4 Marks Questions

Question 14 Marks

Two friends $A$ and $B$ are standing a distance $x$ apart in an open field and wind is blowing from $A$ to $B$.
A beats a drum and $B$ hears the sound $t_1$ time after he sees the event. $A$ and $B$ interchange their positions and the experiment is repeated. This time $B$ hears the drum $t_2$ time after he sees the event. Calculate the velocity of sound in still air $v$ and the velocity of wind $u$. Neglect the time light takes in travelling between the friends.

Answer

Velocity of sound $v,$ Velocity of air $u,$
Distance between $A$ and $B$ be $x$.
In the first case, resultant velocity of sound $\text{= v + u}$

$\Rightarrow(\text{v + u})\text{t}_1=\text{x}$
$\Rightarrow\text{v + u}=\frac{\text{x}}{\text{t}_1} \ ...(1)$
In the second case, resultant velocity of sound $=\text{v}-\text{u}$
$\therefore(\text{v}-\text{u})\text{t}_2=\text{x}$
$\Rightarrow\text{v}-\text{u}=\frac{\text{x}}{\text{t}_2} \ ...(2)$
From $(1)$ and $(2)$
$2\text{v}=\frac{\text{x}}{\text{t}_1}+\frac{\text{x}}{\text{t}_2}=\text{x}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$
$\Rightarrow\text{v}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$
From $(1)$
$\text{u}=\frac{\text{x}}{\text{t}_1}-\text{v}=\frac{\text{x}}{\text{t}_1}-\Big(\frac{\text{x}}{2\text{t}_1}+\frac{\text{x}}{2\text{t}_2}\Big)=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$
$\therefore$ Velocity of air $\text{ V}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$
And Velocity of wind $\text{u}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$

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Question 24 Marks

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Answer

For elevator and coin u = 0 As the elevator descends downward with acceleration a' (say) The coin has to move more distance than 1.8m to strike the floor. Time taken t = 1 sec.$\text{S}_{\text{c}}=\text{ut}+\frac{1}{2}\text{a}'\text{t}^2=0+\frac{1}{2}\text{g}(1)^2=\frac{1}{2}\text{g}$
$\text{S}_{\text{e}}=\text{ut}+\frac{1}{2}\text{at}^2=\text{u}+\frac{1}{2}\text{a}(1)^2=\frac{1}{2}\text{a}$
Total distance covered by coin is given by $=1.8+\frac{1}{2}\text{a}=\frac{1}{2}\text{g}$$\Rightarrow1.8+\frac{\text{a}}{2}=\frac{9.8}{2}=4.9$
$\Rightarrow\text{a}=6.2\text{m/s}^2=6.2\times3.28=20.34\text{ft/s}^2.$

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Question 34 Marks

Complete the following table:

Car Model

Driver $X$
Reaction time $0.20s$

Driver $Y$
Reaction time $0.30s$

$A ($deceleration on hard braking $= 6.0m/s^2)$

Speed $= 54\ km/h$
Braking distance

$a = ...$
Total stopping distance
$b = ...$

Speed $= 72\ km/h$
Braking distance

$c = ...$
Total stopping distance
$d = ...$

$B ($deceleration on hard braking $= 7.5m/s^2)$

Speed $= 54\ km/h$
Breaking distance

$e = ...$
Total stopping distance
$f = ...$

Speed $72\ km/h$
Braking distance

$g = ...$
Total stopping distance
$h = ...$

Answer

Car Model

Driver $X$
Reaction time $0.25s$

Driver $Y$
Reaction time $0.35s$

$A ($deceleration on hard braking $= 6.0m/s^2)$

Speed $= 54\ km/h$
Braking distance

$a = 19m$
Total stopping distance
$b = 22m$

Speed $= 72\ km/h$
Braking distance

$c = 33m$
Total stopping distance
$d = 39m$

$B ($deceleration on hard braking $= 7.5m/s^2)$

Speed $= 54\ km/h$
Breaking distance

$e = 15m$
Total stopping distance
$f = 18m$

Speed $72\ km/h$
Braking distance

$g = 27m$
Total stopping distance
$h = 33m$

$\text{a}=\frac{0^2-15^2}{2(-6)}=19\text{m}$
So$, b = 0.2 \times 15 + 19 = 33m$
Similarly other can be calculated.
Braking distance : Distance travelled when brakes are applied.
Total stopping distance $=$ Braking distance $+$ distance travelled in reaction time.

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Question 44 Marks

Complete the following table:

