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NCERT Exemplar · Class 9 Science Motion

24 questions · 24 still being checked

Long Answer Questions 19–24 (part 3 of 3)

  1. Exercise 19

    An object is dropped from rest at a height of 150\displaystyle 150 m and simultaneously another object is dropped from rest at a height 100\displaystyle 100 m. What is the difference in their heights after 2\displaystyle 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

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    NCERT’s answer
    Initial difference in height = ($\displaystyle 150$ - $\displaystyle 100$) m = $\displaystyle 50$ m Distance travelled by first body in $\displaystyle 2$ s = \(\displaystyle h_{1}\) = $\displaystyle 0$ + $\displaystyle 2$ g ($\displaystyle 2$)$\displaystyle 2$ = $\displaystyle 2$ g Distance travelled by another body in $\displaystyle 2$ s = \(\displaystyle h_{2}\) = $\displaystyle 0$ + $\displaystyle 1$ $\displaystyle 2$ g ($\displaystyle 2$)$\displaystyle 2$ = $\displaystyle 2$ g After $\displaystyle 2$ s, height at which the first body will be = h ′ = $\displaystyle 150$ - $\displaystyle 2$ g After $\displaystyle 2$ s, height at which the second body will be = h ′ = $\displaystyle 100$ - $\displaystyle 2$ g Thus, after $\displaystyle 2$ s, difference in height = $\displaystyle 150$ - $\displaystyle 2$ g - ($\displaystyle 100$ - $\displaystyle 2$ g) = $\displaystyle 50$ m = initial difference in height Thus, difference in height does not vary with time.
    Both objects fall from rest with the same acceleration \(\displaystyle g\), so each covers the same distance in the same time. \[s(t) = \dfrac{1}{2}gt^{2} \qquad (u=0) \] Heights above the ground: \[h_1(t) = 150 - \dfrac{1}{2}gt^{2}, \qquad h_2(t) = 100 - \dfrac{1}{2}gt^{2} \] \[\Delta h(t) = h_1(t)-h_2(t) = 150-100 = 50\ \text{m} \] Taking \(\displaystyle g = 9.8\ \text{m s}^{-2}\), at \(\displaystyle t=2\ \text{s}\): \[s(2) = \dfrac{1}{2}(9.8)(2)^{2} = 19.6\ \text{m} \] \[h_1(2) = 150-19.6 = 130.4\ \text{m}, \qquad h_2(2) = 100-19.6 = 80.4\ \text{m} \] \[h_1(2)-h_2(2) = 50\ \text{m} \] NCERT_Solution_Class9_Science_Exemplar_Ch8_Q19 \(\displaystyle g\) and \(\displaystyle t\) cancel identically out of the subtraction, so the gap never changes with time. Answer: the height difference is $\displaystyle 50$ m at \(\displaystyle t=2\) s and stays fixed at $\displaystyle 50$ m for all time, until the lower object ($\displaystyle 100$ m) lands.

  2. Exercise 20

    An object starting from rest travels 20\displaystyle 20 m in first 2\displaystyle 2 s and 160\displaystyle 160 m in next 4\displaystyle 4 s. What will be the velocity after 7\displaystyle 7 s from the start.

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    NCERT’s answer
    \(\displaystyle s_{1}\) = ut + 2at or $\displaystyle 20$=$\displaystyle 0$+ $\displaystyle 1$ a ($\displaystyle 2$)$\displaystyle 2$ or a= $\displaystyle 10$ m \(\displaystyle s^{-2}\), v = u + at = $\displaystyle 0$ + ($\displaystyle 10$ × $\displaystyle 2$) = $\displaystyle 20$ m \(\displaystyle s^{-1}\) \(\displaystyle s_{2}\) = $\displaystyle 160$ = vt′ + $\displaystyle 2$ a′ (t′)$\displaystyle 2$ = ($\displaystyle 20$ × $\displaystyle 4$) + ( $\displaystyle 2$ a′ × $\displaystyle 16$) ⇒ a′=$\displaystyle 10$ m \(\displaystyle s^{-2}\) Since acceleration is the same, we have v′ =$\displaystyle 0$+ ($\displaystyle 10$ × $\displaystyle 7$) = $\displaystyle 70$ m \(\displaystyle s^{-1}\)
    Starts from rest, \(\displaystyle u=0\). \[s = ut + \dfrac{1}{2}at^{2} \] First $\displaystyle 2$ s: \[20 = \dfrac{1}{2}a(2)^{2} = 2a \ \Rightarrow\ a = 10\ \text{m s}^{-2} \] Check against the next $\displaystyle 4$ s (\(\displaystyle t=2\) to \(\displaystyle t=6\)): \[s(6)-s(2) = \dfrac{1}{2}(10)(6^{2}-2^{2}) = 160\ \text{m} \] matches the given $\displaystyle 160$ m, so the acceleration is the same $\displaystyle 10$ m s\(\displaystyle ^{-2}\) throughout. Velocity after $\displaystyle 7$ s: \[v = u+at = 0+(10)(7) = 70\ \text{m s}^{-1} \] Answer: \(\displaystyle v = 70\ \text{m s}^{-1}\).

