Class 9 Science Chapter 8 Force And Laws Of Motion

Key Concept: Unbalanced Forces, Frictional Force

b) 1.0 m/s$^{2}$

[Solution Description]

The net force acting on the box is the difference between the applied force and the frictional force. So, $F_{net} = F_{applied} – F_{friction} = 30\,N – 20\,N = 10\,N$. According to Newton’s second law, $F_{net} = m \cdot a$, where $m$ is the mass of the box and $a$ is its acceleration. Rearranging for acceleration: $a = \frac{F_{net}}{m} = \frac{10\,N}{10\,kg} = 1\,m/s^2$.

Key Concept: First Law of Motion

d) The net force is zero

[Solution Description]

According to the first law of motion, if an object is moving with a constant velocity, the net force acting on it must be zero. This is because there is no acceleration ($a = 0$), and hence $F = ma = 0$.

Therefore, the net force acting on the object is zero.

Key Concept: Relation Between Mass and Inertia

b) Train, because it has more inertia

[Solution Description] According to Newton’s First Law of Motion, inertia is the tendency of an object to resist changes in its state of motion. The mass of an object is a measure of its inertia. Since the train has a much larger mass compared to the bicycle, it will have greater inertia. Therefore, more force will be required to stop the train than the bicycle.

Key Concept: First Law of Motion, Inertia

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that an object in motion will continue to move with a constant velocity unless acted upon by an external force. This is a direct statement of Newton’s First Law of Motion, which is true. The reason given is that inertia resists any change in the state of motion of the object. Inertia is indeed the property of an object to resist changes in its state of motion, which aligns with Newton’s First Law. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Recoil of a gun

c) 2.5 m/s

[Solution Description]

According to Newton’s Third Law of Motion, the total momentum of the bullets equals the momentum of the gun. The momentum of the first bullet is $0.01 \times 400 = 4 \, \text{kg m/s}$, and the momentum of the second bullet is $0.02 \times 300 = 6 \, \text{kg m/s}$. The total momentum of the bullets is $4 + 6 = 10 \, \text{kg m/s}$. Let $v$ be the recoil velocity of the gun. Then, $4 \times v = 10 \, \Rightarrow \, v = 2.5 \, \text{m/s}$.

Key Concept: Action and Reaction Forces

c) The boat moves backward.

[Solution Description]

According to Newton’s Third Law of Motion, when the sailor exerts a force on the boat by jumping forward, the boat exerts an equal and opposite force on the sailor. This causes the boat to move backward. Therefore, the boat moves in the opposite direction to the sailor’s jump.

Key Concept: Force, Deceleration

c) 4 s

[Solution Description]

To determine the time taken for the object to come to rest, we first calculate the deceleration caused by the opposing force using Newton’s second law: $F = ma$. Given $F = -25$ N (opposing force) and $m = 5$ kg, the deceleration $a$ is:

$a = \frac{F}{m} = \frac{-25}{5} = -5 \text{ m/s}^2$

Using the kinematic equation for final velocity: $v = u + at$, where $v = 0$ (final velocity), $u = 20$ m/s (initial velocity), and $a = -5$ m/s$^2$. Solving for $t$:

$0 = 20 + (-5)t \\ \\ t = \frac{20}{5} = 4 \text{ s}$

Therefore, it will take 4 seconds for the object to come to rest.

Key Concept: Newton’s Third Law of Motion

d) The bullet exerts an equal and opposite force on the gun

[Solution Description] When a gun is fired, it exerts a forward force on the bullet. According to Newton’s Third Law of Motion, the bullet exerts an equal and opposite force on the gun, causing the gun to recoil. This demonstrates the action-reaction pair described by Newton’s Third Law.

Key Concept: Understanding Inertia

b) The tendency of an object to resist changes in its state of motion

[Solution Description]

Inertia is the property of an object that resists any change in its state of motion. It depends on the mass of the object; the greater the mass, the greater the inertia.

Key Concept: Newton’s Third Law of Motion

b) The boat moves backward with a force of 50 N

[Solution Description]

According to Newton’s Third Law of Motion, when the sailor jumps out of the boat with a force of 50 N, the boat experiences an equal and opposite force of 50 N in the opposite direction. This causes the boat to move backward.

