Self Studies
Selfstudy
Selfstudy

Motion in A Straight Line Test - 18

Result Self Studies

Motion in A Straight Line Test - 18
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    An object is thrown from  height of h, its covers height h in time T. After time T/2 where will be the object.

    Solution

    Given,

    Object is free fall from height $$h$$

    Initial velocity, $$u=0$$

    Apply kinematic equation of motion

    $$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$

    $$ h=\dfrac{1}{2}g{{t}^{2}}\,\,........\,(1) $$

    At half time

    $$s=u\dfrac{t}{2}+\dfrac{1}{2}g{{\left( \dfrac{t}{2} \right)}^{2}}=\dfrac{1}{4}\left( \dfrac{1}{2}g{{t}^{2}} \right)=\dfrac{h}{4}$$

    Height from ground, $$h-\dfrac{h}{4}=\dfrac{3h}{4}$$meter

    Hence, at half time height from ground is $$\dfrac{3h}{4}$$m

  • Question 2
    1 / -0
    A man throws balls into the air one after another. He always throws a ball when the previous one thrown has just reached the highest point. The height to which each ball rises, if he throws 5 balls per second is $$(g = 10 ms^{-2})$$:
    Solution

    suppose $$n - balls \to $$

    $$final\,velocity\,,\,$$ $$v = 0$$

    $$u = 0$$

    $$T = {\dfrac{1} {n}}\sec $$

    $$0 = u - g{\dfrac{1}{n}}$$

    $${\dfrac{g} {n}} = u$$

    $${v^2} = {u^2} - 2gh$$

    $$0 = {\dfrac{{g^2}} {{n^2}}} - 2gh$$

    $${\dfrac{{g^2}} {{n^2}}} = 2gh$$

    $$h = {\dfrac{g} {2{n^2}}}$$

    $$n = 5$$

       $$ = {\dfrac{10} {2 \times 25}}$$

       $$ = {\dfrac{1}{5}}m$$

  • Question 3
    1 / -0
    A ball is released from certain height h reaches ground in time T. Where will it be from the ground at time $$ \frac { 3T }{ 4 } ? $$
    Solution

    Given that,

    Height $$=h$$

    Time $$t=\dfrac{3T}{4}$$

    Since, the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.

    If we apply the second equation of motion

      $$ s=ut+\dfrac{1}{2}g{{t}^{2}} $$

     $$ h=\dfrac{g{{T}^{2}}}{2} $$

     $$ {{T}^{2}}=\dfrac{2h}{g} $$

    Now, we apply the same equation to calculate the height covered in time 3T/4

      $$ s=ut+\dfrac{1}{2}g{{t}^{2}} $$

     $$ s=g\times \dfrac{{{\left( \dfrac{3T}{4} \right)}^{2}}}{2} $$

     $$ s=\left( \dfrac{9g}{32} \right)\times {{T}^{2}} $$

     $$ s=\left( \dfrac{9g}{32} \right)\times \dfrac{2h}{g} $$

     $$ s=\dfrac{9h}{16} $$

    Now, the distance from the ground

      $$ s=h-\dfrac{9h}{16} $$

     $$ s=\dfrac{7h}{16} $$

    Hence, the distance from the ground is $$\dfrac{7h}{16}$$ 

  • Question 4
    1 / -0
    A man slides down a snow covered hill along a curved path and falls $$20$$m below his initial position. The velocity in m/sec with which he finally strikes the ground is? $$(g=10 m/sec^2)$$.
    Solution
    The equation of motion is

    $${ v }^{ 2 }={ u }^{ 2 }+2gh$$

     $${v }^{ 2 }=0+2gh$$

     $${v }^{ 2 }=\sqrt { 2gh } $$ 

    $$v=\sqrt { 2\times 10\times 20 } $$

     $$v=20\dfrac { m }{ s } $$
  • Question 5
    1 / -0
    If we use plus and minus signs to indicate the direction of velocity and acceleration in one dimension, in which of the following situation does the object speed up?
    Solution

    $$\vec{v}=\vec{u}+\vec{a}t$$

    $$or\,\,\vec{v}=-\vec{u}-\vec{a}t$$

    Direction of initial velocity and direction of acceleration should be same, either negative or positive

    Hence, at negative velocity and negative acceleration object will speed up

  • Question 6
    1 / -0
    An object is thrown from the height of 125 cm take g = 10 m/s. Find the ratio of distance covered by object in the 1 st and last 1 sec 
    Solution

