5 Marks Questions

Question 15 Marks

The benches of a gallery in a cricket stadium are 1m wide and 1m high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at 35m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110m from the batsman. On which bench will the ball hit?

Answer

$\theta=53^{\circ},$ so $\cos53^{\circ}=\frac{3}{5}$

$\sec^2\theta=\frac{25}{9}$ and $\tan\theta=\frac{4}{3}$

Suppose the ball lands on nth bench

So, y = (n - 1)1 …(1) [ball starting point 1m above ground]

Again $\text{y = x}\tan\theta-\frac{\text{gx}^2\sec^2\theta}{2\text{u}^2}$ [x = 110 + n - 1 = 110 + y]

$\Rightarrow\text{y}=(110+\text{y})\Big(\frac{4}{3}\Big)-\frac{10(110+\text{y})^2\big(\frac{25}{9}\big)}{2\times35^2}$

$\Rightarrow\frac{440}{3}+\frac{4}{3}\text{y}-\frac{250(110+\text{y})^2}{18\times35^2}$

From the equation, y can be calculated.

$\therefore\text{y}=5$

$\Rightarrow\text{n}-1=5$

$\Rightarrow\text{n}=6.$

The ball will drop in sixth bench.

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Question 25 Marks

A person is standing on a truck moving with a constant velocity of 14.7m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8m. Find the speed and the angle of projection:

As seen from the truck.
As seen from the road.

Answer

As seen from the truck the ball moves vertically upward comes back. Time taken = time taken by truck to cover 58.8m.

$\therefore\text{time}=\frac{\text{s}}{\text{v}}=\frac{58.8}{14.7}=4\sec.$ (V = 14.7m/s of truck)

u = ?, v = 0, g = -9.8m/s² (going upward), $\text{t}=\frac{4}{2}=2\sec.$

v = u + at

⇒ 0 = u - 9.8 × 2

⇒ u = 19.6m/s. (vertical upward velocity).

From road it seems to be projectile motion.

Total time of flight = 4 sec

In this time horizontal range covered 58.8m = x

$\therefore\text{X = u}\cos\theta\text{t}$

$\Rightarrow\text{u}\cos\theta=14.7 \ ...(1)$

Taking vertical component of velocity into consideration.

$\text{y}=\frac{0^2-(19.6)^2}{2\times(-9.8)}=19.6\text{m}$ [from(a)]

$\therefore\text{y = u}\sin\theta\text{ t}-\frac{1}{2}\text{gt}^2$

$\Rightarrow19.6=\text{u}\sin\theta(2)-\frac{1}{2}(9.8)2^2$

$\Rightarrow2\text{u}\sin\theta=19.6\times2$

$\Rightarrow\text{u}\sin\theta=19.6 \ ...(2)$

$\Rightarrow\frac{\text{u}\sin\theta}{\text{u}\cos\theta}=\tan\theta$

$\Rightarrow\frac{19.6}{14.7}=1.333$

$\Rightarrow\theta=\tan^{-1}(1.333)=53^{\circ}$

Again $\text{u}\cos\theta=14.7$

$\Rightarrow\text{u}=\frac{14.7}{\text{u}\cos53^{\circ}}=24.42\text{m/s}.$

The speed of ball is 42.42m/s at an angle 53° with horizontal as seen from the road.

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Question 35 Marks

It is 260km from Patna to Ranchi by air and 320km by road. An aeroplane takes 30 minutes to go from Patna to Ranchi whereas a delux bus takes 8 hours.

Find the average speed of the plane.
Find the average speed of the bus.
Find the average velocity of the plane.
Find the average velocity of the bus.

Answer

$\text{V}_{\text{ave}}\text{ of plane}\Big(\frac{\text{Distance}}{\text{Time}}\Big)$

$=\frac{260}{0.5}=520\text{km/hr}.$

$\text{V}_{\text{ave}}\text{ of bus}=\frac{320}{8}=40\text{km/hr}.$
plane goes in straight path

$\text{velocity}=\overrightarrow{\text{V}}_{\text{ave}}=\frac{260}{0.5}=520\text{km/hr}.$

Straight path distance between plane to Ranchi is equal to the displacement of bus.

$\therefore\text{Velocity}=\overrightarrow{\text{V}}_{\text{ave}}=\frac{260}{8}=32.5\text{km/hr}.$

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Question 45 Marks

A stone is thrown vertically upward with a speed of 28m/s.

