Mix Examples - FORCE AND LAWS OF MOTION Questions in English

(N/A) The second law of motion states that, "The rate of change of momentum of an object is directly proportional to the applied unbalanced force and it takes place in the direction of the force." The mathematical expression is $F = ma$.
Since $a = \frac{v-u}{t}$, we have $F = m \frac{v-u}{t}$, which implies $Ft = mv - mu$.
If the external force $F = 0$, then $mv - mu = 0$, which means $mv = mu$, or $v = u$. This implies that if no external force acts on an object, its velocity remains constant (uniform motion). If $u = 0$, then $v = 0$, meaning the object remains at rest. This is the definition of Newton's first law of motion.
$(b)$ Given: $u = 0 \, m \, s^{-1}$, $t = 0.5 \, s$, $g = 10 \, m \, s^{-2}$.
$(i)$ Using $v = u + gt$: $v = 0 + 10 \times 0.5 = 5 \, m \, s^{-1}$.
$(ii)$ Average speed = $\frac{v + u}{2} = \frac{5 + 0}{2} = 2.5 \, m \, s^{-1}$.
$(iii)$ Using $h = ut + \frac{1}{2}gt^2$: $h = 0 + \frac{1}{2} \times 10 \times (0.5)^2 = 5 \times 0.25 = 1.25 \, m$.

(N/A) $(i)$ The property of bodies to resist a change in their state of rest or motion is called Inertia.
$(ii)$ According to Newton's second law of motion,the relationship is $F = ma$,where $F$ is force,$m$ is mass,and $a$ is acceleration.
$(iii)$ The product of mass and velocity of a body is called Momentum $(p = mv)$.
$(iv)$ The rate of change of momentum is equal to the applied Force.
$(v)$ $A$ rocket works on the principle of Newton's third law of motion and the law of conservation of linear momentum.

(N/A) Given: $m_1 = 100\, g = 0.1\, kg$,$m_2 = 200\, g = 0.2\, kg$,$u_1 = 2\, m s^{-1}$,$u_2 = 1\, m s^{-1}$,$v_1 = 1.67\, m s^{-1}$.
Using the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
Substituting the values: $(0.1 \times 2) + (0.2 \times 1) = (0.1 \times 1.67) + (0.2 \times v_2)$.
$0.2 + 0.2 = 0.167 + 0.2 v_2$.
$0.4 = 0.167 + 0.2 v_2$.
$0.2 v_2 = 0.4 - 0.167 = 0.233$.
$v_2 = 0.233 / 0.2 = 1.165\, m s^{-1}$.
$(b)$ This phenomenon is based on Newton's third law of motion and the law of conservation of momentum. When the man jumps forward,he exerts a force on the boat in the backward direction. Since the boat is floating in water and is not fixed,it moves backwards to conserve the total momentum of the system.

(N/A) Newton's third law of motion states that, "To every action, there is always an equal and opposite reaction."
Examples:
$(i)$ Walking: When a person walks, they push the ground backward with their foot (action). The ground exerts an equal and opposite force on the foot (reaction), which pushes the person forward.
$(ii)$ Recoil of a boat: When a person jumps out of a boat, they push the boat backward (action). The boat exerts an equal and opposite force on the person, pushing them forward, while the boat moves backward (reaction).
$(b)$ Using the formula $F = m \times a$:
$(i)$ Given $m = 4 \, kg$ and $a = 2.5 \, m s^{-2}$.
$F = 4 \, kg \times 2.5 \, m s^{-2} = 10 \, N$.
$(ii)$ Given $m = 10 \, kg$ and $a = 3 \, m s^{-2}$.
$F = 10 \, kg \times 3 \, m s^{-2} = 30 \, N$.

(N/A) Momentum is defined as the product of mass and velocity of an object. Its $SI$ unit is $kg \ m s^{-1}$.
$(b)$ The formula for momentum is $p = m v$. Since the momentum $p$ is the same for both,$m_{T} v_{T} = m_{C} v_{C}$. Because the mass of the truck $m_{T}$ is greater than the mass of the car $m_{C}$,the velocity of the car $v_{C}$ must be greater than the velocity of the truck $v_{T}$ to maintain the same momentum.
$(c)$ Given: Mass of bullet $m_{1} = 20 \ g = 0.02 \ kg$,initial velocity $u_{1} = 500 \ m s^{-1}$. Mass of block $m_{2} = 1 \ kg$,initial velocity $u_{2} = 0 \ m s^{-1}$.
Applying the law of conservation of momentum: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$.
$(0.02 \times 500) + (1 \times 0) = (0.02 + 1) v$.
$10 = 1.02 \times v$.
$v = 10 / 1.02 \approx 9.8 \ m s^{-1}$.