Key Concept: Force, Motion, Direction Change

d) 41.23 m/s

[Solution Description]

To solve this problem, we need to consider the effect of the force on both the magnitude and direction of the ball’s velocity.

Step 1: Calculate the acceleration caused by the force.

$F = ma \implies a = \frac{F}{m} = \frac{5 \, \text{N}}{0.5 \, \text{kg}} = 10 \, \text{m/s}^2$

Step 2: Determine the change in velocity in the northward direction.

$\Delta v = a \cdot t = 10 \, \text{m/s}^2 \times 4 \, \text{s} = 40 \, \text{m/s}$

Step 3: The initial velocity in the northward direction is 0 m/s, so the final velocity in the northward direction is 40 m/s.

Step 4: The final velocity in the eastward direction remains unchanged at 10 m/s.

Step 5: Use the Pythagorean theorem to find the magnitude of the final velocity.

$v_{\text{final}} = \sqrt{(10 \, \text{m/s})^2 + (40 \, \text{m/s})^2} = \sqrt{100 + 1600} = \sqrt{1700} \approx 41.23 \, \text{m/s}$

Therefore, the final velocity of the ball is approximately 41.23 m/s, directed northeast.

Key Concept: Balanced and Unbalanced Forces, Friction

b) 1 N to the right

[Solution Description]

To find the net force acting on the block, we need to consider all the forces acting on it. The forces to the right are $F_1$ = 10 N, and the forces to the left are $F_2$ = 6 N and the frictional force = 3 N. The net force is calculated by subtracting the total force to the left from the force to the right:

$\text{Net Force} = F_1 – (F_2 + \text{Friction}) = 10\,\text{N} – (6\,\text{N} + 3\,\text{N}) = 10\,\text{N} – 9\,\text{N} = 1\,\text{N}$

Thus, the net force acting on the block is 1 N to the right.

Key Concept: State of Motion

d) It maintains its state of motion

[Solution Description]

Newton’s first law of motion states that an object will continue to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force. This means that in the absence of such a force, the object will not change its state of motion.

Key Concept: Balanced Forces

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that a wooden block remains at rest when pulled from both sides with equal forces, which is true because balanced forces cancel each other out, resulting in no net force acting on the object. The reason correctly explains that balanced forces do not change the state of rest or motion of an object. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Effect of Force on Elasticity

c) It changes from spherical to oblong

[Solution Description]

When a spherical rubber ball is pressed between two hands, the force applied changes the shape of the ball from spherical to oblong. This deformation occurs because the force acts in opposite directions, causing the material of the ball to redistribute. Once the force is removed, the ball returns to its original shape due to its elasticity, assuming the force was within the elastic limit.

Key Concept: Effect of Force on Motion

c) The box remains stationary

[Solution Description]

When the children push the box with a force that is less than the friction force, the pushing force is balanced by the friction force. Since the forces are balanced, there is no net force acting on the box. According to the concept of balanced forces, the box will not move.

Therefore, the box remains stationary.

Key Concept: Mathematical Formulation of Second Law

b) 25 N

[Solution Description]

To find the force applied, we use Newton’s second law: $F = ma$. First, calculate the acceleration using the formula $a = \frac{v – u}{t}$, where $v = 20$ m/s, $u = 10$ m/s, and $t = 2$ s. Substituting the values, we get $a = \frac{20 – 10}{2} = 5$ m/s$^{2}$. Then, substitute the mass ($m = 5$ kg) and acceleration into the equation for force: $F = 5 \times 5 = 25$ N.

Key Concept: Electrostatic Force

a) $8.44 \times 10^{1} \, \text{N}$

[Solution Description]

The electrostatic force between two charges is given by Coulomb’s law:

$F = k \frac{q_1 q_2}{r^2}$

Here, $q_1 = 3 \times 10^{-6} \, \text{C}$, $q_2 = -5 \times 10^{-6} \, \text{C}$, $r = 4 \times 10^{-2} \, \text{m}$, and $k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$.

Substituting these values into the formula:

$F = 9 \times 10^9 \times \frac{(3 \times 10^{-6}) \times (-5 \times 10^{-6})}{(4 \times 10^{-2})^2}$

Simplifying the equation:

$F = 9 \times 10^9 \times \frac{-15 \times 10^{-12}}{16 \times 10^{-4}}$

$F = 9 \times 10^9 \times \frac{-15}{16} \times 10^{-8}$

$F = -8.4375 \times 10^{1} \, \text{N}$

The magnitude of the electrostatic force is $8.44 \times 10^{1} \, \text{N}$ and the negative sign indicates attraction.