Find the maximum height reached by the stone.
Find its velocity one second before it reaches the maximum height.
Does the answer of part.
Change if the initial speed is more than 28m/s suchas 40m/s or 80m/s?

Answer

u = 28m/s, v = 0, a = -g = -9.8m/s²

$\text{S}=\frac{\text{v}^2-\text{u}^2}{2\text{a}}=\frac{0^2-28^2}{2(9.8)}=40\text{m}$
time $\text{t}=\frac{\text{v}-\text{u}}{\text{a}}=\frac{0-28}{-9.8}=2.85$

t' = 2.85 - 1 = 1.85

v' = u + at' = 28 - (9.8)(1.85) = 9.87m/s.

$\therefore$ The velocity is 9.87m/s.

No it will not change. As after one second velocity becomes zero for any initial velocity and deceleration is g = 9.8m/s² remains same. Fro initial velocity more than 28m/s max height increases.

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Question 55 Marks

An aeroplane has to go from a point A to another point B, 500km away due 30° east of north. A wind is blowing due north at a speed of 20m/s. The air-speed of the
plane is 150m/s.

Find the direction in which the pilot should head the plane to reach the point B.
Find the time taken by the plane to go from A to B.

Answer

In resultant direction $\vec{\text{R}}$ the plane reach the point B.

Velocity of wind $\vec{\text{V}}_{\text{w}}=20\text{m/s}$

Velocity of aeroplane $\vec{\text{V}}_{\text{a}}=150\text{m/s}$

In $\triangle\text{ABC}$ according to sine formula

$\therefore\frac{20}{\sin\text{A}}=\frac{150}{\sin30^{\circ}}$

$\Rightarrow\sin\text{A}=\frac{20}{150}\sin30^{\circ}=\frac{20}{150}\times\frac{1}{2}=\frac{1}{15}$

$\Rightarrow\text{A}=\sin^{-1}\Big(\frac{1}{15}\Big)$

The direction is $\sin^{-1}\Big(\frac{1}{15}\Big)$ east of the line AB.

$\sin^{-1}\Big(\frac{1}{15}\Big)=3^{\circ}48'$

$\Rightarrow30^{\circ}+3^{\circ}48'=33^{\circ}48'$

$\Rightarrow\text{R}\sqrt{150^2+20^2+2(150)20\cos33^{\circ}48'}=167\text{m/s}.$

$\text{Time}=\frac{\text{s}}{\text{v}}=\frac{500000}{167}=2994\text{sec}=49=50\text{min}.$

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Question 65 Marks

A car travelling at 60km/h overtakes another car travelling at 42km/h. Assuming each car to be 5.0m long, find the time taken during the overtake and the total road distance used for the overtake.

Answer

v₁ = 60km/hr = 16.6m/s.

v₂ = 42km/h = 11.6m/s.

Relative velocity between the cars = (16.6 - 11.6) = 5m/s.

Distance to be travelled by first car is 5 + t = 10m.

Time $=\text{t}=\frac{\text{s}}{\text{v}}=\frac{0}{5}=2\sec.$ to cross the 2^nd car.

In 2 sec the 1^st car moved = 16.6 × 2 = 33.2m

H also covered its own length 5m.

$\therefore$ Total road distance used for the overtake = 33.2 + 5 = 38m.

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Question 75 Marks

Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Answer

Velocity of man $\overrightarrow{\text{V}}_{\text{m}}=3\text{km/hr}$

BD horizontal distance for resultant velocity R.

X-component of resultant $\text{R}_{\text{x}}=5+3\cos\theta$

$\text{t}=\frac{0.5}{3\sin\theta}$

which is same for horizontal component of velocity.

