Motion
Multiple Choice Questions
1. A particle is moving in a circular path of radius $r$. The displacement after half a circle would be:
(a) Zero
(b) $\pi r$
(c) $2 {r}$
(d) $2 \pi r$
Answer Answer is (c) $2 r$ Explanation: After half revolution $\text{ Distance travelled }=\dfrac{1}{2} X$ circumference $=\pi r$ Path length Displacement $=$ Final position- Initial Position It comes out to be the diameter of the circle $=2 R$.Show Answer
(a) ${u} / {g}$
(b) ${u}^{2} / 2 {g}$
(c) ${u}^{2} / {g}$
(d) ${u} / 2 {g}$
Answer Answer is (b) $u^{2} / 2 g$ Explanation: $V^{2}=u^{2}+2$ as here $v=0$ $a=-g$ $s=H$ $0=u^{2}-2 gH$ $H=u^{2} / 2 g$Show Answer
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1
Answer Answer is (d) equal or less than 1 Explanation: Shortest distance between initial and end point is called displacement. Distance is the total path length. Displacement is vector and it may be positive or negative whereas Distance is scalar and it can never be negative. Distance can be equal or greater than displacement which means ratio of displacement to distance is always equal to or less than 1 .Show Answer
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration
Answer Answer is (b) uniform acceleration Explanation: Velocity is measured in distance /second and acceleration is measured in Distance/ second ${ }^{2}$. Hence Uniform acceleration is the right answer.Show Answer
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration
Answer Answer is (a) in uniform motion Explanation: From the above given graph it is clear that velocity of the object remain constant throughout hence the object is in uniform motion.Show Answer
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity
Answer Answer is (c) in accelerated motion Explanation: Boy is moving in a circular motion and circular motion is an accelerated motion hence C) is right answer.Show Answer
(a) ${m}^{2}$
(b) ${m}$
(c) ${m}^{3}$
(d) $m s^{-1}$
Answer Answer is (b) $m$ Explanation: Area given in the graph represents Displacement and its unit is meter. Hence the answer is (b) $m$Show Answer
(a) Car $A$ is faster than car $D$.
(b) Car $B$ is the slowest.
(c) Car D is faster than car C.
(d) Car $C$ is the slowest.
Answer Answer is (b) Car B is the slowest. Explanation: Graph shows that Car B covers less distance in a given time than A, C and D cars hence it is the the slowest.Show Answer
a
b
c
d
Answer Answer is (a) Explanation: Distance in graph (a) is uniformly increasing with time hence it represents uniform motion.Show Answer
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed
Answer Answer is (c) the accelerationShow Answer
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun
Show Answer
Answer
Answer is (a) If the car is moving on straight road
Explanation:
In other cases given here displace can be less than distance hence option (a) If the car is moving on straight road is the right answer.
Short Answer Questions
12. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.
Answer Displace zero does not mean zero distance. Distance can be zero when moving object back to the place it started. Displacement is either equal or less than distance but distance is always greater than one and it cannot be a negative value.Show Answer
Answer If object moving in a uniform velocity then $v=\mu$ and $a=0$. In this scenario equation for distance is given below. $S=ut$ and $V^{2}-\mu^{2}=0$Show Answer
AnswerShow Answer
Answer Car Starts from rest hence Initial velocity $u=0$ acceleration $a=5 ms^{2}$ and time $t=8 s$ $v=u+at$ $v=0+5 x 8$ $v=40 ms^{-1}$ From second equation $s=ut+\dfrac{1}{2} a t^{2}$ $s=0 x 8+\dfrac{1}{2} x 5 x(8)^{2}$ $s=+\dfrac{1}{2} x 5 x(8)^{2}$ $s=\dfrac{1}{2} x 5 \times 64$ $s=5 \times 32=160$Show Answer