Key Concept: Change in Motion

b) The force applied by the cyclist increases the speed

[Solution Description]

When the cyclist pedals harder, they apply a greater force to the pedals, which transfers to the wheels of the bicycle. According to the relationship between force and acceleration ($F = ma$), an increase in force results in an increase in acceleration if the mass remains constant. Therefore, the bicycle accelerates due to the increased force applied by the cyclist.

Key Concept: Force Calculation

c) 15 m/s

[Solution Description]

To find the final velocity ($v$) of the object, we can use the equation derived from Newton’s Second Law: $v = u + \frac{Ft}{m}$. Here, $u = 0$ m/s (since the object starts from rest), $F = 10$ N, $t = 3$ s, and $m = 2$ kg. Substituting these values, we get $v = 0 + \frac{10 \times 3}{2} = 15$ m/s.

Key Concept: Inertia of Rest

d) The book moves to the left relative to the table

[Solution Description] According to Newton’s First Law of Motion (Law of Inertia), an object at rest tends to stay at rest unless acted upon by an external force. When the table is pulled to the right, the book initially remains in its state of rest due to its inertia. Thus, the book will appear to move to the left relative to the table.

Key Concept: Force, Momentum

c) 25 m/s

[Solution Description]

First, calculate the acceleration using Newton’s second law: $a = \frac{F}{m} = \frac{10}{2} = 5 \, \text{m/s}^2$.

Next, use the equation of motion: $v = u + at$, where $u = 0 \, \text{m/s}$ (initial velocity), $a = 5 \, \text{m/s}^2$, and $t = 5 \, \text{s}$.

So, $v = 0 + 5 \times 5 = 25 \, \text{m/s}$.

Key Concept: Momentum, Force, Newton’s Second Law

c) $40 \, kg \, m/s$

[Solution Description]

To find the final momentum, we first use Newton’s second law to calculate the acceleration:

$F = ma \implies a = \frac{F}{m} = \frac{10}{5} = 2 \, m/s^2$

Next, we find the final velocity using the equation of motion:

$v = u + at = 0 + 2 \times 4 = 8 \, m/s$

Finally, we calculate the final momentum:

$p = mv = 5 \times 8 = 40 \, kg \, m/s$

Thus, the final momentum of the object is $40 \, kg \, m/s$.

Key Concept: Inertia and Rest

c) The forces acting on the book are balanced.

[Solution Description]

Newton’s First Law of Motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the book is at rest on the table because the forces acting on it are balanced. The gravitational force pulling the book downward is counteracted by the normal force exerted by the table upward. Since there is no unbalanced force acting on the book, it remains at rest.

Key Concept: Unbalanced Forces

c) Accelerate

[Solution Description]

An unbalanced force occurs when the net force on an object is not zero. This results in a change in the object’s state of motion, either by changing its speed or direction. Therefore, an unbalanced force causes the object to accelerate in the direction of the greater force.

Key Concept: Inertia of Direction

b) Due to the inertia of direction

[Solution Description]

When the bus suddenly starts moving forward, the person standing in the bus tends to remain at rest due to inertia. This causes the person to feel a backward push as the bus moves forward.

Key Concept: Impulse and Momentum

b) -5 m/s

[Solution Description]

We are given:

– Mass, $m = 10$ kg

– Initial velocity, $u = 5$ m/s

– Force, $F = -50$ N (opposite direction)

– Time, $t = 2$ s

The acceleration $a$ can be calculated as:

$a = \frac{F}{m} = \frac{-50}{10} = -5 \text{ m/s}^2$

The final velocity $v$ can be calculated using the equation:

$v = u + at = 5 + (-5 \times 2) = -5 \text{ m/s}$

Thus, the final velocity of the object is -5 m/s, indicating it is now moving in the opposite direction.

Key Concept: Balanced and Unbalanced Forces, Friction

c) The frictional force is equal to 10 N

[Solution Description]

Since the block does not move, the forces acting on it must be balanced. The applied force of 10 N to the right is balanced by the frictional force of 10 N acting to the left. Therefore, the frictional force is equal to 10 N.

