Question: 1 . In which of the following examples of motion can the body be considered approximately a point object (a) a railway carriage moving without jerks between two stations

(b) a monkey sitting on the top of a man cycling smoothly on a circular track,

(c) a spinning cricket ball turns sharply on hitting the ground.

(d) a thumbing beaker that has slipped off the edge of a table ?

Solution:

(a) The carriage can be consider a point object if the distance between the two stations is very large compared to the size of railway carriage.

(b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius because in that case the distance covered by the cyclist is quite large as compared to the size of monkey.

Monkey can not be considered as a point object if the cyclist describes a circular track of small radius because in that case the distance covered by the cyclist is not very large as compared to the size of the money.

(c) The cricket ball can not be considered as a point object because the size of the spinning cricket ball is quite appreciable as compared to the distance through which the ball may turn on hitting the ground.

(d) A beaker slipping off the edge of the can not be considered as a point object because the size of the beaker is not negligible as compared to the height of the table.

**Question: 2.**. A woman starts from her home at 9.00 am, walks with a speed 5km/h on straight road up to her office 2.5 km away. Stays at the office up to 5.00 pm and returns home by an auto with a speed of 25 km/h. Choose suitable scales and plot the x – t graph of her motion.

Solution. Time taken in reaching office = distance/speed = 2.5/5 = 0.5 hr.

Time taken in returning from office = 2.5/25 = 0.1 hr = 6minutes. It means the woman reaches the office at 9.30am and returns home at 5.06pm.

Question:3 .. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward. Followed again 5 steps forward and 3 steps backwards, and so on. Each step is 1 m long and requires 1s. Determine how long the drunkard takes to fall in a pit 13 m away from the stars.

Soution. In 8 steps distance covered is only 2 m , hence 1m distance covered in 4 steps . For covering a distance of 8 m in forwards direction the drunkard will have to move 32 steps in all. Now he will have to cover 5 meters more to reach the pit, for which he has to take only 5 forwards steps. Therefore, he will have to take = 32 + 5 = 37 steps to move 13 meters. Thus he will into the pit after taking 37 steps i.e. after 37 seconds from the start.

Question.4. A jet airplane travelling at the speed of 500 khm-1 ejects its products of combustion at the speed of 1500 kmh-1 relative to the jet plane. What is the speed of the latter with respect to observer on the ground?

Solution. Let, v_{P} = velocity of the product w.r.t. ground consider the direction of motion of airplane to be positive direction of x –axis. Here

Speed of jet plane, v_{A} = 500 kmh^{-1}

Relative speed of products of combustion w.r.t. jet plane is, v_{PA} = -1500 kmh^{-1}

Relative speed of products of combustion w.r.t. air plane is, v_{PA} = v_{P} – v_{A} = -1500 kmh-1 or, v_{P} = v_{A} – 1500 = 500 – 1500 = −1000kmh^{-1}

Here –ve sign shows that the direction of products of combustion is opposite to that of the airplane. Thus the magnitude of relative velocity is 1000 kmh-1

Question: 5. A car moving along a straight highway with speed 126 km/h is brought to a brought to a stop with a distance of 200m. what is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?

Solution. Here, u = 126 km/h = 126 × 1000/ (60 × 60) ms^{-1} = 35 ms^{-1} , v = 0, s = 200m

We know, v^{2} = u^{2} + 2as

=> 0 = (35)^{2} + 2 a × (200)

=> a =−49/16 = −3.06 ms-2

As v = u + at

=> 0 = 35 + (−49/16)t

t = 35 × 16/49 = 80/7 = 11.43 s.

Question:6. Two trains A and B of length 400m each are moving on two parallel tracks with a uniform speed of 72 kmh^{-1} in the same direction, with A ahead of B. The driver of B decides to overtake A and acceleration by 1 ms^{-2}. If, after 50s, the guard of B just crosses past the driver of A, what was the original distance between them (the guard of B and the driver of A)?

**Solution:** Originally, both the trains have the same velocities. So the relative velocity of B w.r.t. A is zero.

For train A, u = 72kmh^{-1} = (72 × 1000)/(60 × 60) = 20ms^{-1}, t = 50s, a = 0, S = S_{A}

We know, s = ut +(1/2)at ^{2}

S_{A} = 20 × 50 + (1/2)× 0 × (50)^{2 } = 1000m

For train B, u = 72kmh^{-1} = (72 × 1000)/(60 × 60) = 20ms^{-1}, t = 50s, a = 1 ms^{-2}, S = S_{B}

As, S = ut +(1/2) at^{2}

S_{B} = 20 ×50 +(1/2)×1× 50^{2}

= 2250m

Taking the guard of the train B is the last compartment of the train B, it comes follows that original distance between the two trains + length of the train A and B = S_{A} – S_{B} or, original distance between the two trains + 400 + 400 = 2250 – 1000 = 1250 Original distance between the two trains = 1250 – 800 = 450 m

Question: 7.. On a two lane road, car A is travelling with a speed of 36kmh^{-1}. Two cars B and C approach car A in opposite directions with a speed of 54kmh^{-1 } At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

**Solution.** Velocity of car A = 36 kmh^{-1} = 10ms^{-1}; velocity of car B or C = 54kmh^{-1} = 15 ms^{-1}

Relative velocity of B w.r.t. A = 15 – 10 = 5 ms^{-1}

Relative velocity of C w.r.t A = 15 + 10 = 25ms^{-1}

Time available to B or C for crossing A = 1000/25 = 40s.

If car B acceleration with acceleration a, to cross A before car C does then, u = 5 ms^{-1}. t = 40s. s = 1000m

Using s = ut + (1/2)at^{2} we have 1000 = 5 × 40 + (1/2)× a × (40)^{2} or, a = 1m/s^{2} .