A man runs across the roof, top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is at a lower height than the first. If his speed is 9 m/s, the (horizontal) distance between the two buildings is 10 m and the height difference is 9 m, will be able to land on the next building? (Take $g = 10 \text{ m/s}^2$)
Given, horizontal speed of the man ($u_x$) = 9 m/s
Horizontal distance between the two buildings = 10 m
Height difference between the two buildings = 9 m
and $g = 10 \text{ m/s}^2$
Let the man jumps from point $A$ and land on the roof of the next building at point $B$.
Taking motion in vertical direction,
$y = ut + \frac{1}{2} at^2$
$9 = 0 \times t + \frac{1}{2} \times 10 \times t^2$
$9 = 5t^2$
or $t = \sqrt{ \frac{9}{5}} = \frac{3}{\sqrt{5}}$
Therefore, horizontal distance travelled = $u_x \times t = 9 \times \frac{3}{\sqrt{5}} = \frac{27}{\sqrt{5}} \approx 12$ m
Horizontal distance travelled by the man is greater than 10 m, therefore, he will land on the next building.
A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed of 40 m/s. Calculate the relative speed of the balls as a function of time.
For the ball dropped from the building, $u_1 = 0, u_2 = 40 \text{ m/s}$
Velocity of the dropped ball after time $t$,
$v_1 = u_1 + gt$
$v_1 = gt$ (downward)
For the ball thrown up, $u_2 = 40 \text{ m/s}$
Velocity of the ball after time $t$:
$v_2 = u_2 - gt$
$v_2 = (40 - gt) $ (upward)
∴ Relative velocity of one ball with respect to another ball:
$v_1 - v_2$
$= gt - [40 - gt]$
$= 40 \text{ m/s}$
Note: When we are applying equations for rectilinear motion we should carefully put up the signs for the physical quantities.
The velocity-displacement graph of a particle is shown in the figure.
(a) Write the relation between $v$ and $x$.
(b) Obtain the relation between acceleration and displacement and plot it.
Given, initial velocity $= v_0$
Let the distance traveled in time $t = x_0$.
For the graph:
$ \tan \theta = \frac{v_0}{x_0} = \frac{v_0 - v}{x}$ ...($i$)
where, $v$ is velocity, and $x$ is the displacement at any instant of time $t$.
From Eq. ($i$):
$v_0 - v = \frac{v_0}{x_0} x$
⇒ $v = -\frac{v_0}{x_0} x + v_0$
We know that
$$ \begin{aligned} \text { Acceleration } a & =\frac{d v}{d t}=\frac{-v_0}{x_0} \frac{d x}{d t}+0 \\\\ \Rightarrow a & =\frac{-v_0}{x_0}(v) \\\\ & =\frac{-v_0}{x_0}\left(\frac{-v_0}{x_0} x+v_0\right) \\\\ & =\frac{v_0{ }^2}{x_0{ }^2} x-\frac{v_0{ }^2}{x_0} \end{aligned} $$
Graph of a versus $x$ is given above.
It is a common observation that rain clouds can be at about a kilometer altitude above the ground.
(a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h $(g = 10 m/s^2)$.
(b) A typical rain drop is about 4 mm in diameter. Momentum is mass $\times$ speed in magnitude. Estimate its momentum when it hits ground.
(c) Estimate the time required to flatten the drop.
(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
(e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.
(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through.)
In this problem equation of motion and Newton’s seconds law that is $F_{ext} = \frac{dp}{dt}$ will be used, where dp is change in momentum over time dt.
Given, height $(h) = 1 km = 1000 m$
$g = 10 m/s^2$
(a) Velocity attained by the rain drop in freely falling through a height $h$.
