# Punch numbers about force and the laws of motion. (2023)

Below are the numerical laws of force and motion of class 9.
(a) objective questions
(b) Tabular questions
(c) Very short questions

Movement class 9 part number - 1 |...

Movement type 9 Numeric part - 1 | physical catalyst

(c) Short questions
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## objective type questions

Question 1
If he applies a net force of 3 N to a 0.1 kg box, what is the acceleration of the box?
(a) 2 m/s2
(b) 30 m/s2
c) 10 m/s2
d) None of them.
respondedor

Datos F=3N
m = 0,1 kg
$F=ma$
o
$a = \frac{F}{m}= 30$m/s2

Question 2
If a net force of 7 N is constantly applied to a 400-g object at rest, how long does it take for it to increase its speed to 80 m/s?
El. 0 sec
B 2.23 seconds
w.3.47 seconds
D 4.57 seconds
respondedor

Given F=7 N,m=400g= .4 kgThe acceleration is given by$a= \frac {F}{m}$
a=17,5m/s2
Ágora u=0,v=80 m/s$v=u + at$
$t=\frac{v-u}{a}$
t = 4.57 seconds

Question 3
A body with a mass of 10 g slides with a constant speed of 2 m/s on a horizontal frictionless table. The force required to keep the object moving at the same speed is
(a) 0 north
(b) 5N
(c) 10N
(d) 20N
respondedor

Da m = 0, F = 0
So (a) is correct

question 4
A body with a mass of 2 kg slides with a constant speed of 8 m/s on a horizontal frictionless table. The force required to keep the object moving at the same speed is
(a) 16N
(b) 8N
(c) 2N
(d) 0N
respondedor

## question table type

question 5
A 10 kg particle moves with a constant acceleration of 2 m/s.2starting from rest. What is your swing and speed according to the following table?

respondedor

The speed can be determined with
$v=u+en$
for you = 0
$v=and$
impulses
$P=mv$

question 6
How much net force is required to accelerate a 1000 kg car at 4.00 m/s?2?
respondedor

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

question 7
A body with a mass of 1 kg experiences a change in velocity of 4 m/s in 4 s. What force acts on it?
respondedor

Dado $\Delta v=4 m/s$ ,t=4 s ,m=1kg
The acceleration is given by $a= \frac {\Delta v}{t}$
a=1m/s2
Now the power is given through
$F=ma$
F=1N

question 8
A sedan with a mass of 200 kg moves at a certain speed. It stops braking at a distance of 20 m when the average resistance offered to it is 500 N. How fast did the car go?
respondedor

$F=ma$
o
$a= \frac{F}{m}$
o
a= -500/200=-2,5 m/s2
Now

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

question 9
A driver initially accelerates his car at a speed of 4 m/s.2and then at a speed of 8 m/s2.Calculate the ratio of the forces exerted by the motors?
respondedor

$F_1= em_1$
mi
$F_2=ma_2$
Therefore, the ratio of the 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 with a mass of 0.20 kg is moving at a speed of 1.2 m/s. Find the impulse on the ball and the average force exerted by the player when he is able to stop the ball in 0.10 seconds.
respondedor

Momentum = change in momentum
$I=\Delta p= m \Delta v= .20 \times 1.2=.12$ Kgm/s
Now
Momentum is also defined as
$I=F \vezes t$
o
$F \veces t=.12$
o
$F=\frac {0,12}{0,10}=1,2$ N

question 11
A hockey ball with a mass of 0.2 kg moves at 10 ms.-1is hit by a hockey stick in such a way that it returns to its original trajectory with a speed of 2 m/s. Calculate the change in momentum of the hockey ball's motion due to the force exerted by the hockey stick.
respondedor

$\Delta P= m \times (v - u) = 0,2 \times (-2 - 10) = -2,4$ kg ms-1
(The minus sign indicates a change in the direction of the hockey ball after being hit by the hockey stick.)

question 12
Two objects with masses 100 g and 200 g move along the same line and direction with speeds of 2 ms.-1mi 1 ms-1or they collide and after the collision the first object moves at a speed of 1.67 ms-1. Determine the velocity of the second object?
respondedor

Given $m_1 = 100 g = 0.1 kg$, $m_2 = 200 g = 0.2 kg$
$u_1 = 2$ms-1, $u_2 = 1$ ms-1, $v_1 = 1,67$ms-1, $v_2 = ?$
According to 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 the collision.

question 13
Anand leaves home at 8:30 for school. The school is 2 km away and classes start at 9 in the morning. If he walks the first kilometer at a speed of 3 km/h, how fast does he have to cover the second kilometer to be just on time?

question 14
A body with a mass of 1 kg reaches a speed of 10 m/s when it is pushed forward. What impulse does the object give?
respondedor

Impulse = change in momentum = 10 kg/s

question 15
A bullet with a mass of 10 g is fired from a rifle with a mass of 4 kg with a muzzle velocity of 20 m/s. Calculate the initial recoil velocity of the rifle?
respondedor

