Physics

Kinematics and Dynamics of Motion

121 Questions

Kinematics and dynamics explore projectile motion, inertia, free fall, and kinetic energy. These physics topics frequently appear in general science papers of major competitive exams. Practice these questions to understand the mechanical laws of motion.

Projectile motionFree fall calculationsInertia conceptsKinetic energyImpulse and collision

Kinematics and Dynamics of Motion Questions

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A 20 g ball is fired horizontally toward a 100 g ball that is hanging motionless from a 1.0-m-long string. The balls undergo a head-on, elastic collision, after which the 100 g ball swings out to a maximum angle of 50 degrees. Determine the initial speed of the 20 g ball.

  1. 8.4m/s

  2. 4.2m/s

  3. 2.1m/s

  4. 16.8m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At first we can find the final velocity of the $100\ g$ ball.

$ \dfrac { { m } _{ 2 }{ { v } _{ 2 } }^{ 2 } }{ 2 } =mgh={ m } _{ 2 }gl(1-cos\alpha )$

$ \implies { v } _{ 2 }=\sqrt { 2gl\left( 1-cos\alpha  \right)  } =\sqrt { 2*9.81*1.1(1-cos{ 50 }^{ o }) } =2.78m/s $

Now applying conservation of the momentum principle, we get:

$ { m } _{ 1 }{ V } _{ 1 }+{ m } _{ 2 }{ V } _{ 2 }={ m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 }$

$ { v } _{ 1 }=\dfrac { { m } _{ 1 }{ V } _{ 1 }-{ m } _{ 2 }{ v } _{ 2 } }{ { m } _{ 1 } } ={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } $

Principle of kinetic energy conservation , we get 

$ \dfrac { 1 }{ 2 } { m } _{ 1 }{ { { V } _{ 1 } }^{ 2 } }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { V } _{ 2 } }^{ 2 }=\dfrac { 1 }{ 2 } { m } _{ 1 }{ { v } _{ 1 } }^{ 2 }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { v } _{ 2 } }^{ 2 }$


$ { v } _{ 1 }={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } \ { V } _{ 2 }=0$

$\implies { V } _{ 1 }=\dfrac { { { { { m } _{ 1 } }^{ 2 }v } _{ 1 } }^{ 2 }+{ m } _{ 2 }{ m } _{ 1 }{ { v } _{ 2 } }^{ 2 } }{ 2{ m } _{ 1 }{ v } _{ 2 }{ m } _{ 2 } } =\dfrac { v _{ 2 }\left( { m } _{ 2 }+{ m } _{ 1 } \right)  }{ 2{ m } _{ 1 } } $

$ \implies  \dfrac { 2.78\left( 0.1+0.024 \right)  }{ 2*0.024 } =8.4m/s$

Multiple choice physics properties of a magnetic field earth - a gigantic magnet magnetic field of earth introduction to magnetic field and magnetic flux

A ball 'A' of mass m falls to the surface of the earth from infinity. Another ball 'B' of mass 2m falls to the earth from the height equal to six times radius of the earth then ratio of velocities of 'A' and 'B' on reaching the earth is

  1. $\sqrt (6/5)$
  2. $\sqrt (5/6)$

  3. 1
  4. $\sqrt (7/6)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} { v _{ A } }=\sqrt { \dfrac { { 2GM } }{ R }  } \left( { escape\, \, velocity } \right)  \ { v _{ B } }=\sqrt { \dfrac { { 2gh } }{ { 1+\frac { h }{ R }  } }  } =\sqrt { \dfrac { { 2gh } }{ 7 }  } \left( { h=6R } \right)  \ { v _{ B } }=\sqrt { \dfrac { { 2gh6R } }{ 7 }  } =\sqrt { \dfrac { { 12GM } }{ { 7R } }  }  \ \dfrac { { { v _{ B } } } }{ { { v _{ A } } } } =\sqrt { \dfrac { { 2GM } }{ h }  } =\sqrt { \dfrac { { 7R } }{ { 12GM } }  } =\sqrt { \dfrac { 7 }{ 6 }  }  \ Hence, \ \dfrac { { { v _{ _{ B } } } } }{ { { v _{ A } } } } =\sqrt { \dfrac { 7 }{ 6 }  }  \ \therefore \, option\, \, D\, \, is\, correct\, \, answer. \end{array}$

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A ball rolling off the top of a staicase of each step with height H and width W, with an initial velocity U will just hit nth step. Then n = 

  1. $\frac{2U^2H^2}{gW}$
  2. $\frac{2U^2H^2}{gW^2}$
  3. $\frac{2U^2H}{gW^2}$
  4. $\frac{2UH^2}{gW^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ball hits the nth step when its vertical displacement is nH and horizontal displacement is nW. Using y = (1/2)gt^2 and x = Ut, we get nH = (1/2)g(nW/U)^2. Solving for n gives n = 2U^2H / (gW^2).