$\text{H = BD}=(5+3\cos\theta)\Big(\frac{0.5}{3\sin\theta}\Big)=\frac{5+3\cos\theta}{6\sin\theta}$

For H to be min $\Big(\frac{\text{dH}}{\text{d}\theta}\Big)=0$

$\Rightarrow\frac{\text{d}}{\text{d}\theta}\Big(\frac{5+3\cos\theta}{6\sin\theta}\Big)=0$

$\Rightarrow-18(\sin^2\theta+\cos^2\theta)-30\cos\theta=0$

$\Rightarrow-30\cos\theta=18$

$\Rightarrow\cos\theta=-\frac{18}{30}=-\frac{3}{5}$

$\sin\theta=\sqrt{1-\cos^2\theta}=\frac{4}{5}$

$\therefore\text{H}=\frac{5+3\cos\theta}{6\sin\theta}$

$=\frac{5+3\big(-\frac{3}{5}\big)}{6\times\big(\frac{4}{5}\big)}=\frac{2}{3}\text{km.}$

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Question 85 Marks

A ball is thrown horizontally from a point 100m above the ground with a speed of 20m/s. Find:

The time it takes to reach the ground.
The horizontal distance it travels before reaching the ground.
The velocity (direction and magnitude) with which it strikes the ground.

Answer

It is a case of projectile fired horizontally from a height.

h = 100m, g = 9.8m/s²

Time taken to reach the ground $\text{t}=\sqrt{(2\text{h/g})}$

$=\sqrt{\frac{2\times100}{9.8}}=4.51\text{sec}.$

Horizontal range x = ut = 20 × 4.5 = 90m.
Horizontal velocity remains constant through out the motion.

At A, V_x = 20m/s

A V_y = u + at = 0 + 9.8 × 4.5 = 44.1m/s.

Resultant velocity $\text{V}_{\text{r}}=\sqrt{(44.1)^2+20^2}=48.42\text{m/s}.$

$\tan\beta=\frac{\text{V}_{\text{y}}}{\text{V}_{\text{x}}}=\frac{44.1}{20}=2.205$

$\Rightarrow\beta=\tan^{-1}(2.205)=60^{\circ}.$

The ball strikes the ground with a velocity 48.42m/s at an angle 66° with horizontal.

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Question 95 Marks

A train starts from rest and moves with a constant acceleration of 2.0m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute. Find:

The total distance moved by the train.
The maximum speed attained by the train.
The position(s) of the train at half the maximum speed.

Answer

Initial velocity u = 0

Acceleration a = 2m/s². Let final velocity be v (before applying breaks)

t = 30 sec

v = u + at ⇒ 0 + 2 × 30 = 60m/s

$\text{S}_1=\text{ut}+\frac{1}2{}\text{at}^2=900\text{m}$

when breaks are applied u' = 60m/s

v' = 0, t = 60 sec (1 min)

Declaration $\text{a}'=\frac{(\text{v}-\text{u})}{\text{t}}=\frac{0-16}{60}=-1\text{m/s}^2.$

$\text{S}_2=\frac{\text{v}'^2-\text{u}'^2}{2\text{a}'}=1800\text{m}$

Total S = S₁ + S₂ = 1800 + 900 = 2700m = 2.7km.

The maximum speed attained by train v = 60m/s
Half the maximum speed $=\frac{60}{2}=30\text{m/s}$

Distance $\text{A}=\frac{\text{v}^2-\text{u}^2}{2\text{a}}=\frac{30^2-0^2}{2\times2}=225\text{m}$ from starting point

When it accelerates the distance travelled is 900m. Then again declarates and attain 30m/s

$\therefore$ u = 60m/s, v = 30m/s, a = -1m/s²

Distance $=\frac{\text{v}^2-\text{u}^2}{2\text{a}}=\frac{30^2-60^2}{2(-1)}=1350\text{m}$

Position is 900 + 1350 = 2250 = 2.25km from starting point.

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Question 105 Marks

A ball is dropped from a height of 5m onto a sandy floor and penetrates the sand up to 10cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Answer

Consider the motion of ball form A to B.

B → just above the sand (just to penetrate)

u = 0, a = 9.8m/s², s = 5m

$\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$

$\Rightarrow5=0+\frac{1}{2}(9.8)\text{t}^2$

$\Rightarrow\text{t}^2=\frac{5}{4.9}=1.02$

$\Rightarrow\text{t}=1.01$

$\therefore$ velocity at B, v = u + at = 9.8 × 1.01 (u = 0) = 9.89m/s.

From motion of ball in sand

u₁ = 9.89m/s, v₁ = 0, a = ?, s = 10cm = 0.1m.

$\text{a}=\frac{\text{v}^2_1-\text{u}^2_1}{2\text{S}}=\frac{0-(9.89)^2}{2\times0.1}=-490\text{m/s}^2$

The retardation in sand is 490m/s².

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