Physics NCERT Exemplar Solutions Class 11th Chapter Three Motion in a Straight Line

1. It is a common observation that rain clouds can be at about a kilometre 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 = 10m/s2)

(b) A typical rain drop is about 4mm diameter. Momentum is mass x 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 1m and cloth is not pierced through !!)

Explanation –(a)  velocity attained by a falling rain drop will be = $\sqrt[]{2gh}=\sqrt[]{2\times 10\times 1000}$

= $100\sqrt[]{2}m{s}^{-1}=\frac{510km}{h}$

(b) diameter of the rain drop = 2r=4mm

Radius = 2mm= 2 $\times {10}^{-3}m$

Mass of rain drop = V $\times \rho =\frac{4}{3}\pi r^3\rho =\frac{4}{3}\times \frac{22}{7}\times {\left(2\times {10}^{-3}\right)}^3\times {10}^3=3.4\times {10}^{-5}kg$

Momentum of rain drop= mv= 3.4 $\times {10}^5\times 100\sqrt[]{2}=5\times {10}^{-3}kgm/s$

(c) time ,t = d/v= $\frac{4\times {10}^{-3}}{100\sqrt[]{2}}=0.028\times {10}^{-3}s$

(d) force exerted, F = change in momentum /time=

(e) area = $\pi R^2=\pi \times {\frac{1}{2}}^2=\frac{11}{14}=0.8m^2$

number of drops striking the the umbrella with separation of 5 $\times {10}^{-2}$

so net force = $\frac{0.8}{{(5\times {10}^{-2})}^2}\times 168=53760N$

2. A motor car moving at a speed of 72km/h can not come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a higway the car is behind the truck both moving at 72km/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 0.5s.

Explanation- speed of car and truck = 72km/h = 72(5/18) =20m/s

V= u+at

0=20+a(5) so a=-4m/s²

But retarted acceleration will be v=u+at

0=20+a(3)

So a= $\frac{20}{3}\left(t-0.5\right)$ -20/3m/s²

We also need to consider human response time = 0.5 s

V=u-at (for retarded motion)

V= 20- $\frac{20}{3}\left(t-0.5\right)$ ….1

V_t=20-4t …..2

From 1 and 2

20-=20-4t

After solving we get t= 5/4s

Distance travelled by truck in time t , S=ut+1/2at²

= 20 $\times \frac{5}{4}+\frac{1}{2}\left(-4\right)\times {\left(\frac{5}{4}\right)}^2=21.875m$

To avoid the bump onto the truck car must maintain distance = 23.125-21.875=1.250m

3. 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 < 6 s in m/s. It repeats this cycle till it reaches the height of 20 m.

(a) At what time is its velocity maximum?

(b) At what time is its average velocity maximum?

(c) At what time is its acceleration maximum in magnitude?

(d) How many cycles (counting fractions) are required to reach the top?

Explanation –(a)  for maximum velocity dv/dt=0

d/dt(6t-2t²)=0

6-4t=0 t= 6/4=1.5s

(b) v=6t-2t²

ds/dt=6t-2t²

ds=6t-2t²dt

distance in 3s ,S= ${\int }_0^3\left(6t-{2t}^2\right)dt=[3t^2-\frac{2}{3}{t}^3]$ ³₀

s= 27-18=9m

average velocity = distance /time =9/3 = 3m/s

x= 6t-2t²

3=6t-2t²

After solving we get t= 9/4s approx.

(c) in periodic motion when when velocity is zero

0=6t-2t²

0=t(6-2t)

So t=0, 3 sec

(d) distance covered from 0 to 3s=9m

distance covered in 3 to 6s= ${\int }_3^6\left(18-9t+t^2\right)dt$

S= (18t- $\frac{9t^2}{2}+\frac{t^3}{3}$ )⁶

S= 108-9(18)+ $\frac{6^3}{3}-18\left(3\right)+\frac{9}{2}\left(9\right)-\frac{27}{3}$

S= -4.5m

So total distance covered = 9+(-4.5)=4.5m

No of cycles covered in that distance =20/4.5=4.44approx

4. A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Explanation- let speed of two balls be V₁and V₂

Where v₁=2v and v₂=v and y₁and y₂ be the distance covered

So y₁= $\frac{v_1^2}{2g}=\frac{4v^2}{2g}$ and y₂= $\frac{v_2^2}{2g}=\frac{v^2}{2g}$

So y₁-y₂= 15

$\frac{3v^2}{2g}=15$

V²= $\sqrt[]{5\times 2\times 10}=\frac{10m}{s}$

So clearly we can say v₁=20 and v₂=10

And y₁=20m and y₂=5m

If t₂ is the time taken by ball 2 through a distance of 5m ,y₂=v₂t-1/2gt²

5=10t₂-5t₂² so t₂ will be 15

Then time covered by ball 1 in 2 sec between two throws = t₁-t₂= 2-1=1s