  3. Exercise 21

    Using following data, draw time - displacement graph for a moving object: Time (s) 0\displaystyle 0 2\displaystyle 2 4\displaystyle 4 6\displaystyle 6 8\displaystyle 8 10\displaystyle 10 12\displaystyle 12 14\displaystyle 14 16\displaystyle 16 Displacement (m) 0\displaystyle 0 2\displaystyle 2 4\displaystyle 4 4\displaystyle 4 4\displaystyle 4 6\displaystyle 6 4\displaystyle 4 2\displaystyle 2 0\displaystyle 0 Use this graph to find average velocity for first 4\displaystyle 4 s, for next 4\displaystyle 4 s and for last 6\displaystyle 6 s.

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    NCERT’s answer
    Average velocity for first $\displaystyle 4$ s. Average velocity = Change in displacement Total time taken v = $\displaystyle 4$ − = − = $\displaystyle 1$ m \(\displaystyle s^{-1}\) For next $\displaystyle 4$ s, v = $\displaystyle 4$ − = − = $\displaystyle 0$ m \(\displaystyle s^{-1}\) (or as x remains the same from $\displaystyle 4$ to $\displaystyle 8$ seconds, velocity is zero) For last $\displaystyle 6$ s, v − = − = -$\displaystyle 1$ m \(\displaystyle s^{-1}\)
    Plot \(\displaystyle s\) (m) against \(\displaystyle t\) (s) from the table. NCERT_Solution_Class9_Science_Exemplar_Ch8_Q21 Average velocity over an interval is the chord's slope, \(\displaystyle \bar v=\dfrac{\Delta s}{\Delta t}\). First $\displaystyle 4$ s (\(\displaystyle t=0\) to \(\displaystyle 4\)): \[\bar v=\dfrac{4-0}{4-0}=1\ \text{m s}^{-1} \] Next $\displaystyle 4$ s (\(\displaystyle t=4\) to \(\displaystyle 8\)): \[\bar v=\dfrac{4-4}{8-4}=0 \] Last $\displaystyle 6$ s (\(\displaystyle t=10\) to \(\displaystyle 16\)): \[\bar v=\dfrac{0-6}{16-10}=-1\ \text{m s}^{-1} \] Answer: \(\displaystyle 1\ \text{m s}^{-1}\), \(\displaystyle 0\), \(\displaystyle -1\ \text{m s}^{-1}\).

  4. Exercise 22

    An electron moving with a velocity of 5\displaystyle 5 × 104\displaystyle 10^{4} m s−1\displaystyle s^{-1} enters into a uniform
    electric field and acquires a uniform acceleration of 104\displaystyle 10^{4} m s−2\displaystyle s^{-2} in the direction
    of its initial motion.
    (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?

