2 Marks Questions

Question 12 Marks

A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18m/h while the other has the speed of 27km/h. The bird starts moving from first car towards the other and is moving with the speed of 36km/h and when the two cars were separted by 36km. What is the total distance covered by the bird? What is the total displacement of the bird?

Answer

Relative speed of cars = 27 + 18 = 45Km/hr Time to meet the two cars together (t)$\text{t}=\frac{\text{distance between cars}}{\text{relative speed of cars}}=\frac{36\text{km}}{(27+18)\text{km/hr}}$ $=\frac{36}{45}=\frac{4}{5}\Rightarrow\text{t}=\frac{4}{5}\text{hours}$
$\therefore$ Distance covered by the bird in $\frac{4}{5}\text{hours}=36\times\frac{4}{5}=28.8\text{km.}$

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Question 22 Marks

It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
Estimate the time required to flatten the drop.

Answer

Key concept: This problem can be solved by kinematic equations of motion and Newton’s second law that $\text{F}_\text{ext}=\frac{\text{dp}}{\text{dt}}$ will be used, where dp is change in momentum over time dt. Time required to flatten the drop = Time taken by the drop to travel the distance equal to the diameter of the drop near the ground.$\text{t}=\frac{\text{d}}{\text{v}}=\frac{4\times10^{-3}}{100\sqrt{2}}=0.028\times10^{-3}\text{s}$
$=2.8\times10^{-5}\text{s}=30\text{ms}$

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Question 32 Marks

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 - t); 0 < t < 3 and v(t) = -(t - 3)(6 - t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20m.
At what time is its average velocity maximum?

Answer

Given velocity$\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$
From Eq. (i) $\text{v}=6\text{t}-2\text{t}^2$$\Rightarrow\frac{\text{ds}}{\text{dt}}=6\text{t}-2\text{t}^2$
$\Rightarrow\text{ds}=(6\text{t}-2\text{t}^2)\text{dt}$
where, s is displacement$\therefore$ Distance travelled in time interval 0 to 3s,
$\text{s}=\int^3_0(6\text{t}-2\text{t}^2)\text{dt}$
$=\Big[\frac{6\text{t}^2}{2}-\frac{2\text{t}^3}{3}\Big]^3_0=\Big[3\text{t}^2-\frac{2}{3}\text{t}^3\Big]^3_0$
$=3\times9-\frac{2}{3}\times3\times3\times3$
$=27-18=9\text{m}$
$\text{Average velocity}=\frac{\text{Distance travelled}}{\text{Time}}$
$=\frac{9}{3}=3\text{m/s}$
Given, $\text{x}=6\text{t}-2\text{t}^2$$\Rightarrow3=6\text{t}-2\text{t}^2\Rightarrow2\text{t}^2-6\text{t}-3=0$
$\Rightarrow\text{t}=\frac{6\pm\sqrt{6^2-4\times2\times3}}{2\times2}=\frac{6\pm\sqrt{36-24}}{4}$
$=\frac{6\pm\sqrt{12}}{4}=\frac{3\pm2\sqrt{3}}{2}$
Considering positive sign only$\text{t}=\frac{3+2\sqrt{3}}{2}=\frac{3+2\times1.732}{2}=\frac{9}{4}\text{s}$

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Question 42 Marks

It is a common observation that rain clouds can be at about a kilometre altitude above the ground.A typical rain drop is about 4mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.

Answer

Key concept: This problem can be solved by kinematic equations of motion and Newton’s second law that $\text{F}_\text{ext}=\frac{\text{dp}}{\text{dt}}$ will be used, where dp is change in momentum over time dt. Diameter of the drop (d) = 2r = 4mm Radius of the drop (r) = $2mm = 2 \times 10^{-3}m$ Mass of a rain drop (m) = v × p$=\frac{4}{3}\pi\text{r}^3\text{p}=\frac{4}{3}\times\frac{22}{7}\times(2\times10^{-3})^3\times10^3$
($\because$ Density of water = $10^3kg/m^2) \Rightarrow\text{m}=3.4\times10^{-5}\text{kg}$
Momentum of the rain drop (p) = mv$=3.4\times10^{-5}\times100\sqrt{2}$
$\Rightarrow\text{p}=4.7\times10^{-3}\text{kg-m/s}=5\times10^{-3}\text{kg-m/s}$

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Question 52 Marks

Answer

Given velocity$\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$
For maximum velocity $\frac{\text{dv(t)}}{\text{dt}}=0$$\Rightarrow\frac{\text{d}}{\text{dt}}(6\text{t}-2\text{t}^2)=0$
$\Rightarrow6-4\text{t}=0$
$\Rightarrow\text{t}=\frac{6}{4}=\frac{3}{2}\text{s}=1.5\text{s}$

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