Answer Let the distance from $A$ to $B$ is $D kms$. Distance for the entire journey is $2 D kms$. Time taken to go from A to B is D/30 hr and that of B to A is D/20 hr. So, total time taken T $T=(D / 30)+(D / 20)$. By solving, we will get, $T=D / 12$ hrs. Average speed $=$ Total distance/Total time. Av.speed $=2 D \div D / 12$ $=>2 D \times 12 / D=24 km / h$. Hence Average speed of the motocycle is $24 km / h$.Show Answer
Fig. 8.5
Answer Here Velocity is constant hence $v=20 ms-1$ (iii) $s=vxt$ $=20 \times 15$ $=300 m$Show Answer
AnswerShow Answer
Answer When two objects fall with same acceleration simultaneously. after 2 seconds the difference in their heights will not change and it remain $50 m$. $d_1=h_1-s_1$ $d_1=150-\dfrac{1}{2} a^{2}=150-(\dfrac{1}{2} \times 10 \times 4)$ $d_1=150-20=130 m$ Therefore the height of first object after 2 seconds is $130 m$. In the same way the height of second object is $d_2=h_2-s_2$ $d_2=100-\dfrac{1}{2} at^{2}=100-(\dfrac{1}{2} \times 10 \times 4)$ $d_1=100-20=80 m$ Therefore, the height of second object after 2 seconds is $80 m$. So, the difference is same, i.e. $50 m$. This concludes that the difference in height of the two objects does not depend on time and will always be same.Show Answer
Answer Here Object starts from rest hence initial velocity $u=0 t=2 s$ and $s=20 m$ According to Second equation of motion $s=ut+\dfrac{1}{2} at^{2}$ $S=0+\dfrac{1}{2} a \times 2^{2}$ $20=2+\dfrac{1}{2} a \times 2^{2}=2 a$ $=20 / 2$ $a=10 m / s$ According to first equation of motion velocity after $7 s$ from the start $V=u+at$ $V=0+10 x 7$ $V=70 m / s$Show Answer
| Time(s) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|---|
| Displacement(m) | 0 | 2 | 4 | 4 | 4 | 6 | 4 | 2 | 0 |
Use this graph to find average velocity for first $4 s$, for next $4 s$ and for last ${6} s$.
Answer Average velocity for the first $4 s=\dfrac{\text{ Change in displacement }}{\text{ Total time taken }}$ Average velocity of next $4 s=V=\dfrac{4-4}{8-4}=\dfrac{0}{4}$ Average velocity for last $6 s=\dfrac{(0-6) m}{(16-10) s}=\dfrac{(-6)}{6}=1 ms^{-1}$Show Answer
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?
Answer Given initial velocity, $u=5 \times 104 m s-1$ and acceleration, $a=104 m s-2$ (i) final velocity $=v=2 u=2 \times 5 \times 104 m s-1=10 \times 104 m s-1$ To find $t, \quad use v=$ at $\quad$ or $t=u-u / a$ (ii) Using $s=$ ut $+\dfrac{1}{2}$ at $2=(5 \times 10^{4}) \times 5+1 / 2(10) \times(5) 2$ $
\begin{aligned}
& =\dfrac{25 \times 10^{4}+25}{2 \times 10^{4}} \\
& =37.5 \times 10^{4} m
\end{aligned}
$Show Answer
Answer ${a}={d v} / {d t}$ Assume that air resistanceis nil. We can directly contain it by using Newton’s equations of motion or from the below mentioned method: Thus area under the v-t curve and the x-axis where the slope of the curve is the instantaneous acceleration. In this case acceleration $g$ is constant and due to free fall condition, initial velocity is zero.Therefore the v-t curve is a straight line with a slope equal to g equal to $9.81 m / s$ passing through origin. On dividing the total area under the curve into interval of unit seconds, then we initially obtain a triangle followed by trapeziums of increasing height. The ratio of area of first triangle to second triangle to third triangle is equal to the ratio of displacement in first, second and third second. We get ratio equal to 1:3:5:7:9… and so on. For 4 th & 5th second it is 7:9.Show Answer
Show Answer
Answer
We know for upward motion, $v^{2}=u^{2}-2 g h$ or $h=\dfrac{u^{2}-v^{2}}{2 g}$
But at highest point $v=0$
Therefore, $h=u^{2} / 2 g$
For first ball, $h_1=u_1^{2} / 2 g$
and for second ball, $h_2=u_2^{2} / 2 g$
Thus $ \large\dfrac{h_1}{h_2} = \large\dfrac{\large\dfrac{u^2_1}{\not 2g}}{\large\dfrac{u^2_2}{\not 2g}} = \dfrac{u^2_1}{u^2_2} $ or $ h_1 : h_2 = u^2_1 :u^2_2 $