Car Model

Driver $X$
Reaction time $0.20s$

Driver $Y$
Reaction time $0.30s$

$A\ ($deceleration on hard braking $= 6.0\ m/s^2)$

Speed $= 54\ km/h$
Braking distance

$a = ...$
Total stopping distance
$b = ...$

Speed $= 72\ km/h$
Braking distance

$c = ...$
Total stopping distance
$d = ...$

$B\ ($deceleration on hard braking $= 7.5\ m/s^2)$

Speed $= 54\ km/h$
Breaking distance

$e = ...$
Total stopping distance
$f = ...$

Speed $72\ km/h$
Braking distance

$g = ...$
Total stopping distance
$h = ...$

Answer

Car Model

Driver $X$
Reaction time $0.25s$

Driver $Y$
Reaction time $0.35s$

$A\ ($deceleration on hard braking $= 6.0\ m/s^2)$

Speed $= 54\ km/h$
Braking distance

a = 19m
Total stopping distance
$b = 22m$

Speed $= 72\ km/h$
Braking distance

$c = 33m$
Total stopping distance
$d = 39m$

$B\ ($deceleration on hard braking $= 7.5\ m/s^2)$

Speed $= 54\ km/h$
Breaking distance

$e = 15m$
Total stopping distance
$f = 18m$

Speed $72\ km/h$
Braking distance

$g = 27m$
Total stopping distance
$h = 33m$

$\text{a}=\frac{0^2-15^2}{2(-6)}=19\text{m}$
So$, b = 0.2 \times 15 + 19 = 33m$
Similarly other can be calculated.
Braking distance : Distance travelled when brakes are applied.
Total stopping distance $=$ Braking distance $+$ distance travelled in reaction time.

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Question 54 Marks

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of $72\ km/h.$ The jeep follows it at a speed of $90\ km/h,$ crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Answer

$V_P = 90\ km/h = 25m/s. V_C= 72\ km/h = 20m/s$. In $10\ sec$ culprit reaches at point $B$ from $A.$
Distance converted by culprit $S = vt = 20 \times 10 = 200m.$
At time $t = 10 sec$ the police jeep is $200m$ behind the culprit.Time$=\frac{\text{s}}{\text{v}}=\frac{200}{5}=40\text{s}. ($Relative velocity is considered$).$
In $40s$ the police jeep will move from $A$ to $a$ distance $S,$
where $S = vt = 25 \times 40 = 1000m = 1.0\ km$ away.
$\therefore$ The jeep will catch up with the bike, $1\ km$ far from the turning.

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Question 64 Marks

Two friends $A$ and $B$ are standing a distance $x$ apart in an open field and wind is blowing from $A$ to $B$. $A$ beats a drum and $B$ hears the sound $t_1$ time after he sees the event. $A$ and $B$ interchange their positions and the experiment is repeated. This time $B$ hears the drum $t_2$ time after he sees the event. Calculate the velocity of sound in still air $v$ and the velocity of wind $u$. Neglect the time light takes in travelling between the friends.

Answer

Velocity of sound $v,$ Velocity of air $u,$
Distance between $A$ and $B$ be $x$.
In the first case, resultant velocity of sound $\text{= v + u}$

$\Rightarrow(\text{v + u})\text{t}_1=\text{x}$
$\Rightarrow\text{v + u}=\frac{\text{x}}{\text{t}_1} \ ...(1)$
In the second case, resultant velocity of sound $=\text{v}-\text{u}$
$\therefore(\text{v}-\text{u})\text{t}_2=\text{x}$
$\Rightarrow\text{v}-\text{u}=\frac{\text{x}}{\text{t}_2} \ ...(2)$
From $(1)$ and $(2)$
$2\text{v}=\frac{\text{x}}{\text{t}_1}+\frac{\text{x}}{\text{t}_2}=\text{x}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$
$\Rightarrow\text{v}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$
From $(1)$
$\text{u}=\frac{\text{x}}{\text{t}_1}-\text{v}=\frac{\text{x}}{\text{t}_1}-\Big(\frac{\text{x}}{2\text{t}_1}+\frac{\text{x}}{2\text{t}_2}\Big)=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$
$\therefore$ Velocity of air $\text{ V}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$
And Velocity of wind $\text{u}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$

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Question 74 Marks

A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Answer

Same displacement.

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Question 84 Marks

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Answer

For elevator and coin u = 0 As the elevator descends downward with acceleration a' (say) The coin has to move more distance than 1.8m to strike the floor. Time taken t = 1 sec.$\text{S}_{\text{c}}=\text{ut}+\frac{1}{2}\text{a}'\text{t}^2=0+\frac{1}{2}\text{g}(1)^2=\frac{1}{2}\text{g}$
$\text{S}_{\text{e}}=\text{ut}+\frac{1}{2}\text{at}^2=\text{u}+\frac{1}{2}\text{a}(1)^2=\frac{1}{2}\text{a}$
Total distance covered by coin is given by $=1.8+\frac{1}{2}\text{a}=\frac{1}{2}\text{g}$$\Rightarrow1.8+\frac{\text{a}}{2}=\frac{9.8}{2}=4.9$
$\Rightarrow\text{a}=6.2\text{m/s}^2=6.2\times3.28=20.34\text{ft/s}^2.$

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Question 94 Marks

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of $72\ km/h$. The jeep follows it at a speed of $90\ km/h,$ crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Answer

$V_P = 90\ km/h = 25m/s$.
$V_C= 72\ km/h = 20m/s$.
In $10 \sec$ culprit reaches at point $B$ from $A$.
Distance converted by culprit $S = vt = 20 \times 10 = 200m$.
At time $t = 10 \sec$ the police jeep is $200m$ behind the culprit.
Time $=\frac{\text{s}}{\text{v}}=\frac{200}{5}=40\text{s}.\ ($Relative velocity is considered$)$.
In $40s$ the police jeep will move from $A$ to a distance $S,$ where $S = vt = 25 \times 40 = 1000m = 1.0\ km$ away.
$\therefore$ The jeep will catch up with the bike$, 1\ km$ far from the turning.

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Question 104 Marks

A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Answer

Same displacement.

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