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Force & Laws Of Motion Numericals Chapter 9 Class 9 Physics





Given below are the force and laws of motion class 9 numericals
(a) Objective Type Questions
(b) Table Type Questions
(c) Very Short questions
(c) Short questions
(d) Long answer questions
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Objective Type Questions

Question 1
If you apply a net force of 3 N on .1 kg-box, what is the acceleration of the box
(a) 2 m/s2
(b) 30 m/s2
(c) 10 m/s2
(d) None of these.

Answer

Given F=3 N
m=.1 kg
$F=ma$
or
$a = \frac {F}{m}= 30$ m/s2


Question 2
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

Answer

Given F=7 N,m=400g= .4 kg Acceleration is given by $a= \frac {F}{m}$
a=17.5 m/s2
Now u=0,v=80 m/s $v=u + at$
$t=\frac {v-u}{a}$
t=4.57 sec


Question 3
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

Answer

As m=0, F=0
Hence (a) is correct


Question 4
An object of mass 2 kg is sliding with a constant velocity of 8 m/s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is
(a) 16 N
(b) 8 N
(c) 2 N
(d) 0 N

Answer

(d) as there is no friction force



Table type Question

Question 5.
A particle of 10 kg is moving in a constant acceleration 2m/s2 starting from rest. What is its momentum and velocity per the table given below
numericals of force and laws of motion

Answer

Velocity can find using
$v=u+at$
For u=0
$v=at$
Momentum
$P=mv$
force and law of motion class 9 numericals


Very Short answer type

Question 6
How much net force is required to accelerate a 1000 kg car at 4.00 m/s2?

Answer

$F=ma$
Given a=4.00 m/s2
m=1000kg
Therefore , $F= ma= 1000 \times 4 =4000$ N


Question 7
A body of mass 1 kg undergoes a change of velocity of 4m/s in 4s what is the force acting on it?

Answer

Given $\Delta v=4 m/s$ ,t=4 s ,m=1kg
Acceleration is given by $a= \frac {\Delta v}{t}$
a=1 m/s2
Now force is given by
$F=ma$
F=1 N


Question 8
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?

Answer

$F=ma$
or
$a= \frac {F}{m}$
or
a= -500/200=-2.5 m/s2
Now

$v^2=u^2 +2as$
Now v=0,s=20 m,a=-2.5 m/s2
So, u=10 m/s


Question 9
A driver accelerates his car first at the rate of 4 m/s2 and then at the rate of 8 m/s2 .Calculate the ration of the forces exerted by the engines?

Answer

$F_1= ma_1$
and
$F_2=ma_2$
So,Ratio of force exerted is given by
$=\frac {F_1}{F_2}=\frac {ma_1}{ma_2}=\frac {a_1}{a_2}= 1:2$


Question 10
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?

Answer

The impulse experienced by an object is given by the change in momentum. Mathematically, impulse (I) is defined as: \[ I = \Delta p \] where \( \Delta p \) is the change in momentum. The momentum (p) of an object is given by the product of its mass (m) and velocity (v): \[ p = m \cdot v \] So, the change in momentum is: \[ \Delta p = m \cdot \Delta v \] In this case, the cricket ball is stopped, so the final velocity (\( v_f \)) is 0 m/s. Therefore, the change in velocity (\( \Delta v \)) is: \[ \Delta v = v_f - v_i \] where \( v_i \) is the initial velocity. Now, calculate the impulse: \[ I = m \cdot \Delta v \] \[ I = m \cdot (v_f - v_i) \] \[ I = (0.20 \, \text{kg}) \cdot (0 - 1.2 \, \text{m/s}) \] \[ I = -0.24 \, \text{Ns} \] The negative sign indicates that the direction of the impulse is opposite to the initial velocity. Now, to find the average force (\( F \)) applied by the player, you can use the formula: \[ F = \frac{I}{\Delta t} \] where \( \Delta t \) is the time over which the force is applied. In this case, \( \Delta t = 0.10 \, \text{s} \): \[ F = \frac{-0.24 \, \text{Ns}}{0.10 \, \text{s}} \] \[ F = -2.4 \, \text{N} \] The negative sign indicates that the force is applied in the opposite direction to the initial motion of the ball.


Short Answer type

Question 11
A hockey ball of mass .2 Kg travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 2 m/s . Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer

$ \Delta P= m \times (v - u) = 0.2 \times (-2 - 10) = -2.4$ kg ms-1
(The negative sign indicates a change in direction of hockey ball after it is struck by hockey stick. )


Question 12
Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 ms-1 and 1 ms-1 respectively. They collide and after collision, the first object moves at a velocity of 1.67 ms-1. Determine the velocity of the second object?

Answer

Given $m_1 = 100 gm = 0.1 kg$, $m_2 = 200 gm = 0.2 kg$
$u_1 = 2$ ms-1, $u_2 = 1$ ms-1, $v_1 = 1.67$ ms-1, $v_2 = ?$
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-1
It will move in the same direction after collision


Question 13
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 kilometre, at what speed should he walk the second kilometre to reach just in time?

Question 14
An object of mass 1kg acquires a speed of 10 m/s when pushed forward. What is the impulse given to the object?