Let v be the initial recoil velocity of the weapon. According to the law of conservation of momentum
$0=0.01 \times 20 + 4 \times v$
You $v=-.05 m/s$

question 16
Which would require a greater force to accelerate a 2 kg mass at 5 m/s?2or a mass of 6 kg at 2 m/s2?
respondedor

we have $F = ma$.
Here we have $m_1 = 2$ kg and $a_1 = 5 m/s^2$
e $m_2 = 6$ kg e $a_2 = 2 m/s^2$ .
So,
$F_1 = m_1a_1 = 2 \times 5 = 10 N$
$F_2 = m_2a_2 = 6 \times 2 = 8 N$.
Here $F_1 < F_2$
Then accelerate a mass of 6 kg with 2 m/s2would require more force

question 17.
A force of 5 N generates an acceleration of 8 m/s2with a mass $m_1$ and an acceleration of 24 m/s2in a mass $m_2$. What acceleration would give the same force if both masses were attached?
respondedor

From the formula of Force
$F=ma$
$5=8m_1$ o $m_1= \frac {5}{8}$
y $5= 24m_2$ o $m_2 = \frac {5}{24}$
Now, if the two masses are connected and the same force is exerted, the 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 with a mass of 500 g moving at 50 m/s strikes a nail. The nail hits the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
respondedor

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

question 19
An 8000 kg locomotive pulls a train of 5 carriages of 2000 kg each on a horizontal track. If the motor exerts a force of 40,000 N and the conveyor belt exerts a frictional force of 5,000 N, calculate:
(a) the resultant acceleration force;
(b) the acceleration of the train; Is
(c) The force of car 1 on car 2.
respondedor

Total mass, m = mass of the engine + mass of the wagon
Ou, $m = 8000 + 5 \times 2000 = 18000$ kg.
(a) The net acceleration force, F = engine force - friction force
You, 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 car 1 on car 2
Assume friction force on all cars and engines.
4-car friction drag = $\frac {5000 \times (2000 \times 4)}{18000} =2222 .22 N$
G-force on 4 cars = $2000 \times 4 \times 1.944 = N$15552
Therefore, the total force exerted by the car on the car 2 = force of acceleration + frictional resistance
$=2222 0,22 +15552=17774.22 N$
Only a frictional force on the motor is assumed.
G-force on 4 cars = $2000 \times 4 \times 1.944 = N$15552

question 20.
A 10-g bullet hits a sandbag with a velocity of 103m/s and embeds after a movement of 5 cm. calculation
(i) the drag force that the sand exerts on the bullet.
(ii) the time it takes for the ball to come to rest.
respondedor

I. you=103m/s ,v=0 ,s=5 cm= 0,05 m ,a=?
$v^2=u^2 + 2as$
a=- 107EM
Drag the ball = $0.001 \times 10^7= 10^5 N$
ii. $v=u+en$
t=104S

question 21.
Two objects A and B with masses 100 kg and 75 kg are moving at speeds of 40 km/h and 6 km/h, respectively. Answer the following:
a. Which have greater inertia?
b.Which will have more momentum?
W. Which stops first when the same negative acceleration is applied to both?
d.Which will cover a greater distance?
e. Which gives more momentum when hitting a wall?
respondedor

$M_a=100$kg, $M_b=75$kg, $v_a=40km/h$, $v_b=6km/h$
El. Now $M_a > M_b$, so object A has more inertia
B. $p_a= M_a v_a= 4000 kg km/h$, $p_b= M_b v_b= 450 kg km/h$
Clearly $p_a > p_b$
W. Since the velocity of object B is less than the velocity of object A, object B will stop first if both are subjected to the same negative acceleration.
D. Object A
Is. Object A

question 22
A car starts from a standstill and reaches a speed of 54 km/h in 2 s. Meet
(e) acceleration
(ii) the distance traveled by the car assumes that the motion of the car is uniform
(iii) If the mass of the car is 1000 kg, what is the force acting on it?
respondedor

Given u=0, v= 54 km/h= 15 m/s, t=2 drought. The acceleration is given by
$a=\frac{\Delta v}{t}$
Also a = 7.5 m/s2

B. The distance is given by
$s=ut+ \frac{1}{2}a^2$
s = 15m

W. The force is given by $F=ma=1000 \times 7.5=7500$ N

question 23
Below is the graph of velocity versus time for a 100-g ball rolling on a concrete floor. Calculate the acceleration and the force of friction on the ball from the ground?

respondedor

We can see that in the graph
$\Delta v=-80 m/s$, t=8 seconds
Now
$a= \frac{\Delta v}{t} = -10 m/s^2$
The friction force is given as
$F= ma= .1 \times -10 = -1 N$

question 24.
A 10-g ball moving at a speed of 400 m/s is trapped in a freely floating block of wood with a mass of 900 g. What speed does the block detect?
respondedor

Application of the law of conservation of momentum
$m_1u_1 = (m_1 + m_2)v$
$.01 \ times 400 = .91 en$
v=4,39 m/s

question 25.
A 60 kg man runs down the tracks at a speed of 18 km/h and jumps into a car with a mass of 100 kg parked on the tracks. He calculates the speed at which the car begins to move on the tracks.
respondedor

Hier m = 60 kg,u1= 18 km/h = 5 m/s , M=100 kg ,u2=0
Let v be the speed at which the car starts moving.
Now
$mu_1 + Mu_2 = (M+m)v$
$60 \times 5 = 160 v$
v = 1,875 m/s

## Summary

These Grade 9 Force and Laws of Motion figures were created using the latest CBSE syllabus. This is designed to improve student academic performance. If you find errors, please provide feedback in the email.

• Walnuts
• Forces and Laws of Motion Grade 9 Notes
• assignments
• Numerically on force and the laws of motion Lecture 9
• Additional Questions Lesson 9 Force, Momentum, and Laws of Motion

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