Textbook - FORCE AND LAWS OF MOTION Questions in English

(N/A) Inertia is the measure of the mass of an object. The greater the mass of the object,the greater is its inertia,and vice-versa.
$(a)$ $A$ stone has more mass than a rubber ball of the same size. Therefore,the stone has more inertia.
$(b)$ $A$ train has significantly more mass than a bicycle. Therefore,the train has more inertia.
$(c)$ $A$ five-rupee coin has more mass than a one-rupee coin. Therefore,the five-rupee coin has more inertia.

(N/A) The velocity of the ball changes $4$ times.
$1$. The first player kicks the ball: The velocity changes from $0$ to a certain value. The agent is the first player.
$2$. The second player kicks the ball towards the goal: The direction changes, hence velocity changes. The agent is the second player.
$3$. The goalkeeper collects the ball: The ball comes to rest, so velocity changes from a certain value to $0$. The agent is the goalkeeper.
$4$. The goalkeeper kicks the ball to his teammate: The velocity changes from $0$ to a certain value. The agent is the goalkeeper.

(N/A) When we shake the branch of a tree vigorously,the branch comes into motion. However,the leaves tend to remain in their state of rest due to the inertia of rest. Because the leaves resist the change in their state of motion,they experience a force that causes them to detach and fall from the tree.

(A) This phenomenon occurs due to the $inertia$ of the passenger.
$1$. When a bus is moving, the passenger is also in a state of motion. When the driver applies brakes, the lower part of the passenger's body, which is in contact with the bus, comes to rest. However, the upper part of the body continues to be in a state of motion due to the $inertia$ of motion, causing the passenger to fall forward.
$2$. When the bus accelerates from rest, the lower part of the passenger's body starts moving with the bus, but the upper part tends to remain at rest due to the $inertia$ of rest. This causes the passenger to fall backward relative to the bus.

(B) According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
When a horse pulls a cart,it pushes the ground backward with its feet at an angle.
This action force exerted by the horse on the ground results in an equal and opposite reaction force exerted by the ground on the horse in the forward direction.
This reaction force allows the horse to move forward,and since the cart is attached to the horse,the cart is pulled along with it.

(N/A) The difficulty arises due to the backward reaction force of the water being ejected.
When a fireman holds a hose that ejects a large volume of water at a high velocity,the water exerts an equal and opposite reaction force on the hose in the backward direction,according to $Newton's$ third law of motion.
This backward force pushes the fireman,making it difficult for him to maintain his balance and stability while holding the hose.

(C) Mass of the rifle,$m_1 = 4\, kg$.
Mass of the bullet,$m_2 = 50\, g = 0.05\, kg$.
Initial velocity of the bullet,$v_2 = 35\, m/s$.
Let the recoil velocity of the rifle be $v_1$.
Initially,the system is at rest,so the initial momentum is $0$.
According to the law of conservation of momentum,the total momentum after firing must be equal to the total momentum before firing.
$m_1 v_1 + m_2 v_2 = 0$.
$4(v_1) + (0.05 \times 35) = 0$.
$4 v_1 + 1.75 = 0$.
$4 v_1 = -1.75$.
$v_1 = -1.75 / 4 = -0.4375\, m/s$.
The negative sign indicates that the rifle recoils in the direction opposite to the bullet,with a magnitude of $0.4375\, m/s$.

(D) Mass of the first object,$m_1 = 100 \, g = 0.1 \, kg$.
Mass of the second object,$m_2 = 200 \, g = 0.2 \, kg$.
Velocity of $m_1$ before collision,$u_1 = 2 \, m/s$.
Velocity of $m_2$ before collision,$u_2 = 1 \, m/s$.
Velocity of $m_1$ after collision,$v_1 = 1.67 \, m/s$.
Let the velocity of $m_2$ after collision be $v_2$.
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$(0.1 \times 2) + (0.2 \times 1) = (0.1 \times 1.67) + (0.2 \times v_2)$
$0.2 + 0.2 = 0.167 + 0.2 v_2$
$0.4 = 0.167 + 0.2 v_2$
$0.2 v_2 = 0.4 - 0.167$
$0.2 v_2 = 0.233$
$v_2 = 0.233 / 0.2 = 1.165 \, m/s$.
Thus,the velocity of the second object after the collision is $1.165 \, m/s$.

(N/A) Yes. According to Newton's First Law of Motion,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.
If the net external unbalanced force is zero,the object will maintain its state of motion.
Therefore,if the object is already moving with a non-zero velocity,it will continue to move with the same constant velocity (i.e.,constant magnitude and constant direction) in a straight line.
Thus,it is possible for an object to travel with a non-zero velocity under zero net external force,provided the velocity remains constant in both magnitude and direction.

