# Physics 9702 Doubts | Help Page 162

__Question 806: [Dynamics > Conservation of Linear momentum]__
An ice-hockey puck slides along
horizontal, frictionless ice-rink surface. It collides inelastically with wall
at right angles to its path, and then rebounds along its original path.

Which graph shows variation with
time t of the momentum p of the puck?

For J92 / I / 4:

For November 2011 Paper 11 Q12 & Paper 13 Q10:

**Reference:**

*Past Exam Paper – J92 / I / 4 & November 2011 Paper 11 Q12 & Paper 13 Q10*

__Solution 806:__**Answer: A.**

A frictionless ice-rink surface
means that the speed of the puck is not affected (by friction) during its
motion. Thus, it can be assumed that it moves with constant speed before
collision.

Momentum p = mv

Since the speed is constant before
the collision, the momentum should also be constant. [B
and C are incorrect]

After the collision, the puck
rebounds along its original path.

Momentum is a vector, so if the
original direction of the momentum is taken as positive, the momentum after
collision is negative. [D is incorrect]

The collision is inelastic, so the
puck loses some of its energy. The puck moves with a lower speed after the
collision, and thus has a smaller momentum. Note that the wall gains some
momentum so that the total momentum is conserved in the system.

__Question 807: [Current of Electricity > e.m.f.]__
When a battery is connected to a
resistor, battery gradually becomes warm. This causes the internal resistance
of the battery to increase whilst its e.m.f. stays unchanged.

As internal resistance of the
battery increases, how do the terminal potential difference and the output
power change, if at all?

terminal
potential difference output power

A decrease
decrease

B decrease
unchanged

C unchanged
decrease

D unchanged
unchanged

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q33*

__Solution 807:__**Answer: A.**

Let the e.m.f. of the battery be E.

Due to its internal resistance (r),
there will be some lost volts (V

_{L}) in the battery.
V

_{L}= Ir where I is the current provided by the battery
Let the terminal potential
difference across the battery be V.

V = E – V

_{L}
As the internal resistance (r) of the
battery increases V

_{L}also increases causes the terminal p.d. V to decrease.
Output power P of battery = VI

Since the terminal p.d. V across the
battery decreases, the output power also decreases.

Note that the output power is NOT P
= EI because some volts are lost inside the battery and not provided to the
external circuit.

__Question 808: [Measurement > Uncertainty]__
Young modulus of the material of a
wire is to be found. Young modulus E is given by the equation below.

E = 4F

*l*/ Ï€d^{2}x
The wire is extended by a known
force and the following measurements are made.

Which measurement has the largest
effect on uncertainty in the value of the calculated Young modulus?

**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q5*

__Solution 808:__**Answer: B.**

The percentage uncertainty in the
diameter needs to be doubled since it is squared in the equation.

E = 4F

*l*/ Ï€d^{2}x
Î”E / E = Î”F/F + Î”l/l + 2(Î”d/d) + Î”x/x

Î”F / F = 0.01 / 19.62 = 0.000 51

Î”

*l*/*l*= 0.002 / 2.043 = 0.000 98
2 (Î”d / d) = 2 (0.02 / 0.54) = 0.074
07

Î”x / x = 0.2 / 5.2 = 0.038 46

So the diameter has the largest
effect on the uncertainty in the value of the calculated Young modulus.

__Question 809: [Waves > Intensity]__
A plane wave of amplitude A is
incident on surface of area S placed so that it is perpendicular to the
direction of travel of the wave. The energy per unit time reaching the surface
is E.

Amplitude of the wave is increased
to 2 A and the area of the surface is reduced to ½ S.

How much energy per unit time reaches
this smaller surface?

A 4E B 2E C E D ½ E

**Reference:**

*Past Exam Paper – June 2004 Paper 1 Q26*

__Solution 809:__**Answer: B.**

Intensity = Power / Surface Area

‘Energy per unit time’ is the power.
So, the power is E.

Let the initial intensity be I.

I = E / S giving E = IS

The intensity is proportional to the
amplitude squared. Increasing the amplitude to 2A causes the intensity to
become 4I.

The area of the surface is reduced
to ½ S.

New Power = New Intensity × New Area = 4I × ½ S = 2IS = 2E

__Question 810: [Electric field + Kinematics]__
The path of an electron with initial
speed v in uniform electric field between two parallel plates is shown.

Vertical deflection x is measured at
the right-hand edge of the plates.

Distance between the plates is
halved. The potential difference between the plates remains the same.

What will be the new deflection of
the electron with the same initial speed v?

A x B
√(2x)
C 2x D 4x

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q32*

__Solution 810:__**Answer: C.**

Electric field strength, E = V / d

where V is the p.d between the
plates and d is the separation

Electric force F = Eq

The distance d is halved while V
remains the same.

New electric field strength = V /
0.5d = 2 (V/d) = 2E

New electric force = (2E)q = 2F

In both cases, the electron enters
the field horizontally – so its initial vertical velocity is zero in both cases.

F = ma

The electric force causes a constant
acceleration vertically, it does not affect the horizontal motion – so we need
to only consider the vertical motion of the particle.

The deflection is measured at the
end of the plates. Even if the separation of the plates have been halved, the
horizontal length of the plate is still the same. In both cases, the initial speed
is the same, so the time taken (t) to travel the length of the plates is the
same in both cases.

Consider the vertical motions in
both cases.

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

^{2}
In the 1

^{st}case, [The force F produces an acceleration]
Deflection s = 0 + ½ at2 = ½ at

^{2}
Deflection x = ½ at

^{2}
In the 2

^{nd}case, [The acceleration is now 2a since the force is 2F {F = ma, so 2F = m(2a) since the mass of the electron is constant}]
Deflection s = 0 + ½ (2a)t

^{2}= at^{2}= 2(½ at^{2}) = 2x

__Question 811: [Kinematics]__
A particle moves in the manner shown
by velocity-time graph.

The displacement of particle has
been measured so that it is zero at t = 0. Point Q refers to a point in its
motion.

**Reference:**

*Past Exam Paper – June 2011 Paper 12 Q7*

__Solution 811:__**Answer: D.**

Gradient of a velocity-time graph
gives the acceleration

Gradient (acceleration) is zero at the
maxima and minima. [A and B are incorrect]

Displacement (NOT distance) is taken
to be zero at t = 0.

In a velocity-tome graph, the
displacement is given by the area under the graph. If the area is taken above
the time axis, then the displacement is positive and if it is below the time
axis, the displacement is negative.

As long as it is above the time
axis, the displacement is positive – even if the curve is increasing or
decreasing on the upper part of the time axis. So, the maximum displacement occurs
when the curve touches the time axis. This is at times t = 5s and t = 15s.

in solution 811 why is c incorrect

ReplyDeleteAt t = 2.5s, the velocity has a maximum value. But even after t = 2.5s up to t = 5s, the velocity is still positive, so the displacement keeps on increasing.

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