- Force
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- Law's Of motion
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- First Law of Motion
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- Momentum
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- Second Law of Motion
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- Third Law of Motion
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- Law Of conservation of Momentum

How much net force is required to accelerate a 1000 kg car at 4.00 m/s

Given a=4.00 m/s

m=1000kg

Therefore , $F= ma= 1000 \times 4 =4000$ N

If you apply a net force of 3 N on .1 kg-box, what is the acceleration of the box

(a) 2 m/s

(b) 30 m/s

(c) 10 m/s

(d) None of these.

Given F=3 N

m=.1 kg

$F=ma$

or

$a = \frac {F}{m}= 30$ m/s

A body of mass 1 kg undergoes a change of velocity of 4m/s in 4s what is the force acting on it?

Given $\Delta v=4 m/s$ ,t=4 s ,m=1kg

Acceleration is given by $a= \frac {\Delta v}{t}$

a=1 m/s

Now force is given by

$F=ma$

F=1 N

A particle of 10 kg is moving in a constant acceleration 2m/s

S.No |
time |
Momentum |
Velocity |

1 |
1sec |
||

2 |
1.5 sec |
||

3 |
2 sec |
||

4 |
2.5 sec |

Velocity can find using

$v=u+at$

For u=0

$v=at$

Momentum

$P=mv$

S.No |
time |
Momentum |
Velocity |

1 |
1sec |
20 Kg m/s |
2 m/ |

2 |
1.5 sec |
30 kg m/s |
3 m/s |

3 |
2 sec |
40 kg m/s |
4 m/s |

4 |
2.5 sec |
50 kg m/s |
5 m/s |

If a net force of 7 N was constantly applied on 400 g object at rest, how long will it take to raise its velocity to 80 m/s?

a. 0 s

b. 2.23 s

c. 3.47 s

d. 4.57 s

Given F=7 N,m=400g= .4 kg Acceleration is given by $a= \frac {F}{m}$

a=17.5 m/s

Now u=0,v=80 m/s $v=u + at$

$t=\frac {v-u}{a}$

t=4.57 sec

A sedan car of mass 200kg is moving with a certain velocity . It is brought to rest by the application of brakes, within a distance of 20m when the average resistance being offered to it is 500N.What was the velocity of the motor car?

$F=ma$

or

$a= \frac {F}{m}$

or

a= -500/200=-2.5 m/s

Now

$v^2=u^2 +2as$

Now v=0,s=20 m,a=-2.5 m/s

So, u=10 m/s

A driver accelerates his car first at the rate of 4 m/s

$F_1= ma_1$

and

$F_2=ma_2$

So,Ratio of force exerted is given by

$=\frac {F_1}{F_2}=\frac {ma_1}{ma_2}=\frac {a_1}{a_2}= 1:2$

An object of mass 10 g is sliding with a constant velocity of 2 m/ s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is

(a) 0 N

(b) 5 N

(c) 10 N

(d)20 N

As m=0, F=0

Hence (a) is correct

A cricket ball of mass 0.20 kg is moving with a velocity of 1.2m/s . Find the impulse on the ball and average force applied by the player if he is able to stop the ball in 0.10s?

Impulse= Change in momentum

$I=\Delta p= m \Delta v= .20 \times 1.2=.12$ Kgm/s

Now

Impulse is also defined as

$I=F \times t$

or

$F \times t=.12$

or

$F=\frac {.12}{.10}=1.2$ N

A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find

(i) the acceleration

(ii) distance travelled by car assume motion of car is uniform

(iii) If the mass of the car is 1000 Kg,what is the force acting on it?

Given u=0 ,v= 54 km/hr= 15 m/s , t=2 sec a. Acceleration is given by

$a=\frac {\Delta v}{t}$

So, a=7.5 m/s

b. Distance is given by

$s=ut+ \frac {1}{2}at^2$

s= 15 m

c. Force is given by $F=ma=1000 \times 7.5=7500$ N

A hockey ball of mass .2 Kg travelling at 10 ms

$ \Delta P= m \times (v - u) = 0.2 \times (-2 - 10) = -2.4$ kg ms

(The negative sign indicates a change in direction of hockey ball after it is struck by hockey stick. )

Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 ms

Given $m_1 = 100 gm = 0.1 kg$, $m_2 = 200 gm = 0.2 kg$

$u_1 = 2$ ms

By the law of conservation of momentum,

$m_1 u_1+m_2 u_2=m_1 v_2 + m_2 v_2$

$0.1 \times 2+ 0.2 \times 1=0.1 \times 1.67 + 0.2 v_2$

$v_2 = 1.165$ ms

It will move in the same direction after collision

Anand leaves his house at 8.30 a.m. for his school. The school is 2 km away and classes start at 9.00 a.m. If he walks at a speed of 3 km/h for the first kilometer, at what speed should he walk the second kilometer to reach just in time?

An object of mass 1kg acquires a speed of 10 m/s when pushed forward. What is the impulse given to the object?

Impluse=Change in Momentum= 10 Kgm/s

A bullet of mass 10 gm is fired with an initial velocity of 20 m/s from a rifle of mass 4 kg. Calculate the initial recoil velocity of the rifle?

Let v be the initial recoil velocity of the rifle From law of conservation of Momentum

$0=.01 \times 20 + 4 \times v$

Or $v=-.05 m/s$

Velocity versus time graph of a ball of mass 100 g rolling on a concrete floor is shown below. Calculate the acceleration and the frictional force of the floor on the ball?

From the graph ,we can see that

$\Delta v=-80 m/s$, t=8 sec

Now

$a= \frac {\Delta v}{t} = -10 m/s^2$

Frictional force will be given as

$F= ma= .1 \times -10 = -1 N$

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Class 9 Maths Class 9 Science

Given below are the links of some of the reference books for class 9 science.

These books by S.Chand Publications are detailed in their content and are must have books for class 9 students.- Science for Ninth Class Part1 Physics
- Science for Ninth Class Part2 Chemistry
- Science for Ninth Class Part3 Biology

- Oswaal CBSE Question Bank Class 9 Science Chapterwise & Topicwise We all know importance of practicing what we have studied. This book is perfect for practicing what you have learned and studied in the subject.
- Pearson Foundation Series (IIT-JEE/NEET) Physics, Chemistry, Maths & Biology for Class 9 (Main Books) | PCMB Combo These Foundation books would be helpful for students who want to prepare for JEE/NEET exams. Only buy them if you are up for some challenge and have time to study extra topics. These might be tough for you and you might need extra help in studying these books.
- Foundation Science Physics for Class - 9 by H.C. Verma This is one of my favorite Physics book for class 9. Most of the book is within the limits of CBSE syllabus. It might overwhelm you with its language but I feel if you can understand the content is authentic with plenty of problems to solve.

You can use above books for extra knowledge and practicing different questions.

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