(N/A) Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick,the carpet is set into motion. However,the dust particles present in the carpet tend to resist this change and remain in their state of rest due to the inertia of rest. According to Newton's first law of motion,an object at rest remains at rest unless acted upon by an external force. As the carpet moves away,the dust particles remain behind due to their inertia,causing them to fall out of the carpet.

(N/A) When the bus accelerates and moves forward,it acquires a state of motion. However,the luggage kept on the roof,owing to its inertia,tends to remain in its state of rest. Hence,with the forward movement of the bus,the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this,it is advised to tie any luggage kept on the roof of a bus with a rope.

(D) batsman hits a cricket ball,which then rolls on a level ground. After covering a short distance,the ball comes to rest because there is a frictional force acting on the ball that opposes its motion.
Frictional force always acts in the direction opposite to the direction of motion. Hence,this force is responsible for slowing down and eventually stopping the cricket ball.

$(A)$ Initial velocity of the truck,$u = 0 \, m/s$.
Time taken,$t = 20 \, s$.
Distance covered,$s = 400 \, m$.
According to the second equation of motion,$s = ut + \frac{1}{2}at^2$.
Substituting the values: $400 = 0 \times 20 + \frac{1}{2} \times a \times (20)^2$.
$400 = 200a$.
Therefore,acceleration $a = 2 \, m/s^2$.
Mass of the truck,$m = 7 \, \text{tonnes} = 7000 \, kg$.
Force is given by $F = m \times a$.
$F = 7000 \times 2 = 14000 \, N$.

(B) Initial velocity of the stone,$u = 20\, m/s$.
Final velocity of the stone,$v = 0\, m/s$.
Distance covered by the stone,$s = 50\, m$.
According to the third equation of motion,$v^2 = u^2 + 2as$.
Substituting the values: $(0)^2 = (20)^2 + 2 \times a \times 50$.
$0 = 400 + 100a$.
$100a = -400$.
$a = -4\, m/s^2$.
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone,$m = 1\, kg$.
From Newton's second law of motion,Force $F = m \times a$.
$F = 1\, kg \times (-4\, m/s^2) = -4\, N$.
Hence,the force of friction between the stone and the ice is $-4\, N$.

(C) Force exerted by the engine,$F = 40000 \,N$.
Frictional force offered by the track,$F_f = 5000 \,N$.
The net accelerating force is the difference between the applied force and the opposing frictional force.
Net accelerating force,$F_a = F - F_f$.
$F_a = 40000 \,N - 5000 \,N = 35000 \,N$.
Hence,the net accelerating force is $35000 \,N$.

(D) Total mass of the train $(M)$ = Mass of engine + Mass of $5$ wagons.
Mass of engine = $8000 \,kg$.
Mass of $5$ wagons = $5 \times 2000 \,kg = 10000 \,kg$.
Total mass $(M)$ = $8000 + 10000 = 18000 \,kg$.
Net force $(F_{net})$ = Applied force - Frictional force.
$F_{net} = 40000 \,N - 5000 \,N = 35000 \,N$.
Using Newton's second law of motion,$F_{net} = M \times a$.
$a = \frac{F_{net}}{M} = \frac{35000}{18000} \,m/s^2$.
$a = 1.944 \,m/s^2$.

(A) Mass of the automobile vehicle,$m = 1500 \, kg$.
Acceleration of the automobile,$a = -1.7 \, m \, s^{-2}$.
From Newton's second law of motion,the force $F$ is given by the product of mass and acceleration:
$F = m \times a$
$F = 1500 \, kg \times (-1.7 \, m \, s^{-2})$
$F = -2550 \, N$.
Therefore,the force between the vehicle and the road is $-2550 \, N$,acting in the direction opposite to the motion of the vehicle.

(B) The momentum of an object is defined as the product of its mass and its velocity.
Given:
Mass of the object $= m$
Velocity of the object $= v$
Therefore, Momentum $(p) = \text{Mass} \times \text{Velocity} = mv$.

(C) When an object moves at a constant velocity,the net force acting on it is zero.
This means that the applied force must be perfectly balanced by the opposing frictional force.
Given that the applied horizontal force is $200\, N$,the frictional force must also be $200\, N$ in the opposite direction to maintain a constant velocity.
Therefore,the friction force exerted on the cabinet is $200\, N$.

