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Motion Test - 40

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Motion Test - 40
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  • Question 1
    1 / -0
    The two ends of a train moving with constant acceleration pass a certain point with velocities $$u$$ and $$v$$. The velocity with which the middle point of the train passes the same point is
    Solution
    Let  $$s$$ be the length of train using $${ v }\displaystyle^{ 2 }={ u }\displaystyle^{ 2 }+2a\times s$$  $$or\quad { 2as }={ v }^{ 2 }-{ u }^{ 2 }$$                   ......(1)
    Where $$v$$ is final velocity after travelling a distance $$s$$ with an acceleration $$a$$ and $$u$$ is initial velocity as per question
    Let velocity of middle point of train at same point is $${ v }\displaystyle^{ \prime  }$$, then
    $${ \left( { v }\displaystyle^{ \prime  } \right)  }\displaystyle^{ 2 }={ u }\displaystyle^{ 2 }+2a\times \left( { s }/{ 2 } \right) $$ $$={ { u }^{ 2 }+as }$$                      .....(2)
    By equations (1) and (2), we get $${ v }^{ \prime  }=\sqrt { \displaystyle\frac { { v }\displaystyle^{ 2 }+{ u }\displaystyle^{ 2 } }{ 2 }  }$$
  • Question 2
    1 / -0
    A stone is thrown vertically upwards. When the particle is at one-half its maximum height, its speed is $$10  { m }/{ s }$$, then maximum height attained by particle is $$\left( g=10  { m }/{ { s }^{ 2 } } \right)$$:
    Solution
    From third equation of motion
    $${v}^{2}={u}^{2}-2gh$$
    Given v = 10m/s at h/2.     But v=0, when particle attains maximum height h.
    Therefore,  $${10}^{2}={u}^{2}- 2g(\dfrac{h}{2})$$
    Now, $${u}^{2}-0=2gh \quad or \quad {u}^{2}=2gh$$
    $$\Rightarrow 100=2gh-gh=gh$$
    $$\Rightarrow   h=10m$$

  • Question 3
    1 / -0
    A stone is thrown upwards with a velocity $$v$$ from the top of a tower. It reaches the ground with a velocity $$3v$$. What is the height of the tower?
    Solution
    $${ v }^{ 2 }-{ u }^{ 2 }=2as\\ \Rightarrow { (3v) }^{ 2 }-{ v }^{ 2 }=2gh\\ \Rightarrow h=\dfrac { { 4v }^{ 2 } }{ g } $$
  • Question 4
    1 / -0
    A ball is dropped downwards, after $$1$$ sec another ball is dropped downwards from the same point. What is the distance between them after $$3$$ sec? (Take $$g=10ms^{-2}$$)
    Solution
    Distance at a given time is given by equation,
    $$ S = ut +\dfrac{1}{2}at^{2} $$
    For first body, $$ u_{1} = 0 ,\  S = S_{1}, a = g ,\  t_{1} = 3 sec $$
    $$ \therefore  S_1 = \dfrac{1}{2}g\times 9 $$
    For second body, $$ u_{2} = 0,\  S = S_{2},\  a = g,\  t_{2} = 2 sec $$
    $$ \therefore  S_{2} = \dfrac{1}{2}g\times 4$$
    So difference between them after $$ 3 sec = S_{1} -S_{2} $$
                                                           $$ =    \dfrac{1}{2}g\times 5 $$
    If$$\  g = 10\  m/s^{2}$$ then $$  S_{1}-  S_{2} = 25 m $$

  • Question 5
    1 / -0
    A stone thrown vertically upwards with a speed of $$5{ m }/{ s }$$ attains a height $${ H }_{ 1 }$$. Another stone thrown upwards from the same point with a speed of $$10{ m }/{ s }$$ attains a height $${ H }_{ 2 }$$. The correct relation between $${ H }_{ 1 }$$ and $${ H }_{ 2 }$$ is
    Solution
    From third equation of motion, $$ v^{2} = u^{2} + 2ah $$
    In first case initial velocity $$u_{1}  = 5\ ms^{-1}  $$
    final velocity$$ v_{1} = 0,\ a=g $$
    So, $$ 0 = 25 - 2gH_{1} \Rightarrow  H_{1} = \dfrac{25}{2g} $$
    and maximum height obtained is H1, then,$$ H_{1} = \dfrac{25}{2g} ...........(1)1$$
    In second case $$ u_{1} = 10 ms^{-1} ,\ v_{1} = 0,\ a = g $$
    So, $$0 = 100 - 2gH_{2} \Rightarrow  H_{2} = \dfrac{100}{2g} $$
    and maximum height obtained is H_2, then,$$ H_2 = \dfrac{100}{2g} .............(2)$$
    from $$(1)$$ and $$(2),$$ we get,
    $$ H_2=4H_1 $$
  • Question 6
    1 / -0
    Similar balls are thrown vertically each with a velocity $$20\ { m }/{ s }$$, one on the surface of earth and the other on the surface of moon. What will be ratio of the maximum heights attained by them? $$($$Acceleration on moon $$=\dfrac{g}{6}\ m / s^2$$ approximately, where $$g$$ is gravitational acceleration due to earth$$)$$
    Solution
    Since $${ v }\displaystyle^{ 2 }={ u }\displaystyle^{ 2 }+2as$$ .......(1)

