Tag: physics

Questions Related to physics

A force of 100 dynes acts on mass of 5 gm for 10 sec. The velocity produced is:

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

  1. 2 cm/sec

  2. 20 cm/sec

  3. 200 cm/sec

  4. 2000 cm/sec

Show answer
Correct Option: C
Explanation:

Given that,

Force $F=100\,dyn$

Mass $m=5\,g$

Time $t=10\,s$

Initial velocity $u=0$

Now, the acceleration is

  $ F=ma $

 $ a=\dfrac{F}{m} $

 $ a=\dfrac{100\times {{10}^{-5}}}{5\times {{10}^{-3}}} $

 $ a=0.20\,m/{{s}^{2}} $

Now, from equation of motion

  $ v=u+at $

 $ v=0+0.20\times 10 $

 $ v=2\,m/s $

 $ v=200\,cm/s $

Hence, the velocity produced is $200\ cm/s$

Jahnvi is walking at $1.63 m/s$. If she weighs $583 N$, what is the magnitude of her momentum?

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

  1. 951 kg m/s951 kg m/s

  2. $68.8 \ kg \ m/s$

  3. $137 \ kg \ m/s$

  4. $672 \ kg \ m/s$

Show answer
Correct Option: A
Explanation:
$P = mv$
$ = 1.63 \times 583$
$ = 951\,kg\,m/s$
Hence, 
No any option is match so answer is $ 951\,kg\,m/s.$

The moment P (in $kg\, ms^{-1}$) of a particle is varying with time t (in second) as $p=2+3t^2$. The force acting on the particle at $t=3s$ will be 

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

  1. 18 N

  2. 54 N

  3. 9 N

  4. 15 N

Show answer
Correct Option: A
Explanation:

Given,

$p=2+3t^2$

From Newton's second law,
$F=\dfrac{dp}{dt}$

$F=\dfrac{d(2+3t^2)}{dt}$

$F=0+6t=6t$

At $t=3s$

$F=6\times 3=18N$
The correct option is A.

A homogeneous disc of mass 2 kg and radius 15 cm is rotating about its axis (which is fixed) with an angular velocity of 4 rad/s. The linear momentum of the disc is :

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

  1. 1.2 kg-m/s

  2. 1.0 kg-m/s

  3. 0.6 kg-m/s

  4. None of the above

Show answer
Correct Option: C
Explanation:
The formula for angular momentum is:
$L=Iw$, where is the angular velocity and $I$  is the moment of inertia.
For a disk$,$
$I = \dfrac{1}{2}M{R^2},$ where $M$ is the mass of the disk and $R$ is the radius of the disk.
For this disk, 
$I = \frac{1}{2}M{R^2} = \dfrac{1}{2}\left( 2 \right) \times {\left( {0.15} \right)^2} = 0.0225\,kg\,{m^2}$
$L = Iw = 0.0225\,kg\,{m^2} \times \left( {4\,rad/s} \right) = 0.09\,kg\,{m^2}\,rad/s$
The linear momentum is related to the the angular momentum  by the formula; $L=pr.$
We can solve for p, the linear momentum in this equation to get,
$p = \dfrac{L}{r} = \dfrac{{0.09}}{{0.15}} = 0.6kgs$
Hence,
option $(C)$ is correct answer.

Which of the following must be true for the sum of the magnitude of the momenta of the individual particles in the system?

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

  1. It must be zero

  2. It could be non-zero, but it must be a constant

  3. It could be non zero, and it might not be a constant

  4. It could be zero, even if the magnitude of the total momentum is not zero

Show answer
Correct Option: C
Explanation:

The sum of the magnitude of the momenta of the individual particles in the system is equal to the total mass times the velocity of centre of mass.
Velocity of COM depends upon the net external force, and as nothing is given about it we can't tell anything strictly about it.
Hence It could be non zero, and it might not be a constant

A force of $10 \,N$ acts on a body of mass $0.5 \,kg$ for $0.25 \,sec$ starting from rest. What is its momentum now?

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

  1. $2.5 \,N/sec$

  2. $0.25 \,N/sec$

  3. $0.5 \,N/sec$

  4. $0.75 \,N/sec$

Show answer
Correct Option: A
Explanation:

$Ft = mV _2 - mV _1 \Longrightarrow 10 \times 0.25 = P _t - P _i, P _t = 2.5$

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle. 

There is a change in:

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

  1. momentum of the rocket

  2. mass of the rocket

  3. both A & B

  4. None of the above

Show answer
Correct Option: C
Explanation:

As the rocket is moving upward by ejecting gases, its mass is getting decreased gradually. As momentum is the product of the mass and velocity that is $p = m\times v$. Hence momentum decreases as mass is decreasing.

A car of mass $600 kg$ is moving with a speed of $10 m s^{-1}$ while a scooter of mass $80 kg$ is moving with a speed of $50 m s^{-1}$. Compare their momentum.

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

  1. $2 : 3$

  2. $1 : 2$

  3. $3 : 1$

  4. $3 : 2$

Show answer
Correct Option: D
Explanation:

As we know that momentum is nothing but product of mass and velocity. Let us substitute the given values.


Given that mass of the car as 600 kg and velocity as 10 m/sec.


Hence momentum of the car is $P=600\times 10=6000 kg m/sec$.

We have mass of the scooter as 80 kg where as velocity is 50 m/sec.

Hence momentum is $P=80\times 50=4000 kg m/sec$.

Therefore, momentum are in the ratio $3:2$.

What is the value of $p _1$ and $m _2$ ?

m (kg) v (m/s) p = mv (kgm/s)
85 60 $p _1$
$m _2$ 2.5 6.25
physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. 5100, 2.5

  2. 5.1 , 2.5

  3. 2500, 5

  4. 55, 2.6

Show answer
Correct Option: A
Explanation:
m (kg) v (m/s) p = mv (kgm/s)
85 60 5100
2.5 2.5 6.25

In first case:
$ m=85kg $
$v =60 m/s$
$ p =mv = 85 \times 60 = 5100\ kg m/s$
In second  case:
$ m=? $
$v =2.5 m/s$
$ p =mv = 6.25 $
$\therefore m = \dfrac{p}{v} = \dfrac{6.25}{2.5} = 2.5\ kg$

A bullet of mass $50  g$ moving with an initial velocity $100  m  {s}^{-1}$, strikes a wooden block and comes to rest after penetrating a distance $2  cm$ in it. Calculate: (i) initial momentum of the bullet, (ii) final momentum of the bullet. 

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

  1. $5$ $kg$ $m$ ${s}^{-1}$, $5$ $kg$ $m$ ${s}^{-1}$

  2. $5$ $kg$ $m$ ${s}^{-1}$, $0$ $kg$ $m$ ${s}^{-1}$

  3. $5$ $kg$ $m$ ${s}^{-1}$, $2.5$ $kg$ $m$ ${s}^{-1}$

  4. $2.5$ $kg$ $m$ ${s}^{-1}$, $0$ $kg$ $m$ ${s}^{-1}$

Show answer
Correct Option: B
Explanation:

Mass, $m=50\ g=0.05\ kg$

Initial velocity, $u=100\ m/s$
Final velocity, $v=0$
(i) Initial momentum, $p _i=0.05 \times 100=5\ kgm/s$
(ii) Final momentum, $p _f=0.05 \times 0 = 0$