# Physics 9702 Doubts | Help Page 11

__Question 60: [Dynamics > Momentum]__
An air rifle pellet of mass 0.001kg travelling at
a speed of 80ms

^{-1}is stopped by a wooden block. The pellet goes 25mm into the block. Calculate**(a)**the time of impact,

**(b)**the loss of momentum of the bullet,

**(c)**the average force of the block on the bullet.

**Reference:**

*“New Understanding Physics for Advanced Level” by Jim Breithaupt, pg 40 Qu 2.12 – Force and Energy – Theme A: Mechanics*

__Solution 60:__**(a)**

Pellet hits the block at a speed, v = 80ms

^{-1}
But speed of the pellet
decreases as it goes in the block until it becomes zero at a distance of 25mm
in the block.

Average speed = (v + u) /2 = (80 + 0) / 2 = 40ms

^{-1}
Distance travelled in block = 25mm = 0.025m

[Average speed = Total
distance / Total time]

Time of impact, t (= total distance / average
speed) = 0.025 / 40 =

**0.625x10**^{-3}s**(b)**

Loss of momentum of bullet, Δp = mΔv = 0.001 (80 –
0) =

**0.08kgms**^{-1}**(c)**

Force is the rate of change of momentum. Since the momentum of the bullet changes, it exerts a force
on the block. From Newton’s third law, the block also exerts a force on the
ball, equal in magnitude but in opposite direction.

Average force of block on bullet = Δp / t = 0.08
/ (0.625x10

^{-3}) =**128N**

__Question 61: [Current of Electricity > Potential difference]__Motor does not work so, to find the fault, negative terminal of voltmeter is connected to negative terminal of power supply and its other end is connected in turn to terminals X and Y at the motor.

Which row represents two readings and a correct
conclusion?

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q36*

__Solution 61:__**Answer: D.**

From the diagram, X is connected to
the positive terminal of the power supply and Y is connected to the negative
terminal of the power supply. So, if the potential at X is 24 V (as present in all the given choices), then there is
no break in the positive wire. [A is eliminated]

If there was a break in the
connection within the pump motor (it is considered that
there is no break in the wires of the cable. So, the p.d. across the motor
would be 24V), X would be at 24V and Y would be at 0V. [C is
eliminated]

Now, consider a break in the
negative wire of the cable (as concluded from the 2
remaining choices). The circuit is not complete and so, there is

__no current__passing in the circuit. Therefore, there is__no drop__in potential across the pump motor (that is, the potential at Y is the same as the potential at X.). [B is eliminated]. So, a break in the negative wire of the cable would give a potential of 24V at Y. [D is correct]

__Question 62: [Kinematics > Linear motion > Graph]__
The graph shows the speeds of two cars A and B
which are travelling at the same direction over a period of time of 40s. Car A,
travelling at a constant speed of 40ms

^{-1}, overtakes car B at time t = 0. In order to catch up with car A, car B immediately accelerates uniformly for 20s to reach a constant speed of 50ms^{-1}.**(a)**How far does car A travel during the first 20s?

**(b)**Calculate the acceleration of car B in the first 20s.

**(c)**How far does car B in this time?

**(d)**What additional time will it take for car B to catch up with car A?

**(e)**How far will each car have then travelled since t = 0?

**(f)**What is the maximum distance between the cars before car B catches up with car A?

**Reference:**

*Past Exam Paper – November 1993 / II / 2*

__Solution 62:__
[These questions can either be solved through the relevant
formulae or using the graph. Details involving using the graph is given between
curly brackets for each part.]

**(a)**

Speed of car A = 40ms

^{-1}
Distance travelled during first 20s (= Speed x
Time) = 40 x 20 = 800m

{Area under graph gives
the distance travelled}

**(b)**

Acceleration, a = (v – u) / t = (50 – 25) / 20 =
1.25ms

^{-2}
{Gradient of graph gives
the acceleration}

**(c)**

Distance travelled, s = ut + ½at

^{2}
Distance travelled, s = 25(20) + 0.5(1.25)(20

^{2}) = 750m
{Area under graph [area
of parallelogram = 0.5 (sum of parallel sides) (height)] gives distance
travelled}

**(d)**

Let time at which car B catches up with car A = t

Distance travelled by car A at time t = 40t

Distance travelled by car B at time t = 750 +
50(t – 20)

When car B catches up
with car A, their distance travelled (position) are the same.

40t = 750 + 50(t – 20)

10t = 250

Time t = 25s

So, additional time = 25 – 20 = 5s

**(e)**

Distance travelled by car A (it’s the same for
each of them) = 40 x 25 = 1000m

{Distance travelled =
area under graph between t = 0 and t = 25s.}

**(f)**

Distance travelled by car A, s

_{A}= 40t
Distance travelled by car B, s

_{B}(= ut + ½at^{2}) = 25t + 0.5(1.25)t^{2}= 25t + 0.625t^{2}
Let the separation between the 2 cars = y

y = s

_{A}– s_{B}= 40t – (25t + 0.625t^{2}) = 15t – 0.625t^{2}
For the separation between the 2 cars to be
maximum, dy / dt = 0

dy / dt = 15 – 2(0.625)t = 0

Time at which separation, y is maximum, t = 12s

Maximum separation, y

_{max}= 15(12) – 0.625(12^{2}) = 90m
{Before car B catches up
with car A (before t = 25s), when car A is travelling faster than car B, it is
actually increasing the distance between the cars. From the graph, the distance
is represented by the area.

So, maximum separation
can be identified by the difference in area under graph for car A and car B.
This area is the region shaded as shown in the diagram. The time at which
maximum separation occurs is at the cross-over point. On a full-scale diagram,
this can be calculate by proportional and found to be 12s.

The maximum separation is
the difference in area under the graph between time t = 0 and t = 12s. Area of
triangle = ½ (40 – 25) (12) = 90m}

__Question 63: [Kinematics > Linear motion]__^{–1}can bring his car to rest in braking distance of 10 m. In what distance could he bring car to rest from speed of 30 m s

^{–1}using same braking force?

A 17m B
30m C 52m D 90m

**Reference:**

*Past Exam Paper – June 2006 Paper 1 Q17*

__Solution 63:__**Answer: D.**

The braking force produces an
acceleration a.

Equation for uniformly accelerated
motion: v

^{2}= u^{2}+ 2as
Initial speed, u = 10ms

^{-1}.
Final speed, v = 0ms

^{-1}.
Braking distance, s = 10m. a = – 10

^{2}/ 2(10) = – 5ms^{-2}.
Since the same braking force is
used, the same acceleration is produced.

For second case, v = 0ms

^{-1}, u = 30ms^{-1}, a = – 5ms^{-2}.
v

^{2}= u^{2}+ 2as.
Braking distance, s = - 30

^{2}/ 2(-5) = 90m.

ReplyDeletewhy cant we use the formula v=d/t? since distance and speed is given and with that find the accleration using v=u+at? would that be wrong?

For which question?

DeleteAnyway, if the speed is changing (since there is an acceleration), we cannot use that formula.

Deletesorry the question was 63

so we use this formula(v=u+at) when the acceleration is constant and that means that the change in velocity would also be a constant. Am I right?