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    NCERT’s answer
    Given initial velocity, u = $\displaystyle 5$ × \(\displaystyle 10^{4}\) m \(\displaystyle s^{-1}\) and acceleration, a = \(\displaystyle 10^{4}\) m \(\displaystyle s^{-2}\) (i) final velocity = v = $\displaystyle 2$ u = $\displaystyle 2$ × $\displaystyle 5$ ×$\displaystyle 10$ $\displaystyle 4$ m \(\displaystyle s^{-1}\) =$\displaystyle 10$ × $\displaystyle 10$ $\displaystyle 4$ m s -$\displaystyle 1$ To find t, use v = u + at or v u t a - =       $\displaystyle 10$ ×$\displaystyle 10$ - $\displaystyle 5$×$\displaystyle 10$ $\displaystyle 5$ ×$\displaystyle 10$ = = $\displaystyle 5$ s (ii) Using s = ut + 1at = ($\displaystyle 5$ ×\(\displaystyle 10^{4}\)) × $\displaystyle 5$ + $\displaystyle 1$ ($\displaystyle 10$ ) × ($\displaystyle 5$) = $\displaystyle 25$ ×\(\displaystyle 10^{4}\) + $\displaystyle 25$ ×$\displaystyle 10$ = $\displaystyle 37.5$×\(\displaystyle 10^{4}\) m
    \[u = 5\times10^{4}\ \text{m s}^{-1}, \qquad a = 10^{4}\ \text{m s}^{-2}, \qquad v = 2u = 1\times10^{5}\ \text{m s}^{-1} \]
    (i)
    First equation of motion:
    \[v = u+at \]
    \[1\times10^{5} = 5\times10^{4} + (10^{4})t \]
    \[t = \dfrac{1\times10^{5}-5\times10^{4}}{10^{4}} = 5\ \text{s} \]
    (ii)
    Second equation of motion:
    \[s = ut+\dfrac{1}{2}at^{2} \]
    \[s = (5\times10^{4})(5) + \dfrac{1}{2}(10^{4})(5)^{2} \]
    \[s = 2.5\times10^{5} + 1.25\times10^{5} = 3.75\times10^{5}\ \text{m} \]
    Answer: \(\displaystyle t = 5\ \text{s}\), \(\displaystyle s = 3.75\times10^{5}\ \text{m}\).

  5. Exercise 23

    Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th\displaystyle 4^{th} and 5th\displaystyle 5^{th} seconds.

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    NCERT’s answer
    Using the equation of motion s = ut + at $\displaystyle 2$ Distance travelled in $\displaystyle 5$ s s = u × $\displaystyle 5$ + a × or s = $\displaystyle 5$ u + $\displaystyle 25$ $\displaystyle 2$ a ——(i) Similarly, distance travelled in $\displaystyle 4$ s, s′ = $\displaystyle 4$ u + $\displaystyle 2$ a——(ii) Distance travelled in the interval between \(\displaystyle 4^{th}\) and \(\displaystyle 5^{th}\) second = (s - s′) = (u + a ) m
    Distance covered up to time \(\displaystyle t\): \[s(t) = ut+\dfrac{1}{2}at^{2} \] Distance in the \(\displaystyle n\)th second is \(\displaystyle s(n)-s(n-1)\): \[s_n = \left[un+\dfrac{1}{2}an^{2}\right] - \left[u(n-1)+\dfrac{1}{2}a(n-1)^{2}\right] \] \[s_n = u + \dfrac{a}{2}(2n-1) \] The interval between the 4th and 5th seconds is the 5th second, \(\displaystyle n=5\): \[s_5 = u + \dfrac{a}{2}(2\times5-1) = u + \dfrac{9a}{2} \] Answer: \(\displaystyle s = u + \dfrac{9a}{2}\), the general \(\displaystyle n\)th-second relation evaluated at \(\displaystyle n=5\).

  6. Exercise 24

    Two stones are thrown vertically upwards simultaneously with their initial velocities u1\displaystyle u_{1} and u2\displaystyle u_{2} respectively. Prove that the heights reached by them would be in the ratio of u 2\displaystyle 2 1\displaystyle 1: u ( Assume upward acceleration is -g and downward acceleration to be +g ).

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    NCERT’s answer
    We know for upward motion, \(\displaystyle v^{2}\) = \(\displaystyle u^{2}\) - $\displaystyle 2$ g h or h = - u v g But at highest point v = $\displaystyle 0$ Therefore, h = u g For first ball, \(\displaystyle h_{1}\) = $\displaystyle 1$ $\displaystyle 2$ u g and for second ball, \(\displaystyle h_{2}\)= $\displaystyle 2$ $\displaystyle 2$ u g Thus h h = u 2g u 2g = u u or \(\displaystyle h_{1}\) : \(\displaystyle h_{2}\) = : u u
    Take the upward direction as positive, so \(\displaystyle a = -g\); at the top of the path the velocity of each stone is zero. Third equation of motion for the first stone: \[v_{1}^{2} = u_{1}^{2} + 2(-g)h_{1} \] \[0 = u_{1}^{2} - 2gh_{1} \] \[h_{1} = \frac{u_{1}^{2}}{2g} \] For the second stone, in the same way: \[h_{2} = \frac{u_{2}^{2}}{2g} \] Dividing the two results, the common factor \(\displaystyle 2g\) cancels: \[\frac{h_{1}}{h_{2}} = \frac{u_{1}^{2}}{u_{2}^{2}} \] Answer: \(\displaystyle h_{1} : h_{2} = u_{1}^{2} : u_{2}^{2}\).