    For first case $${ u }_{ 1 }=20  { m }/{ s },{ v }_{ 1 }=0,{ a }_{ 1 }=g=-10\ m/s^2,\ { s }_{ 1 }=?$$
    From equation (1)     
    $$s _ 1=\dfrac { u_1^2 }{ 2\times g } $$

    So $$s _ 1=\dfrac { 400 }{ 2\times 10 } =20\   m$$

    For second case (at moon) $$u_2=20  { m }/{ s },\  v _2=0,\ { a }_{ 2 }=\displaystyle\dfrac { g }{ 6 }$$

    Similarlly,
    $$s _ 2=\dfrac { u_2^2 }{ 2\times g } $$

    $${ s }_{ 2 }=\displaystyle\dfrac { 400 }{ 2\times { 10 }/{ 6 } }$$

    Now, 

    So $$\displaystyle\dfrac { { s }_{ 1 } }{ { s }_{ 2 } } =\displaystyle\dfrac { 1 }{ 6 }$$
  • Question 7
    1 / -0
    In uniform circular motion
    Solution
    In a Uniform Circular Motion, the magnitude of velocity, which is speed remains constant. But the direction of velocity, which is tangential to the circle, keeps changing. So velocity also changes. 
  • Question 8
    1 / -0
    If you were to throw a ball vertically upward with an initial velocity of $$50 \; ms^{-1}$$, approximately how long would it take for the ball to return to your hand? Assume air resistance is negligible. (Given $$g=10\; ms^{-2}$$)
    Solution
    The only force acting on the ball is the force of gravity. The ball will ascend until gravity reduces its velocity to zero and then it will descend. Find the time it takes for the ball to reach its maximum height and then double the time to cover the round trip.
    Using $$\displaystyle{ v }=u+at=u-gt$$, we get:
    $$0  { m }/{ s }=50  { m }/{ s }-\left( 10  { m }/{ { s }^{ 2 } } \right) t$$
    Therefore,
    $$t={ \left( 50  { m }/{ s } \right)  }/{ \left( 10  { m }/{ { s }^{ 2 } } \right)  }=5 s$$
    This is the time it takes the ball to reach its maximum height. The total round trip time is $$2t=10s$$.
  • Question 9
    1 / -0
    A stone thrown upward with a speed $$u$$ from the top of the tower reaches the ground with a velocity $$3u$$. The height of the tower is
    Solution
    The stone rises up till its vertical velocity is zero and again reached the top of the tower with a speed $$ u $$(downward). The speed of the stone at the base is $$ 3u $$
    WKT, $$ v^{2} = u^{2} + 2gh $$
    where, $$ v = $$ final velocity
    $$ u = -u $$ initial velocity, -ve because it is thrown upward against gravity
    $$ a = $$ acceleration due to gravity
    $$ h = $$ height or distance
    $$ \Rightarrow \left ( 3u \right )^{2} = \left ( -u \right )^{2} + 2gh   $$
    $$ \Rightarrow 9u^2 - u^2 = 2gh $$
    $$ \Rightarrow h = \dfrac{4u^{2}}{g} $$
  • Question 10
    1 / -0
    A ball released from a height falls $$5 m$$ in one second. In 4 seconds it falls through (Take $$g=10ms^{-2}$$):
    Solution
    Since $$ s=ut+\displaystyle\frac { 1 }{ 2 } g{ t }\displaystyle^{ 2 }$$, where $$u$$ is initial velocity $$\&   a$$ is acceleration.
    Given: $$u=0  \ \&\    a=g$$
    So, distance travelled in 4 sec is,
    $$s=\displaystyle\frac { 1 }{ 2 } \times 10\times 16=80m$$
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