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1000} = 100 \sqrt{2}$ m/s
$= 100 \sqrt{2} \times \frac{60 \times 60}{1000}$ km/h
$= 360 \sqrt{2}$ km/h $\approx 510$ km/h
(b) Diameter of the drop $(d) = 2r = 4 mm$
$\therefore$ Radius of the drop $(r) = 2$ mm $= 2 \times 10^{-3}$ m
Mass of a rain $\operatorname{drop}(m)=V \times \rho$
$$ \begin{aligned} & =\frac{4}{3} \pi r^3 \rho \\ & =\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3 \quad\left(\because \text { Density of water }=10^3 \mathrm{~kg} / \mathrm{m}^3\right) \\ & \approx 3.4 \times 10^{-5} \mathrm{~kg} \end{aligned} $$
Momentum of the rain $\operatorname{drop}(p)=m v$
$$ \begin{aligned} & =3.4 \times 10^{-5} \times 100 \sqrt{2} \\ & =4.7 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\ & \approx 5 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \end{aligned} $$
(c) 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
$$ \begin{aligned} t & =\frac{d}{V}=\frac{4 \times 10^{-3}}{100 \sqrt{2}}=0.028 \times 10^{-3} \mathrm{~s} \\ & =2.8 \times 10^{-5} \mathrm{~s} \approx 30 \mathrm{~ms} \end{aligned} $$
(d) Force exerted by a rain drop
$$ \begin{aligned} F & =\frac{\text { Change in momentum }}{\text { Time }}=\frac{p-0}{t} \\ & =\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}} \approx 168 \mathrm{~N} \end{aligned} $$
(e) Radius of the umbrella $(R)=\frac{1}{2} \mathrm{~m}$
$\therefore$ Area of the umbrella $(A)=\pi R^2=\frac{22}{7} \times\left(\frac{1}{2}\right)^2=\frac{22}{28}=\frac{11}{14} \approx 0.8 \mathrm{~m}^2$
Number of drops striking the umbrella simultaneously with average separation of 5 cm $=5 \times 10^{-2} \mathrm{~m}$.
$$ =\frac{0.8}{\left(5 \times 10^{-2}\right)^2}=320 $$
$\therefore$ Net force exerted on umbrella $=320 \times 168=53760 \mathrm{~N} \approx 54000 \mathrm{~N}$
Note In practice, the velocity of the drops decreases due to air friction.
A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway, the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5 s.
In this problem equations related to one dimensional motion will be applied for acceleration positive sign will be used and for retardation negative sign will be used.
Given, speed of car as well as truck = 72 km/h
$= 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$
Retarded motion for truck
$v = u + a_t t$
$0 = 20 + a_t \times 5$
or $a_t = -4 \text{ m/s}^2$
Retarded motion for the car
$$ \begin{aligned} v & =u+a_c t \\\\ 0 & =20+a_c \times 3 \\\\ \text{or}\,\, a_c & =-\frac{20}{3} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $$
Let car be at a distance $x$ from truck, when truck gives the signal and $t$ be the time taken to cover this distance.
As human response time is 0.5 s , therefore, time of retarded motion of car is $(t-0.5) \mathrm{s}$. Velocity of car after time $t$,
$$ \begin{aligned} v_c & =u-a t \\ & =20-\left(\frac{20}{3}\right)(t-0.5) \end{aligned} $$
Velocity of truck after time $t$,
$$ v_t=20-4 t $$
To avoid the car bump onto the truck,
$$ \begin{aligned} 20-\frac{20}{3} & (t-0.5)=20-4 t \\ 4 t & =\frac{20}{3}(t-0.5) \\ t & =\frac{5}{3}(t-0.5) \\ 3 t & =5 t-2.5 \\ \Rightarrow \quad t & =\frac{2.5}{2}=\frac{5}{4} \mathrm{~s} \end{aligned} $$
Distance travelled by the truck in time $t$,
$$ \begin{aligned} s_t & =u_t t+\frac{1}{2} a_t t^2 \\ & =20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2 \\ & =21.875 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} & \text { Distance travelled by car in time } t=\text { Distance travelled by car in } 0.5 \mathrm{~s} \text { (without retardation) } \\ & + \text { Distance travelled by car in }(t-0.5) \mathrm{s} \text { (with retardation) } \\\\ & s_c=(20 \times 0.5)+20\left(\frac{5}{4}-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(\frac{5}{4}-0.5\right)^2 \\\\ & =23.125 \mathrm{~m} \\\\ & s_c-s_t=23.125-21.875=1.250 \mathrm{~m} \\\\ & \end{aligned} $$
Therefore, to avoid the bump onto the truck, the car must maintain a distance from the truck more than 1.250 m .