(D) Mass of one object,$m_1 = 1.5\, kg$.
Mass of the other object,$m_2 = 1.5\, kg$.
Velocity of $m_1$ before collision,$v_1 = 2.5\, m/s$.
Velocity of $m_2$ before collision,$v_2 = -2.5\, m/s$ (since it moves in the opposite direction).
According to the law of conservation of momentum,the total momentum before collision equals the total momentum after collision.
$m_1v_1 + m_2v_2 = (m_1 + m_2)v$
Substituting the values:
$1.5(2.5) + 1.5(-2.5) = (1.5 + 1.5)v$
$3.75 - 3.75 = 3v$
$0 = 3v$
$v = 0\, m/s$.
Thus,the velocity of the combined object after the collision is $0\, m/s$.

(N/A) The truck has a large mass,which results in a very high static friction between the tires and the road.
To move the truck,the applied force must exceed the maximum static frictional force.
When a person pushes the truck and it does not move,the applied force is balanced by the static frictional force acting in the opposite direction.
Therefore,the student's logic is correct in the context that the net force on the truck is zero because the applied force and the frictional force cancel each other out.

(B) Mass of the hockey ball,$m = 200 \,g = 0.2 \,kg$.
Initial velocity of the hockey ball,$u = 10 \,m \,s^{-1}$.
Initial momentum,$p_i = m \times u = 0.2 \,kg \times 10 \,m \,s^{-1} = 2 \,kg \,m \,s^{-1}$.
Since the ball returns along its original path,the final velocity is in the opposite direction,$v = -5 \,m \,s^{-1}$.
Final momentum,$p_f = m \times v = 0.2 \,kg \times (-5 \,m \,s^{-1}) = -1 \,kg \,m \,s^{-1}$.
Change in momentum,$\Delta p = p_f - p_i = -1 \,kg \,m \,s^{-1} - 2 \,kg \,m \,s^{-1} = -3 \,kg \,m \,s^{-1}$.
The magnitude of the change in momentum is $|\Delta p| = 3 \,kg \,m \,s^{-1}$.

(C) Given:
Initial velocity of the bullet,$u = 150\, m/s$
Final velocity of the bullet,$v = 0\, m/s$ (as it comes to rest)
Time taken,$t = 0.03\, s$
Mass of the bullet,$m = 10\, g = 0.01\, kg$
Using the first equation of motion,$v = u + at$:
$0 = 150 + a(0.03)$
$a = -150 / 0.03 = -5000\, m/s^2$
Using the second equation of motion to find the distance of penetration $(s)$:
$s = ut + (1/2)at^2$
$s = (150 \times 0.03) + (1/2 \times -5000 \times (0.03)^2)$
$s = 4.5 - (2500 \times 0.0009)$
$s = 4.5 - 2.25 = 2.25\, m$
Using Newton's second law of motion to find the force $(F)$:
$F = ma$
$F = 0.01\, kg \times 5000\, m/s^2 = 50\, N$
Thus,the distance of penetration is $2.25\, m$ and the force exerted is $50\, N$.

(D) Mass of the object,$m_1 = 1\, kg$.
Velocity of the object before collision,$v_1 = 10\, m/s$.
Mass of the stationary wooden block,$m_2 = 5\, kg$.
Velocity of the wooden block before collision,$v_2 = 0\, m/s$.
Total momentum before collision $= m_1 v_1 + m_2 v_2 = 1(10) + 5(0) = 10\, kg\, m\, s^{-1}$.
According to the law of conservation of momentum,the total momentum remains conserved.
Therefore,total momentum after collision $= 10\, kg\, m\, s^{-1}$.
Total mass of the combined system $= m_1 + m_2 = 1 + 5 = 6\, kg$.
Let the velocity of the combined object be $v$.
$10 = (m_1 + m_2)v$
$10 = 6v$
$v = 10/6 = 5/3\, m/s$.

(D) Initial velocity of the object, $u = 5 \,m \,s^{-1}$.
Final velocity of the object, $v = 8 \,m \,s^{-1}$.
Mass of the object, $m = 100 \,kg$.
Time taken, $t = 6 \,s$.
Initial momentum $= m \times u = 100 \times 5 = 500 \,kg \,m \,s^{-1}$.
Final momentum $= m \times v = 100 \times 8 = 800 \,kg \,m \,s^{-1}$.
Force exerted on the object, $F = \frac{mv - mu}{t}$.
$F = \frac{800 - 500}{6} = \frac{300}{6} = 50 \,N$.
Thus, the initial momentum is $500 \,kg \,m \,s^{-1}$, the final momentum is $800 \,kg \,m \,s^{-1}$, and the force exerted is $50 \,N$.