(N/A) The force experienced by a mass is given by Newton's second law of motion,$F = ma$,where $a$ is the acceleration calculated from the slope of the $v-t$ graph.
$1$. For portion $OA$:
Acceleration $a = \frac{v_f - v_i}{t_f - t_i} = \frac{20 - 0}{10 - 0} = 2 \ m \ s^{-2}$.
Force $F = ma = 2 \ kg \times 2 \ m \ s^{-2} = 4 \ N$.
$2$. For portion $AB$:
The velocity is constant,so acceleration $a = 0 \ m \ s^{-2}$.
Force $F = ma = 2 \ kg \times 0 \ m \ s^{-2} = 0 \ N$.
$3$. For portion $BC$:
Acceleration $a = \frac{v_f - v_i}{t_f - t_i} = \frac{0 - 20}{40 - 30} = -2 \ m \ s^{-2}$.
Force $F = ma = 2 \ kg \times (-2) \ m \ s^{-2} = -4 \ N$.
The negative sign indicates a retarding force.

(N/A) The rain drop is falling with a constant speed,which means its acceleration $a = 0$. According to Newton's second law,the net force $F = ma = m(0) = 0$. Thus,the net force is zero.
$(b)$ $A$ cork floating on water is in equilibrium. The downward gravitational force (weight) is balanced by the upward buoyant force (upthrust). Therefore,the net force acting on the cork is zero.
$(c)$ Since the kite is held stationary,its velocity is zero and its acceleration is zero. According to Newton's first law,if an object is at rest,the net external force acting on it must be zero.

(A) Given:
Initial velocity, $u = 0 \, m/s$
Distance traveled, $S = 400 \, m$
Time taken, $t = 20 \, s$
Mass of the truck, $m = 7 \, \text{metric tons} = 7000 \, kg$
$(a)$ To find the acceleration $(a)$:
Using the second equation of motion, $S = ut + \frac{1}{2}at^2$
$400 = (0)(20) + \frac{1}{2} \times a \times (20)^2$
$400 = \frac{1}{2} \times a \times 400$
$a = \frac{400 \times 2}{400} = 2 \, m/s^2$
$(b)$ To find the force $(F)$:
Using Newton's second law of motion, $F = m \times a$
$F = 7000 \, kg \times 2 \, m/s^2 = 14000 \, N$

(B) Given: Mass $m = 500 \, g = 0.5 \, kg$,Time $t = 10 \, s$,Force $F = 10^{-2} \, N = 0.01 \, N$,and Initial velocity $u = 0 \, m/s$.
First,calculate the acceleration $a$ using Newton's second law of motion,$F = ma$:
$a = \frac{F}{m} = \frac{0.01 \, N}{0.5 \, kg} = 0.02 \, m/s^2$.
Next,use the second equation of motion to calculate the distance $S$ travelled:
$S = ut + \frac{1}{2}at^2$
$S = 0 \times 10 + \frac{1}{2} \times 0.02 \times (10)^2$
$S = 0 + 0.01 \times 100 = 1 \, m$.
Therefore,the distance travelled by the body is $1 \, m$.

(N/A) Given: Mass $m = 5\, g = 5 \times 10^{-3}\, kg$,Initial velocity $u = 120\, m s^{-1}$,Final velocity $v = 0\, m s^{-1}$,Time $t = 0.01\, s$.
$(a)$ First,calculate the acceleration $a$ using the equation $v = u + at$:
$0 = 120 + a \times 0.01$
$a = -\frac{120}{0.01} = -12000\, m s^{-2}$ (The negative sign indicates retardation).
Now,calculate the distance of penetration $S$ using the equation $S = ut + \frac{1}{2}at^2$:
$S = (120 \times 0.01) + \frac{1}{2} \times (-12000) \times (0.01)^2$
$S = 1.2 - 0.6 = 0.6\, m$.
$(b)$ Calculate the average force $F$ using Newton's second law $F = ma$:
$F = (5 \times 10^{-3}\, kg) \times (12000\, m s^{-2})$
$F = 60\, N$.

(3000 N) Given:
Mass of the car $(m) = 1000 \, kg$
Initial velocity of the car $(u) = 15 \, m/s$
Final velocity of the car $(v) = 0 \, m/s$ (since it comes to rest)
Time taken $(t) = 5 \, s$
Using Newton's second law of motion,the force exerted by the tree on the car is given by:
$F = m \times a = m \times \frac{(v - u)}{t}$
Substituting the values:
$F = 1000 \times \frac{(0 - 15)}{5}$
$F = 1000 \times (-3) = -3000 \, N$
The negative sign indicates that the force exerted by the tree on the car is in the opposite direction of motion.
According to Newton's third law of motion,the force exerted by the car on the tree is equal in magnitude but opposite in direction to the force exerted by the tree on the car.
Therefore,the force exerted by the car on the tree is $3000 \, N$.

