +2mv

The final momentum of the particle needs to be negative since its direction has changed.

If the change in momentum Δm = mv,
Final momentum of particle = +2mv – mv = +mv
The direction is still positive. This means that the particle is still moving towards the wall after collision. We know, from the question, that this is incorrect.

If Δm = 2mv,
Final momentum of particle = +2mv – 2mv = 0
This means that the particle remains stationary after collision. This is incorrect since we know that the particle moves in the opposite direction.

If Δm = 3mv,
Final momentum of particle = +2mv – 3mv = – mv
The particle moves away from the wall. Mass of the particle = 2m, so the velocity of the particle is now –v/2 so that the momentum = (2m) (–v/2) = – mv
Final kinetic energy of particle = ½ (2m) (v/2)2 = mv2 / 4
The energy is less than before since this is an inelastic collision.

If Δm = 4mv,
Final momentum of particle = +2mv – 4mv = – 2mv
The particle moves away from the wall. Mass of the particle = 2m, so the velocity of the particle is now –v so that the momentum = (2m) (–v) = – 2mv
Final kinetic energy of particle = ½ (2m) (v)2 = mv2
Here, the energy is conserved, which is incorrect for an inelastic collision.









Question 724: [Nuclear Physics]
It was once thought that the mass of an atom is spread uniformly through volume of the atom. When α-particles are directed at a piece of gold foil, the results led scientists to believe instead that nearly all the mass of the gold atom is concentrated at a point inside the atom.

Which effect is possible only if nearly all the mass of the gold atom is concentrated at a point?
A a few α-particles bounce back
B most α-particles are only slightly deflected
C some α-particles pass through without any deflection
D some α-particles are absorbed

Reference: Past Exam Paper – June 2013 Paper 11 Q39



Solution 724:
Answer: A.
We need to say which if the effects given is ONLY possible if nearly all the mass is concentrated at one point in the atom.

If nearly all the mass is concentrated at a point, a few α-particles would bounce back. These are the α-particles that collide with that point (of concentrated mass) and their direction are reverse.

The α-particles that are slightly deflected are due to the charges of the α-particles and that of the point being the same (like charges repel). The reason is NOT because nearly all the mass is concentrated at the centre. [B is incorrect]

The fact that some α-particles pass through without any deflection is because most of the golf atom is empty space. By passing at a relatively longer distance than the other α-particles, these α-particles do not experience the electric repulsion sue to the charges. [C is incorrect]

The absorption of some α-particles occurs due to nuclear reactions. This is not ONLY due to the fact that most of the mass is concentrated at a point. [D is incorrect]


2 comments:

  1. in solution 723 isn't the final momentum = initial + change in momentum( as change in momentum = final - initial)?? please help me

    ReplyDelete
    Replies
    1. the change in momentum is negative, as explained above. The values given as the choices are only the magnitudes. Only we need to take the negative.

      Delete

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