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

**Question: 3.**. 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 = 6 minutes.

It means the woman reaches the office at 9.30 am and returns home at 5.06 pm.

Question:4. 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 1 m 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 fall into the pit after taking 37 steps i.e. after 37 seconds from the start.

Question.5. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h 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 km h^{-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 = −1000 kmh^{-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: 6. A car moving along a straight highway with speed 126 km/h is brought to a brought to a stop with a distance of 200 m. 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 = 200 m

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:7. Two trains A and B of length 400 m 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 50 s, 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 = 72 kmh^{-1} = 72 ×(5/18) = 20 ms^{-1}, t = 50 s, a = 0, S = S_{A}

We know, $\large S = u t + \frac{1}{2} a t^2 $

$\large S_A = 20 \times 50 + \frac{1}{2} \times 0 \times (50)^2 $

S_{A} = 1000 m

For train B, u = 72 km h^{-1} = 72 × (5/18) = 20 ms^{-1} , t = 50 s , a = 1 ms^{-2}, S = S_{B}

As, $\large S = u t + \frac{1}{2} a t^2 $

$\large S_B = 20 \times 50 + \frac{1}{2} \times 1 \times (50)^2 $

S_{B} = 2250 m

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: 8. On a two lane road, car A is travelling with a speed of 36 kmh^{-1}. Two cars B and C approach car A in opposite directions with a speed of 54 kmh^{-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 km h^{-1} = 10 ms^{-1};

velocity of car B or C = 54 km h^{-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 = 25 ms^{-1}

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

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

Using , $\large S = u t + \frac{1}{2} a t^2 $

we have, $\large 1000 = 5 \times 40 + \frac{1}{2} a (40)^2 $

or, a = 1 m/s^{2} .

Question.9. Two town A and B are connected by a regular bus service bus service with a bus leaving in either in either direction every T min. A man cycling with a speed of 20 kmh^{-1} in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

**Solution.** Let, v Kmh^{-1} = constant speed with which the buses ply between the towns A and B.

Relative velocity of the bus (for motion A to B) w.r.t. the cyclist (i.e. in the direction in which the cyclist is going = (v – 20) kmh^{-1}

Relative velocity of the bus (for motion B to A) w.r.t. the cyclist = (v + 20) kmh^{-1}

The distance travelled by the bus in time T (minutes) = vT

As per question , $\large \frac{v T}{v-20} = 18 $

or, vT = 18v – 18 × 20 …. (i)

and, $\large \frac{v T}{v+20} = 6 $

or, vT = 6v + 20 × 6 …. (ii)

Solving (i) and (ii), we get, v = 40 kmh^{-1}

Putting this value of v in (i) we get,

40 T = 18 × 40 – 18 × 20

40 T = 18 × 20

or, T = 9 mins.

Question:10.. A player throws a ball upwards a ball upwards with an initial speed of 29.4 ms^{-1}.

(a) What is the direction of acceleration during the upward motion of the ball ?

(b) What are the velocity and acceleration of the ball at the highest point of its motion.

(c) Choose the x = 0, t = 0 be the location and time at its highest point, vertically downwards direction to be the positive direction x axis and given the signs of position, velocity and acceleration of the ball during its upwards, and downwards motion.

(d) To what height does the ball rise and after how long does the ball return to the player’s hands. (g = 9.8 ms^{-2 } and air resistance is negligible).

**Solution.**

(a) Since the ball is moving under the effect of gravity, the direction of acceleration due to gravity to the acceleration vertically downward.

(b) At the highest point, the vertical velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity = 9.8 ms^{-2} in vertically downwards direction.

(c) When the highest point is chosen at the location x = 0 and t = 0 and vertically downward direction to be the positive direction of x axis and upward direction as negative direction of x- axis.

During upward motion, sign is negative, sign of velocity is negative and sign of acceleration is positive. During downward motion, sign is positive, sign of velocity is negative and sign of acceleration is positive.

(d)Let, t = time taken by the ball to reach the highest point where height from the ground is S.

Taking the vertical motion of the ball we have,

u = -29.4 ms^{-1}, a = 9.8 ms^{-2}, v = 0.

As v = u + at, we get

0 = 29.4 – gt

t =29.4/9.8 = 3s

It means time of ascent = 3s.

When an object moves under the effect of gravity alone the time of ascent is always equal to the time of decent. Therefore, total time after which the ball returns to the player’s hand = 3 + 3 = 6s.