Physics

Kinematics and Dynamics of Motion

128 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 force concept of force and its unit force - push or pull forces

A 200 g of putty ball falls vertically into a 2.5 kg cart that is rolling freely at 2 m/s on a horizontal surface. What is the final speed of the cart?

  1. $11.85 m/s$
  2. $1.85 m/s$
  3. $10.85 m/s$
  4. $2.85 m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using conservation of momentum: m1 * v1 + m2 * v2 = (m1 + m2) * v_final. Here, 2.5 * 2 + 0.2 * 0 = (2.5 + 0.2) * v_final. 5 = 2.7 * v_final. v_final = 5 / 2.7 = 1.8518 m/s.

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

A rubber ball is dropped from a height of $10$ meters. If the ball always rebounds $\dfrac {4}{5}$ the distance it has fallen, calculate, how far, in meters, will the ball have travelled at the moment it hits the ground for the fourth time?

  1. $4.10$
  2. $5.12$
  3. $29.52$
  4. $43.92$
  5. $49.04$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation
  • for the first time it hits the ground , it travels $10$ m
  • for the second time it hits the ground , it travels $2 \times 4/5 \times 10 = 16+10 =26$
  • for the third time it hits the ground , it travels $ 2\times 4/5 \times 4/5 \times 10 +26 =64/5+26 = 38.8$
  • for the fourth time it hits the ground , it travels $2 \times 4/5 \times 4/5 \times 4/5 \times 10 +38.8 = 256/25 + 38.8 =49.04$

Multiple choice physics lever common machines terms related to machines introduction to simple machines

A rubber ball dropped from a height of $20\ m (g = 10\ m/s^{2})$ falls on the roof of an elevator going up at $2\ m/s$. If the collision of the ball with the elevator is elastic, the ball rebounds with a velocity of

  1. $20\ m/s$
  2. $22\ m/s$
  3. $24\ m/s$
  4. $26\ m/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Velocity of ball before impact: v = sqrt(2gh) = sqrt(2 * 10 * 20) = 20 m/s (downward). Elevator velocity = 2 m/s (upward). Relative velocity of ball w.r.t elevator = 20 - (-2) = 22 m/s. After elastic collision, it rebounds at 22 m/s relative to the elevator. Absolute velocity = 22 + 2 = 24 m/s.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A metal ball of mass $2kg$ is allowed to fall freely from rest from a height of $5m$ above the ground (Take $g=10ms^{-2}$). After the ball hits the ground and comes to rest, mechanical energy is converted to:

  1. heat energy

  2. sound energy

  3. electrical energy

  4. Both A & B
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When ball is at rest it will have only P.E that is
$P.E = mgh$


$m = 2\  kg; h = 5\  m$

$P.E = 2\times 10\times 5 = 100 J$

This is the total energy as K.E is zero when ball is at maximum height that is when it is at 5m. Hence when ball reaches the ground P.E becomes zero  because height is zero. At this point total energy is equal to K.E that is $K.E = 100 J$.

This $100\  J$ will convert to heat and sound energy as the ball comes to rest.

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

An ice block is falling from a height. If $75\%$ of the block melts then $h = (g = 10\ ms^{-2})$.

  1. $10.2\ Km$
  2. $12.8\ Km$
  3. $25.2\ Km$
  4. $26.8\ Km$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When an ice block falls and 75 percent of it melts due to friction and thermal conversion, the heat required for melting (Q = m_melt * L_f) comes from the loss of potential energy (m * g * h). Since 75 percent melts, 0.75 * m * L_f = m * g * h. Using the latent heat of fusion of ice (approx 3.34 * 10^5 J/kg) and g = 10, solving for height h yields approximately 25.2 Km.

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 $720-400=320\ mtrs$