# Physics 9702 Doubts | Help Page 115

__Question 584: [Vectors]__
Forces of 3 N, 4 N and 5 N act at
one point on an object. Angles at which the forces act can vary.

What is the value of

**minimum**resultant force of these forces?
A 0

B between 0 and 2 N

C 2 N

D between 2 N and 4 N

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q13 & Paper 12 Q12 & Paper 13 Q11*

__Solution 584:__**Answer: A.**

For a minimum resultant force, we
will need to include subtraction as much as possible so that the resultant has
the smallest value possible. This is done by making the forces opposes each
other in some way. If all the forces point in the same direction, the resultant
force would become even bigger than the individual forces.

Consider the resultant magnitude of
3N + 4N such that the 2 forces are perpendicular to each other. Say the 3N
force is vertical and the 4N is horizontal. The 3N does not have any component
along the horizontal that could add to the horizontal 4N vector. Similarly, the
horizontal 4N force does not have add any vertical component to the 3N force.

Resultant (is the hypotenuse of the
triangle formed) = (3

^{2}+ 4^{2})^{0.5}= 5N.
Put the 5N force in the opposite
direction of this resultant calculated above to obtain zero.

__Question 585: [Dynamics > Collisions]__
Two identical, perfectly elastic
spheres have same mass m. They travel towards each other with same speed v
along a horizontal frictionless surface.

Which statement about the sum of
kinetic energies of the spheres is correct?

A The sum of their kinetic energies
before impact is zero.

B The sum of their kinetic energies
before impact is ½ mv

^{2}.
C The sum of their kinetic energies
after impact is zero.

D The sum of their kinetic energies
after impact is mv

^{2}.**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q13*

__Solution 585:__**Answer: D.**

Kinetic energy = ½ mv

^{2}
Since the 2 spheres are moving with
speed v before the impact, the sum of their kinetic energies is not zero. [A is incorrect]

Kinetic energy of EACH sphere = ½ mv

^{2}
Sum of KE of the 2 spheres = 2 (½ mv

^{2}) = mv^{2}[B is incorrect]
The 2 spheres are perfectly elastic
and the surface is frictionless, so the collision is also perfectly elastic. In
an elastic collision, energy is conserved. So, the sum of their kinetic energy
after the impact is equal to the sum before the impact. [C is incorrect]

__Question 586: [Simple harmonic motion]__
A mass of 78 g is suspended from a
fixed point by means of a spring, as illustrated in Fig.1.

The stationary mass is pulled
vertically downwards through a distance of 2.1 cm and then released.

Mass is observed to perform simple
harmonic motion with a period of 0.69 s.

**(a)**Mass is released at time t = 0.

For oscillations of the mass,

(i) calculate angular frequency Ï‰

(ii) determine numerical equations
for variation with time t of

1. displacement x in cm

2. speed v in ms

^{-1}**(b)**Calculate total energy of oscillation of mass

**Reference:**

*Past Exam Paper – June 2013 Paper 42 Q3*

__Solution 586:__**(a)**

(i) Angular
frequency Ï‰ = 2Ï€/T = 2Ï€/0.69 = 9.1rads

^{-1}
(ii)

1. Displacement x = 2.1cos(9.1t)

2.

{Maximum speed v

_{0}= Ï‰a where a is the amplitude (maximum displacement). This is a known formula.}
Maximum speed v

_{o}= Ï‰a = 9.1 x (2.1x10^{-2}) = 0.19ms^{-1}
{The displacement is x.
Speed v = dx/dt. By differentiating the equation for displacement, we can
obtain the speed. OR some students may learn it in this way: if the displacement
contains cosine, then speed will contain sine and vice versa.}

Speed v = v

_{o}sin(9.1t)**(b)**

Energy = either ½ mv

_{o}^{2}or = ½ mÏ‰^{2}x_{o}^{2}
=
either ½ (0.078) (0.19)

^{2 }or = ½ (0.078) (9.1)^{2}(2.1x10^{-2})^{2}
Energy = 1.4x10

^{-3}J

__Question 587: [Kinematics + Energy]__
Two planks of wood AB and BC are
inclined at angle of 15° to the horizontal. The two wooden planks are joined at
point B, as shown in Fig.1.