Multiple choice physics types of energy law of conservation of energy the law of conservation of energy work, energy and machines

A tennis ball dropped from a height of 2 m rebounds only 1.5 m after hitting the ground. What fraction of energy is lost in the impact?

  1. $\dfrac { 1 }{ 4 } $
  2. $\dfrac { 1 }{ 8 } $
  3. $\dfrac { 1 }{ 2 } $
  4. $\dfrac { 1 }{ 16 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation


We know that the potential energy is dependent on the height. Let us consider the tennis ball dropped from height $H$ has P.E of $mgH$.

Then $mgh$ will be the potential energy while the ball rebounds.

Here,$H = 2m$ and $h = 1.5m$

Then, loss of energy  $ = \dfrac{mg(H-h)}{mgH}$

$=\dfrac { H-h }{ H } =\dfrac { 2-1.5 }{ 2 } =\dfrac { 1 }{ 4 } $

Therefore, fraction of energy lost in the impact $ = \dfrac { 1 }{ 4 } $

Multiple choice physics types of energy law of conservation of energy the law of conservation of energy work, energy and machines

A ball is dropped from a height of $10\ m$. If the energy of the ball reduces by $40\%$ after striking the ground, how much high can the ball bounce back? $(g = 10 m s^{-2})$ :

  1. $6\ m$
  2. $4\ m$
  3. $10\ m$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$M g h = M \times 10 \times 10 = 100 M\  J.$
Energy is reduced by $40\%$ then the remaining energy is $60M\ J$.
Then, $60M=M\times 10\times {h}'$

or, ${h}' = 6\ m$

Multiple choice physics force and newton's laws of motion concept of inertia galileo's law and inertia mass and inertia

A football has lesser inertia than a stone of the same size because:

  1. Football has more air inside than the stone.

  2. Football has less air inside than the stone.

  3. Football has less mass than the stone.

  4. Football has more mass than the stone.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Inertia is the property of mass. Mass of stone $>$ mass of football.

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients
Height Long Jump High Jump
A 158 324 175
B 165 365 185
C 162 380 180
D 170 400 184
E 175 350 199
F 163 350 172
G 178 425 189

Find rank correlation between height & long jump and height & high jump.

  1. $0.9 \ and \ 0.25$
  2. $0.86 \ and\ 0.52$
  3. $0.52 \ and \ 0.86$
  4. $0.89 and \ 0.63$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$Height$  $Rank(Height)$  $Long\,Jump$  $Rank(Long\,jump)$  $d$  $d^2$ 
$158$  $7$  $324$  $6$  $1$  $1$ 
$165$  $4$  $365$  $4$  $0$  $0$ 
$162$  $6$  $380$  $3$  $3$  $9$ 
$170$  $3$  $400$  $2$  $1$  $1$ 
$175$  $2$  $350$  $6$  $4$  $16$ 
$163$  $5$  $350$  $5$  $0$  $0$ 
$178$ $1$ $425$  $1$  $0$  $0$ 
          $\sum d^2=26$ 

$\rho=1-\dfrac{6\sum d^2}{n(n^2-1)}$


    $=1-\dfrac{6\times 26}{7((7)^2-1)}$

    $=0.52$

$Height$  $Rank(Height)$  $High\,jump$  $Rank(High\,jump)$  $d$  $d^2$ 
$158$  $7$  $175$  $6$  $1$  $1$ 
$165$  $4$  $185$  $3$  $1$  $1$ 
$162$  $6$  $180$  $5$  $1$  $1$ 
$170$  $3$  $184$  $4$  $1$  $1$ 
$175$  $2$  $199$  $1$  $1$  $1$ 
$163$  $5$  $172$  $7$  $2$  $4$ 
$178$  $1$  $189$  $2$  $1$  $1$ 
          $\sum d^2=10$ 

$\rho=1-\dfrac{6\sum d^2}{n(n^2-1)}$


    $=1-\dfrac{6\times 10}{7((7)^2-1)}$

    $=0.86$

Multiple choice viscosity option b: engineering physics properties of matter physics

When a ball is released from rest in a very long column of viscous liquid, its down ward acceleration is $a'$ (just after released). Its acceleration when it has acquired to third of the maximum velocity is $a/X$. Find the value of $X$.

