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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^-3s

(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]

Diagram shows electric motor for garden pump connected to 24 V power supply by insulated two-core cable.

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 + ½at² 

Distance travelled, s = 25(20) + 0.5(1.25)(20²) = 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 + ½at²) = 25t + 0.5(1.25)t² = 25t + 0.625t²

Let the separation between the 2 cars = y

y = s_A – s_B = 40t – (25t + 0.625t²) = 15t – 0.625t²   

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(12²) = 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]

Motorist travelling at 10 m s^–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: v² = u² + 2as

Initial speed, u = 10ms^-1.

Final speed, v = 0ms^-1.

Braking distance, s = 10m. a = – 10² / 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.  

v² = u² + 2as.

Braking distance, s = - 30² / 2(-5) = 90m.