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 the essence of change different forms of energy energy conversions energy transformations and energy transfers

A steel ball of mass $5$ ${ g }$ is thrown downward with velocity $10$ ${ ms } ^ { - 1 }$ from height $19.5$ ${ m }$ . It penetrates sand by $50$ ${ cm }$ . The change in mechanical energy will be ( ${ g } = 10$ ${ ms } ^ { - 2 }$ )

  1. $1$ ${J}$
  2. $1.25$ ${J}$
  3. $1.5$ ${J}$
  4. $1.75$ ${J}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} The\, \, change\, \, in\, \, mechanic\, \, energy\, \, \Delta U=mg\left( { h+x } \right) +\frac { 1 }{ 2 } m{ v^{ 2 } } \ here\, \, m=5g=0.00\, 5kg\cdot h=19.5\, mx=50cm=0.5m,v=10\, m/s \ So,\, \Delta U=0.005\times 10919.5+0.5+\frac { 1 }{ 2 } \times 0.005\times { \left( { 10 } \right) ^{ 2 } }=0.005\times 10\times 20+\frac { 1 }{ 2 } \times 0.005\times 100=1.25J \end{array}$

Hence,
option $(B)$ is correct answer.

Multiple choice physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

A disc of mass 10 g is kept floating horizontally by throwing 10 marbles per second against it from below. The marbles strike the disc normally and rebound downwards with the same speed. If the mass of each marble is 5 g, the velocity with which the marbles are striking the disc is $\displaystyle \left( g=9.8{ m }/{ { s }^{ 2 } } \right) $

  1. $\displaystyle 0.98m/s$
  2. $\displaystyle 9.8m/s$
  3. $\displaystyle 1.96m/s$
  4. $\displaystyle 19.6m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given that:
Mass of disc,$M=10g=0.010kg$
Mass of each marble,$m=5g=0.005kg$
Suppose the number of marbles striking the disc per second=n and velocity with which marbles strike disc is v.
Then Weight of disc acting downward$=Mg$
Change in velocity of marbles when they rebound $=v-(-v)=2v$
Therefore the change in momentum of each marble when it strikes the disc=$m * 2v=2mv$
and total momentum imparted per second to the disc$=2mnv $= The force exerted by the marbles in the upward direction
The disc will remain at rest if the net force acting on it is zero provided it is initially at rest.
Therefore for the disc to remain at rest,
Weight of disc=Upward force on it
$= Mg=2mnv$
$v=\dfrac{Mg}{2mn}=\dfrac{(0.010 *9.8)}{(2*0.005*10)}$
$=0.98 m/s$
Multiple choice physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

A cricket ball of mass 500 g is moving with speed of $36 \,km\,h^{-1}$. It is reflected back with the same speed. What is the impulse applied on it?

  1. 20 Kg m/s

  2. 10 Kg m/s

  3. 5 Kg m/s

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Impulse $I=$ change in momentum $=mv _2-mv _1$

Here, $v _1=-v _2=36 km/h$
So, $I=(500/1000)[-36-36]=-36 kg.m/s$
The magnitude of impulse $=36 kg.m/s$

Multiple choice

A girl riding on a skateboard at 3 m/s suddenly hits the curb. The girl will fall:

  1. backward with a speed greater than 3 m/s.

  2. backward at a speed equal to 3 m/s.

  3. forward with a speed greater than 3 m/s.

  4. forward at a speed equal to 3 m/s.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to Newton's First Law of Motion (the law of inertia), an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. When the skateboard is abruptly stopped by the curb, the girl's body continues moving forward at her original speed of 3 m/s.

Multiple choice maths polynomials linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

A body falling from rest under gravity passes a certain point $P$.It was a distance of $400m$ from P and $4$ sec prior to passage through $P$ If $g=10m/sec^2$,then the height above the point $"P"$ from where the body began to fall is ?

  1. $900m$
  2. $320m$
  3. $680m$
  4. $720m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Distance travelled $=400\ m$.
Time$=4\ sec$
$B=10m/s^{2}$
$s=ut+1/2 at^{2}$
$400=4u-1/2\times 10\times 16\times 5$
$400=4u-80$
$4u=480$
$u=120$
At highest point
$V=0$
${u}^{2}=2\times g\times h$
$120\times 120=2\times 10\times h$
$h=720$
This height is from $400\ mtr$ below $P$ 
So height above $P$ is $720-400=320\ mtrs$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A rubber ball is bounced on the floor of a room which has its ceiling at a height of  $3.2{ m }$  from the floor. The ball hits the floor with a speed of  $10 m / { s },$  and rebounds vertically up. If all collisions simply reverse the velocity of the ball, without changing its speed, then how long does it take the ball for a round trip, from the moment it bounces from the floor to the moment it returns back to it ? Acceleration due to gravity is  $10 m / s ^ { 2 }.$

  1. $4 s$
  2. $2 s$
  3. $0.8 s$
  4. $1.2 s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A ball is dropped from height h on the ground. If the coefficient of restitution is e, the height to which the ball goes up after it rebounds for the nth time is :

  1. $\dfrac{h}{e^{2n}}$
  2. $\dfrac{e^{2n}}{h}$
  3. $he^{2n}$
  4. $he^n$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $v _0$ be the velocity with which the ball strikes the Earth first time and $v _n$ after the nth rebound.
The coefficient of restitution is
$e=\dfrac{v _1}{v _0}=\dfrac{v _2}{v _1}=\dfrac{v _3}{v _2}=$.....$=\dfrac{v _n}{v _{n-1}}$
$\therefore e^n=\dfrac{v _1}{v _0}\times \dfrac{v _2}{v _1}\times\dfrac{v _3}{v _2}\times $.....$\times \dfrac{v _n}{v _{n-1}}=\dfrac{v _n}{v _0}$
Here, $v _0=\sqrt{2gh}$ and $v _n=\sqrt{2gH}$
$e^n=\dfrac{\sqrt{2gH}}{\sqrt{2gh}}=\dfrac{\sqrt{H}}{\sqrt{h}}$
$e^{2n}=\dfrac{H}{h}\Rightarrow H=he^{2n}$
Multiple choice

Tiffany throws a ball up in the air. Why does the ball come back down?

  1. Tiffany is pulling the ball down.

  2. Air is pushing the ball down.

  3. Gravity is pulling the ball down.

  4. Gravity is pushing the ball away from Tiffany.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Gravity is the fundamental force of attraction that pulls objects toward the center of the Earth. When a ball is thrown upward, gravity acts to decelerate it and eventually pull it back down to the ground.

Multiple choice

A soccer player kicks a ball with an initial velocity of 20 m/s at an angle of 45 degrees to the horizontal. What is the maximum height reached by the ball?

  1. 10 meters

  2. 15 meters

  3. 20 meters

  4. 25 meters
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum height reached by the ball can be calculated using the formula: h = (v^2 * sin^2(theta)) / (2 * g), where v is the initial velocity, theta is the angle of projection, and g is the acceleration due to gravity. Plugging in the values, we get: h = (20^2 * sin^2(45)) / (2 * 9.81) = 15 meters.