  1. $2$
  2. $3$
  3. $23$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of motion for a falling sphere is m*a = mg - F_buoyant - F_viscous. At terminal velocity v_t, the net force is zero, so F_viscous = mg - F_buoyant. When v = v_t/3, the viscous force is F_viscous' = (1/3) * F_viscous. Substituting this, the acceleration becomes a' = (1 - 1/3) * g_effective = (2/3) * a_initial. The question asks for a/X, where a' = a/3, implying X=3.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

A ball is thrown upwards from a rooftop, $28$m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time $t$ is $h$, which is given by, $h =$ $(-t^{2}+2t +35) +28$

How long will it take before hitting the ground?

  1. $6$ seconds
  2. $7$ seconds
  3. $8$ seconds
  4. $9$ seconds
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$h =$ $-t^{2}+2t +35 + 28=0$
$=>$ $t^{2}-2t - 63=0$
$=>(t - 9)(t + 7) = 0$
$=>t = 9$ or $-7$
The time cannot be negative, so the time is $9$ seconds.

Multiple choice physics motion and measurement of distances motion around us motion and rest moving things around us

A free falling body travels ___ of total distance in 5th second.

  1. 8 %

  2. 12 %

  3. 25 %

  4. 36 %
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The correct option is D

Given,

A free falling body 

Distance traveled in 5s

$S=ut+\dfrac{1}{2}gt^2$ since $u=0,g=10m/s$

$=\dfrac{1}{2}\times10\times5^2$

$=12m$

Distance travelled in 5^th sis:

Distance travelled in 5s- Distance travelled in 4s

Distance travelled in $4s=\dfrac{1}{2}\times10\times4^2$

$=80m$

So,

$s _5-s _4=125-80=45$

Therefore,

$\dfrac{s _5-s _4}{s _5}\times100=\dfrac{45}{125}\times100=36\%$

Multiple choice physics propagation of sound waves longitudinal vs transverse wave sound and light comparison of speed of sound with speed of light

 A stone is dropped from the top of a tower $500$m high into a pond of water at the base of the tower. When is the splash heard at the top ? (Given $g = 10 ms^{-2}$ and speed of sound =$ 340 ms^{-1}$)

  1. $10s$
  2. $11.47s$
  3. $1.10s$
  4. $20s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given -  $h=500m   ,  g10m/s^{2} ,  v=340m/s$ ,

by ,  $h=ut+(1/2)gt^{2}$ ,
        $500=0+(1/2)10t^{2}$  ,   (initial velocity , $u=0$) ,
or     $t^{2}=1000/10=100$ ,
or     $t=10s$ ,
it is the time taken by stone to reach the water level , after that a sound is produced due to strike of stone on water , and sound travels upwards .Let t' be the time taken by sound to reach the base of tower ,
then , $t'=h/v=500/340=1.47s$ ,
therefore time taken by splash to hear at the top ,
        $T=t+t'=10+1.47=11.47s$

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

1000 small balls each weighing 1 g strike one sq. cm of a area per second with a velocity of 100 m/sec in a normal direction and rebound with the same velocity. The pressure on the surface (in N/m$^2$) is :

  1. $2 \times 10^3$
  2. $4 \times 10^6$
  3. $10^7$
  4. $2 \times 10^6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Force = change in momentum per second. For one ball, delta p = 2mv. For 1000 balls, F = 1000 * 2 * (0.001 kg) * (100 m/s) = 200 N. Area = 1 cm^2 = 10^-4 m^2. Pressure = 200 / 10^-4 = 2 * 10^6 N/m^2.

Multiple choice physics gravitation tides movements of ocean and sea water circulation of ocean water

A stone drop from height 'h' reaches at earth surface in 1 sec. If the some stone taken to moon and drop freely from height h then it will reaches at the surface of moon in the time:

  1. $\sqrt { 6 } sec$

  2. 9 sec
  3. $\sqrt { 3 } sec$

  4. 6 sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using h = 0.5 * g * t^2, we have h = 0.5 * g_earth * (1)^2, so h = 0.5 * g_earth. On the moon, g_moon = g_earth / 6. Thus, h = 0.5 * (g_earth / 6) * t_moon^2. Substituting h, we get 0.5 * g_earth = 0.5 * (g_earth / 6) * t_moon^2, so t_moon^2 = 6, t_moon = sqrt(6).