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NCERT Exemplar · Class 9 Science Force and Laws of Motion

20 questions · 20 still being checked

Long Answer Questions 18–20 (part 3 of 3)

  1. Exercise 18

    Using second law of motion, derive the relation between force and
    acceleration. A bullet of 10\displaystyle 10 g strikes a sand-bag at a speed of 103\displaystyle 10^{3} m s−1\displaystyle s^{-1} and
    gets embedded after travelling 5\displaystyle 5 cm. Calculate
    (i)
    the resistive force exerted by the sand on the bullet
    (ii)
    the time taken by the bullet to come to rest.

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    NCERT’s answer
    (i)
    m =$\displaystyle 10$ g = kg $\displaystyle 1000$ u =\(\displaystyle 10^{3}\) m/s v = $\displaystyle 0$ s = $\displaystyle 5$ m \(\displaystyle v^{2}\)- \(\displaystyle u^{2}\) = $\displaystyle 2$ a s $\displaystyle 0$ - (\(\displaystyle 10^{3}\))$\displaystyle 2$ = 2.a. a = $\displaystyle 1000$ $\displaystyle 1000$ − × . $\displaystyle 5$ × = - \(\displaystyle 10^{7}\) m \(\displaystyle s^{-2}\) F = m.a = $\displaystyle 10$ $\displaystyle 5$ N (ii) v = u + a t $\displaystyle 0$ = \(\displaystyle 10^{3}\)- $\displaystyle 10$ 7t \(\displaystyle 10^{7}\)t =\(\displaystyle 10^{3}\) t = =\(\displaystyle 10^{-4}\) s
    Second law: force is the rate of change of momentum.
    \[F \propto \frac{mv-mu}{t} = m\left(\frac{v-u}{t}\right) = ma \]
    The newton fixes the constant to $\displaystyle 1$, so \(\displaystyle F=ma\).
    \[m=0.010\ \text{kg},\ u=10^{3}\ \text{m s}^{-1},\ v=0,\ s=0.050\ \text{m} \]
    (i)
    \[v^{2}=u^{2}+2as \Rightarrow a=-\frac{u^{2}}{2s}=-\frac{10^{6}}{0.10}=-1\times10^{7}\ \text{m s}^{-2} \]
    \[F=ma=0.010\times1\times10^{7}=1\times10^{5}\ \text{N} \]
    (ii)
    \[v=u+at \Rightarrow t=\frac{u}{|a|}=\frac{10^{3}}{10^{7}}=1\times10^{-4}\ \text{s} \]
    Answer: resistive force \(\displaystyle 1\times10^{5}\ \text{N}\), opposing the bullet's motion; time to rest \(\displaystyle 1\times10^{-4}\ \text{s}\).

  2. Exercise 19

    Derive the unit of force using the second law of motion. A force of 5\displaystyle 5 N produces an acceleration of 8\displaystyle 8 m s−2\displaystyle s^{-2} on a mass m1\displaystyle m_{1} and an acceleration of 24\displaystyle 24 m s−2\displaystyle s^{-2} on a mass m2\displaystyle m_{2}. What acceleration would the same force provide if both the masses are tied together?

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    NCERT’s answer
    F = m a = kg m s -$\displaystyle 2$ This unit is also called newton. Its symbol is N. \(\displaystyle m_{1}\) = F a = $\displaystyle 8$ kg, \(\displaystyle m_{2}\) = F a = $\displaystyle 24$ kg, M = + kg = kg Acceleration produced in M, a = F M = = $\displaystyle 6$ m \(\displaystyle s^{-2}\)
    Second law gives \(\displaystyle F=kma\); fixing \(\displaystyle k=1\) defines the newton. \[1\ \text{N} = (1\ \text{kg})(1\ \text{m s}^{-2}) = 1\ \text{kg m s}^{-2} \] \[m_1=\frac{F}{a_1}=\frac{5}{8}\ \text{kg}, \quad m_2=\frac{F}{a_2}=\frac{5}{24}\ \text{kg} \] Tied together, the same force acts on \(\displaystyle m_1+m_2\): \[m_1+m_2=\frac{5}{8}+\frac{5}{24}=\frac{20}{24}=\frac{5}{6}\ \text{kg} \] \[a=\frac{F}{m_1+m_2}=\frac{5}{5/6}=6\ \text{m s}^{-2} \] Answer: \(\displaystyle 1\ \text{N}=1\ \text{kg m s}^{-2}\); combined acceleration \(\displaystyle 6\ \text{m s}^{-2}\).

  3. Exercise 20

    What is momentum? Write its SI unit. Interpret force in terms of momentum.
    Represent the following graphically
    (a)
    momentum versus velocity when mass is fixed.
    (b)
    momentum versus mass when velocity is constant.

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    NCERT’s answer
    Momentum = mass × velocity SI unit of momentum is kg m \(\displaystyle s^{-1}\) Force = Rate of change in momentum
    Momentum is the product of a body's mass and its velocity,
    \[p = mv \]
    it is a vector, direction along \(\displaystyle v\). SI unit: \(\displaystyle \text{kg m s}^{-1}\).
    Newton's second law gives force as the rate of change of momentum:
    \[F = \frac{\Delta p}{\Delta t} \]
    For constant mass,
    \[F = \frac{mv - mu}{t} = ma \]
    so force reduces to \(\displaystyle F = ma\) only because mass is fixed; in general it is the rate of change of momentum.
    (a)
    At fixed mass, \(\displaystyle p = mv\) is a straight line through the origin, slope \(\displaystyle m\) -- shown below.
    (b)
    At fixed velocity, \(\displaystyle p = mv\) is the same straight line through the origin, now with slope \(\displaystyle v\) instead of \(\displaystyle m\).
    NCERT_Solution_Class9_Science_Exemplar_Ch9_Q20
    Answer: \(\displaystyle p = mv\), unit \(\displaystyle \text{kg m s}^{-1}\); \(\displaystyle F = \Delta p/\Delta t\); both graphs are straight lines through the origin.