(C) According to the law of conservation of momentum, the total momentum of the system remains conserved during the collision.
$1$. Kiran's suggestion: The insect experiences a much larger change in velocity compared to the car. Since $p = mv$, the change in momentum $(\Delta p = m \Delta v)$ for the insect is significant relative to its mass. Because the car's mass is massive, its change in velocity is negligible, meaning the insect experiences a greater change in momentum relative to its own state, but the magnitude of momentum change for both is equal.
$2$. Akhtar's suggestion: Akhtar is partially correct that the car exerts a large force, but the force exerted by the car on the insect is equal to the force exerted by the insect on the car (Newton's $3^{rd}$ Law).
$3$. Rahul's suggestion: Rahul is correct. According to Newton's $3^{rd}$ Law, the force exerted by the car on the insect is equal and opposite to the force exerted by the insect on the car. Since the time of impact is the same for both, the change in momentum $(F \times t = \Delta p)$ is equal in magnitude for both the car and the insect.

(C) Given: Mass of the dumbbell,$m = 10\, kg$.
Distance covered,$s = 80\, cm = 0.8\, m$.
Acceleration,$a = 10\, m \,s^{-2}$.
Initial velocity,$u = 0\, m \,s^{-1}$.
Using the third equation of motion,$v^2 = u^2 + 2as$:
$v^2 = 0^2 + 2 \times 10 \times 0.8$
$v^2 = 16$
$v = 4\, m \,s^{-1}$.
Momentum $(p)$ is given by the product of mass and velocity: $p = m \times v$.
$p = 10\, kg \times 4\, m \,s^{-1} = 40\, kg\, m \,s^{-1}$.
Therefore,the dumbbell will transfer $40\, kg\, m \,s^{-1}$ of momentum to the floor.

(N/A) The distance covered by the object in equal intervals of time is unequal (it follows the cube of time, $d = t^3$). This indicates non-uniform motion. Since the velocity (rate of change of distance) is increasing at an increasing rate, the acceleration is increasing.
$(b)$ According to Newton's second law of motion, $F = ma$. Since the acceleration $(a)$ of the object is increasing over time, the net force $(F)$ acting on the object must also be increasing.

(A) Mass of the motorcar $(m)$ $= 1200 \, kg$.
When two persons push the car at a uniform velocity,the net external force is zero. This means the force applied by the two persons is exactly balanced by the frictional force $(f)$.
So,$2F = f$,where $F$ is the force applied by each person.
When three persons push the car,the total applied force is $3F$. The net force $(F_{net})$ causing acceleration is the total applied force minus the frictional force:
$F_{net} = 3F - f = 3F - 2F = F$.
According to Newton's second law of motion,$F_{net} = m \times a$.
Given $a = 0.2 \, m \, s^{-2}$,we have:
$F = 1200 \, kg \times 0.2 \, m \, s^{-2} = 240 \, N$.
Therefore,each person pushes the motorcar with a force of $240 \, N$.

(B) Mass of the hammer,$m = 500 \,g = 0.5 \,kg$.
Initial velocity of the hammer,$u = 50 \,m/s$.
Time taken by the nail to stop the hammer,$t = 0.01 \,s$.
Final velocity of the hammer,$v = 0 \,m/s$ (since the hammer comes to rest).
According to Newton's second law of motion,the force $F$ exerted on the hammer is given by $F = m \times a = m \times \frac{v - u}{t}$.
Substituting the values: $F = 0.5 \times \frac{0 - 50}{0.01} = 0.5 \times (-5000) = -2500 \,N$.
The negative sign indicates that the force exerted by the nail on the hammer is in the opposite direction of the hammer's motion.
According to Newton's third law of motion,the magnitude of the force exerted by the nail on the hammer is $2500 \,N$.

(C) Mass of the motor car,$m = 1200\, kg$.
Initial velocity,$u = 90\, km/h = 90 \times (5/18) = 25\, m/s$.
Final velocity,$v = 18\, km/h = 18 \times (5/18) = 5\, m/s$.
Time taken,$t = 4\, s$.
Acceleration $a = (v - u) / t = (5 - 25) / 4 = -20 / 4 = -5\, m/s^2$.
Change in momentum = $m(v - u) = 1200 \times (5 - 25) = 1200 \times (-20) = -24000\, kg\, m/s$.
Force $F = m \times a = 1200 \times (-5) = -6000\, N$.
The magnitude of the force required is $6000\, N$.