    Given,

    Initial velocity, $$u=0$$

    Time period of free fall from height $$125\,m$$

    $$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 125}{10}}=5\,\sec $$

    Apply kinematic equation of motion

    $$s=ut+\dfrac{1}{2}a{{t}^{2}}=\dfrac{1}{2}g{{t}^{2}}$$

    Ratio of displacement for $${{1}^{st}}$$ sec and last $$1\,\sec $$

    $$\dfrac{{{s}_{1}}-{{s}_{0}}}{{{s}_{4}}-{{s}_{5}}}=\dfrac{\dfrac{1}{2}g\left( {{1}^{2}}+0 \right)}{\dfrac{1}{2}g\left( {{5}^{2}}+{{4}^{2}} \right)}=\dfrac{1}{9}$$

    Hence , ratio is $$1:9$$  

  • Question 7
    1 / -0
    The motion of a particle of mass $$m$$ is describe by $$y = ut + \dfrac{1}{2} gt^2$$. Find the force acting on the particle.
    Solution

     $$v=ut+\dfrac{1}{2}\times gt^2$$

    Differentiate y with respect to t which gives velocity as dy/dt = v

    $$\dfrac{dv}{dt}$$ = $$u + \dfrac{1}{2}g (2t)$$

    v=u + gt -------------------Eqn (1) 

    Differentiate v with respect to t which gives velocity as d2y/dt2 = dv/dt = a

    $$\dfrac{dv}{dt}$$ = 0 + g

    a = g  ----------------------Eqn (2)We know that force acting on a mass m is given by F = ma

    Now from Eqn (2), substitute a =g, Hence, F = mg

  • Question 8
    1 / -0
    The rate of change of velocity is:
    Solution
    C. Acceleration
  • Question 9
    1 / -0
    A car initially travelling north at 5 m/s has a constant acceleration of $$ 2 m/s^2 $$ northward. How far does the car travel in the first 10 s?
    Solution

    Given,

    Initial velocity, $$u=5\,m{{s}^{-1}}$$

    Acceleration, $$a=2\,m{{s}^{-2}}$$

    Apply kinematic equation of motion

    $$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$

    $$ s=5\times 10+\dfrac{1}{2}\times 2\times 10^2=150\,m $$

    Hence in first 10 second card travel $$150\,m$$

  • Question 10
    1 / -0
    A ball dropped from one metre above the top of a window, crosses the window in $$t_{1}\;s$$ . If the same ball is dropped from $$2\;m$$ above the top of the same window, time taken by it to cross the window is $$t_{2}\; s$$ . Then
    Solution
    Solution 1
    The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:
    $$v^2=u^2+2aS$$,  where $$u=0$$

    The ball with higher speed will take less time to cross the window.

    $$\because$$ $$H_2 > H_1$$  
    $$\implies$$ $$v_2 > v_1$$
     $$\therefore$$ $$t_2 < t_1$$
    Hence Option D is correct

    Solution 2
    Statement D is correct
    Situation 1:
    We want time to cross x.
    $$v^{2}=u^{2}+2as$$
    $$u=0, a=-g, s=-1 $$
    $$v^{2}=2g\Rightarrow v_{1}=\sqrt{2g}$$
    $$(1+x)=\dfrac{1}{2} gt^{2}\Rightarrow

    T_{2}=\sqrt{\dfrac{2(1+x)}{g}}$$
    $$1=\dfrac{1}{2}gt^{2}=T_{1}=\sqrt{\dfrac{2}{g}}$$
    So time taken to cross $$x=t_{1}=T_{2}-T_{1}

    =\sqrt{\dfrac{2}{g}}(\sqrt{1+x}-1)$$
    Situation 2:
    $$(2+x)=\dfrac{1}{2}gt^{2}\Rightarrow

    T_{2}=\sqrt{\dfrac{2(2+x)}{g}}$$
    $$2=\dfrac{1}{2}gt^{2}\Rightarrow T_{1}=\sqrt{\dfrac{2\times 2}{g}}$$
    So time taken to cross $$x=t_{2}=T_{2}-T_{1}=\sqrt{\dfrac{2}{g}}(\sqrt{2+x}-\sqrt{2})

    $$
    Clearly $$t_{2}< t_{1} $$

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now