Key Concept: Balanced Forces, Application in Real-World Scenario

b) The forces are equal in magnitude and opposite in direction

[Solution Description] For the car not to move, the net force acting on it must be zero. This means that the forces applied from both sides must be equal or balanced. Hence, the difference between the applied forces must be zero, indicating that the forces are balanced.

Key Concept: Balanced and Unbalanced Forces, Direction of Motion

b) It will move to the right

[Solution Description]

The block will move in the direction of the greater force. Here, the force to the right (20 N) is greater than the force to the left (15 N). Therefore, the block will move to the right.

Key Concept: Force, Momentum, Second Law of Motion

c) 6 m/s

[Solution Description]

First, calculate the final velocity of the 3 kg object after the force is applied using $v = u + \frac{Ft}{m}$. Here, $u = 0$, $F = 12$ N, $t = 4$ s, and $m = 3$ kg. Substituting the values, we get:

$v = 0 + \frac{12 \times 4}{3} = 16 \text{ m/s}$

Now, since the collision is perfectly inelastic, the momentum before the collision equals the momentum after the collision. Let $v_f$ be the final velocity of the combined object system. The total mass after the collision is $3 + 5 = 8$ kg. Using the conservation of momentum:

$3 \times 16 + 5 \times 0 = 8 \times v_f$

Solving for $v_f$:

$v_f = \frac{48}{8} = 6 \text{ m/s}$

Key Concept: Inertia and Mass

b) Stone, because it has more mass

[Solution Description]

A stone of the same size as a football will move less when kicked with equal force because it has more mass and thus more inertia. This makes it harder to change its state of motion, as described in the syllabus.

Key Concept: Newton’s Second Law of Motion

b) 20 N

[Solution Description]

We are given:

Mass ($m$) = $10 kg$,

Initial velocity ($u$) = $5 m s^{-1}$,

Final velocity ($v$) = $11 m s^{-1}$,

Time ($t$) = $3 s$.

The formula for force according to Newton’s second law is:

$F = \frac{m(v – u)}{t}$

Substituting the values:

$F = \frac{10 kg \times (11 m s^{-1} – 5 m s^{-1})}{3 s} = \frac{10 kg \times 6 m s^{-1}}{3 s} = \frac{60 kg m s^{-1}}{3 s} = 20 N$

The magnitude of the force applied is $20 N$.

Key Concept: Balanced and Unbalanced Forces, Net Force

a) 15 N to the right

[Solution Description]

The net force can be calculated using Newton’s second law: $F_{\text{net}} = ma$. Given the mass $m = 5$ kg and acceleration $a = 3$ m/s$^2$, the net force is $F_{\text{net}} = 5 \times 3 = 15$ N. The individual forces acting on the object are $F_1 = 25$ N to the right and $F_2 = 10$ N to the left. The net force without considering acceleration would be $25 – 10 = 15$ N to the right, which matches the calculated net force. Thus, the net force acting on the object is 15 N to the right.

Key Concept: Newton’s Third Law of Motion

c) Both players feel forces that are equal in magnitude but opposite in direction.

[Solution Description]

According to Newton’s Third Law of Motion, when Player A exerts a force on the ball, the ball exerts an equal and opposite force on Player A. Similarly, when the ball hits Player B, Player B exerts a force on the ball, and the ball exerts an equal and opposite force on Player B. Thus, both players experience forces that are equal in magnitude but opposite in direction.

Key Concept: Inertia

c) Due to inertia

[Solution Description]

According to Newton’s First Law of Motion (Law of Inertia), an object in motion tends to stay in motion unless acted upon by an external force. When a moving bus brakes suddenly, your body, due to inertia, tends to continue moving forward at the same speed. Since the bus slows down, your body moves forward until another force (like the seat or handle) stops it.

Key Concept: Conservation of Momentum

b) 24000 kg m/s

[Solution Description]

Initial velocity $u = 90$ km/h = 25 m/s. Final velocity $v = 18$ km/h = 5 m/s. Change in momentum $\Delta p = m(v – u)$. Here, $m = 1200$ kg. So, $\Delta p = 1200(5 – 25) = 1200 \times (-20) = -24000$ kg m/s. The magnitude of the change in momentum is 24000 kg m/s.