Answer

Impulse=Change in Momentum= 10 Kgm/s


Question 15
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?

Answer

Let v be the initial recoil velocity of the rifle From law of conservation of Momentum
$0=.01 \times 20 + 4 \times v$
Or $v=-.05 m/s$


Question 16
Which would require a greater force �� accelerating a 2 kg mass at 5 m/s2 or a 6 kg mass at 2 m/s2?

Answer

we have $F = ma$.
Here we have $m_1 = 2$ kg and $a_1 = 5 m/s^2$
and $m_2 = 6$ kg and $a_2 = 2 m/s^2$ .
So,
$F_1 = m_1a_1 = 2 \times 5 = 10 N$
$F_2 = m_2a_2 = 6 \times 2 = 8 N$.
Here $F_1 < F_2$
Thus, accelerating a 6 kg mass at 2 m/s2 would require a greater force


Question 17.
A force of 5 N produces an acceleration of 8 m/s2 on a mass $m_1$ and an acceleration of 24 m/s2 on a mass $m_2$. What acceleration would the same force provide if both the masses are tied together?

Answer

From Force formula
$F=ma$
$5=8m_1$ or $m_1= \frac {5}{8}$
and $5= 24m_2$ or $m_2 = \frac {5}{24}$
Now when the two masses are tied and same force is applied, acceleration will be
$a = \frac {F}{m_1+ m_2} = \frac {5}{\frac {5}{8} + \frac {5}{24}} = 6 m/s^2$


Question 18.
A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. what is the force of the nail on the hammer?

Answer

Acceleration of the Hammer is given by
$a = \frac {v-u}{t} = \frac {0- 50}{.01}= -5000 m/s^2$
Force is given by
$F=ma = .5 \times 5000 = 2500N$


Long Answer type

Question 19
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) The net accelerating force;
(b) The acceleration of the train; and
(c) The force of wagon 1 on wagon 2.

Answer

Total mass, m = mass of engine + mass of wagons
Or, $m = 8000 + 5 \times 2000 = 18000$ kg.
(a) The net accelerating force, F = Engine force - Frictional force
Or, F = 40000 - 5000 = 35000 N
(b) The acceleration of the train
$a= \frac {F}{m}=\frac {35000}{18000}=1.94$ m/s2
(c) The force of wagon 1 on wagon 2
Assuming Frictional Force on all the wagons and engine
Frictional resistance of 4 wagons = $ \frac {5000 \times (2000 \times 4)}{18000} =2222 .22 N$
Accelerating force on 4 wagons =$2000 \times 4 \times 1.944 =15552 N$
Hence total force exertred by wagon on wagon 2= Accelerating force +Frictional resistance
$=2222 .22 +15552=17774.22 N$
Assumining frictional force on engine only
Accelerating force on 4 wagons =$2000 \times 4 \times 1.944 =15552 N$


Question 20.
A bullet of 10 g strikes a sand bag at a speed of 103 m/s and gets embedded after travelling 5 cm. Calculate
(i) the resistive force exerted by the sand on the bullet.
(ii) the time taken by the bullet to come to rest.

Answer

i. u=103 m/s ,v=0 ,s=5 cm= .05 m ,a=?
$v^2=u^2 + 2as$
a=- 107 m/s
Resistive force on bullet = $.001 \times 10^7= 10^5 N$
ii. $v=u+at$
t=104s


Question 21.
Two objects A and B, having mass 100 kg and 75 kg, moving with velocity 40 km/hr and 6 km/hr respectively. Answer the following:
a.Which will have greater inertia?
b.Which will have greater momentum?
c.Which will stop first if equal negative acceleration is applied on both?
d.Which will travel greater distance?
e.Which will impart greater impulse if collides with a wall?

Answer

$M_a=100$ kg, $M_b=75$ kg , $v_a =40 km/hr$, $v_b=6 km/hr$
a. Now $M_a > M_b$, So Object A has more inertia
b. $p_a= M_a v_a= 4000 kg km/hr $, $p_b= M_b v_b= 450 kg km/hr $
Clearly $p_a > p_b$
c. Since velocity of object B is less then velocity of object A, Object B will stops first if equal negative acceleration is applied on both
d. Object A
e. Object A


Question 22
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?

Answer

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/s2

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


Question 23
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?
force and laws of motion class 9 numericals

Answer

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


Question 24.
A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. what is the velocity acquired by the block?

Answer

By using law of conservation of momentum
$m_1u_1 = (m_1 + m_2)v$
$.01 \times 400 = .91 v$
v=4.39 m/s


Question 25.
A man weighing 60 kg runs along the rails with a velocity of 18 km/h and jumps into a car of mass 1 quintal (100 kg) standing on the rails. Calculate the velocity with which car will start travelling along the rails.

Answer

Here m= 60 kg ,u1= 18 km/hr = 5 m/s , M=100 kg ,u2=0
Let v be the velocity with car start travelling
Now
$mu_1 + Mu_2 = (M+m)v$
$ 60 \times 5 = 160 v$
v= 1.875 m/s



Summary

This numericals on force and laws of motion for class 9 is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.


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