Small block of metal M is released
from rest at point A. It slides down the slope to B and up the opposite side to
C. Points A and C are 0.26 m above B. Assume frictional forces are negligible.

**(a)**

(i) Describe and explain
acceleration of M as it travels from A to B and from B to C.

(ii) Calculate time taken for M to
travel from A to B.

(iii) Calculate speed of M at B.

**(b)**Plank BC is adjusted so that the angle it makes with the horizontal is 30°. M is released from rest at point A and slides down slope to B. It then slides a distance along the plank from B towards C.

Use law of conservation of energy to
calculate this distance. Explain your working.

**Reference:**

*Past Exam Paper – November 2012 Paper 23 Q2*

__Solution 587:__**(a)**

(i) The accelerations (from A to B
and from B to C) have the same magnitudes {due to the
same component of the weight as the angles are similar} but are opposite
in directions {from A to B, the acceleration is the
same direction as motion but from B to C, the acceleration opposes motion}
OR both accelerations are toward B.

In both cases (from A to B and B to
C), the component of the weight down the slope provides the acceleration.

(ii)

Acceleration {= component of weight along slope} = g sin15°

Equation for uniformly accelerated
motion: s = ut + ½ at

^{2}= 0 + ½ at^{2}
From the triangle formed: s = 0.26 /
sin 15 ° = 1.0

{Acceleration a = g sin15°

10 = ½ (9.8sin15°) t

^{2}}
t

^{2}= (1.0 × 2) / (9.8 × sin15°)
Time t = 0.89 s

(iii)

{EITHER v = u + at OR v

^{2}= u^{2}+ 2as}
Speed v = 0 + (g sin15)t or v

^{2}= 0 + 2(g sin15) × 1.0
Speed v = 2.26 m s

^{–1}**(b)**

{From the conservation of
energy, the loss of gravitational potential energy (GPE) at A is equal to the
gain in GPE at C (which is now at a greater height since the angle has been
increased).

Alternatively, B is the
lowest position in the system – so the GPE is B is less. B is at a higher
position than B, so the GPE at C is more than at B. From the conservation of
energy, the loss of kinetic energy (KE) at B is equal to the gain in GPE at C.}

Loss of GPE at A = gain in GPE at C
OR loss of KE at B = gain in GPE at C

{For Loss of GPE at A =
gain in GPE ,

As stated in the question,
A is higher than B by h

_{1}= 0.26m. So, the loss of GPE at A = mgh_{1}. Or put in another way, the initial energy at A is mgh_{1}. This is the maximum energy of M. From the conservation of energy, this amount of energy cannot increase by itself (unless work is done to increase the energy, which is not the case here). So, the gain in energy at C must be equal to mgh_{1}. That is, the height h_{2}at C is equal to h_{1}at A, which is 0.26m.}
EITHER Height h

_{1}= h_{2}= 0.26 m
OR ½ mv

^{2}= mgh_{2}
{As calculated above, the
speed at B is 2.26 ms

^{-1}.}
giving Height h

_{2}= 0.5 × (2.26)^{2}/ 9.81 = 0.26 m
From the triangle formed: Slope x =
0.26 / sin 30° = 0.52 m

Can you explain about internal energy? June 2003 Paper 4 Q2(a). I don't understand, in the question "stone falling under gravity in vacuum", how being in a vacuum affect the internal energy? And why does stretching a wire at constant temperature means it's internal energy is increasing? If it's at constant temp, shouldn't ∆U = 0, and W = -Q (ie work is done on wire, and wire loses heat)? Thank you! :)

ReplyDeleteSee question 588 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-116.html

Can u explain how to find the direction and magnitude of the resultant of 3 vectors? Thanks:-)

ReplyDeletefollow the steps of solution 703 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-142.html

just add one more vector to it

Hi, can you explain question 587 b)ii) why is the weight component gsin15 instead of mgsin15? Thank you! :D

ReplyDeletewe are considering the component of acceleration, not the weight

Delete