(A) Given:
Mass of bullet $m_{B} = 50 \, g = 50 \times 10^{-3} \, kg = 0.05 \, kg$.
Velocity of bullet $v_{B} = 150 \, m s^{-1}$.
Mass of stone $m_{S} = 60 \, kg$.
Velocity of stone $v_{S} = 10 \, m s^{-1}$.
Let $n$ be the number of bullets required to stop the stone.
According to the law of conservation of momentum,the total momentum of the system must be zero to stop the stone.
$n(m_{B} v_{B}) + m_{S} v_{S} = 0$.
Taking the direction of the stone as positive,the bullets are fired in the opposite direction,so their velocity is negative relative to the stone's motion.
$n(0.05 \times 150) + 60 \times (-10) = 0$.
$n(7.5) = 600$.
$n = \frac{600}{7.5} = 80$.
Therefore,$80$ bullets must be fired to stop the stone.

(D) Given: Mass $m = 5 \ kg$,initial velocity $u = 3 \ m s^{-1}$,final velocity $v = 7 \ m s^{-1}$,time $t = 2 \ s$.
Using Newton's second law of motion,$F = m \times a = m \times \frac{(v - u)}{t}$.
Substituting the values: $F = \frac{5 \times (7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \ N$.
Now,for the second part,the force $F = 10 \ N$ is applied for $t' = 5 \ s$.
The acceleration $a = \frac{F}{m} = \frac{10}{5} = 2 \ m s^{-2}$.
Using the first equation of motion,$v' = u + a \times t'$.
$v' = 3 + (2 \times 5) = 3 + 10 = 13 \ m s^{-1}$.

(N/A) Given:
Mass of engine $M_{E} = 3 \times 10^{4} \text{ kg}$
Mass of each wagon $M_{w} = 2 \times 10^{4} \text{ kg}$
Acceleration $a = 0.2 \text{ m s}^{-2}$
$(i)$ The total mass being pulled by the engine is $M_{total} = M_{E} + 2M_{w} = 3 \times 10^{4} + 4 \times 10^{4} = 7 \times 10^{4} \text{ kg}$.
The force exerted by the engine is $F = M_{total} \times a = (7 \times 10^{4} \text{ kg}) \times (0.2 \text{ m s}^{-2}) = 1.4 \times 10^{4} \text{ N}$.
$(ii)$ The force experienced by each wagon is the force required to accelerate that specific wagon. For one wagon,$F_{w} = M_{w} \times a = (2 \times 10^{4} \text{ kg}) \times (0.2 \text{ m s}^{-2}) = 0.4 \times 10^{4} \text{ N} = 4000 \text{ N}$.

(N/A) Initial momentum is calculated as $p = m \times u = 0.5 \times 25 = 12.5\, kg\, m s^{-1}$.
$(b)$ First,find the maximum height $h$ using the equation $v^2 = u^2 - 2gh$. At maximum height,$v = 0$,so $0 = (25)^2 - 2 \times 10 \times h$,which gives $h = 625 / 20 = 31.25\, m$.
The halfway mark is $H = h / 2 = 31.25 / 2 = 15.625\, m$.
Now,find the velocity $v$ at this height using $v^2 = u^2 - 2gH$:
$v^2 = (25)^2 - 2 \times 10 \times 15.625 = 625 - 312.5 = 312.5$.
$v = \sqrt{312.5} \approx 17.68\, m s^{-1}$.
The momentum at the halfway mark is $p = m \times v = 0.5 \times 17.68 = 8.84\, kg\, m s^{-1}$.

(N/A) Given: Mass $m = 5 \, g = 5 \times 10^{-3} \, kg$,initial velocity $u = 120 \, m s^{-1}$,final velocity $v = 0$,time $t = 0.01 \, s$.
$(a)$ To find the distance of penetration $(S)$,we first find the acceleration $(a)$ using the first equation of motion: $v = u + at$.
$0 = 120 + a \times 0.01$
$a = -120 / 0.01 = -12000 \, m s^{-2}$.
(The negative sign indicates retardation).
Now,using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
$S = (120 \times 0.01) + \frac{1}{2} \times (-12000) \times (0.01)^2$
$S = 1.2 - 0.6 = 0.6 \, m$.
$(b)$ The average force $(F)$ exerted on the bullet is calculated using Newton's second law: $F = ma$.
$F = (5 \times 10^{-3} \, kg) \times (12000 \, m s^{-2})$
$F = 60 \, N$ (The force is retarding in nature).

(A) Given:
Mass of man $m_{1} = 60 \ kg$
Initial velocity of man $u_{1} = 18 \ km \ h^{-1} = 18 \times (5/18) \ m \ s^{-1} = 5 \ m \ s^{-1}$
Mass of car $m_{2} = 1 \ quintal = 100 \ kg$
Initial velocity of car $u_{2} = 0 \ m \ s^{-1}$
Let $V$ be the common velocity of the car and the man after the man jumps into the car.
Applying the law of conservation of momentum:
Total momentum after jump = Total momentum before jump
$(m_{1} + m_{2}) V = m_{1} u_{1} + m_{2} u_{2}$
$(60 + 100) V = (60 \times 5) + (100 \times 0)$
$160 V = 300$
$V = 300 / 160 = 1.875 \ m \ s^{-1}$
Thus, the car will start travelling with a velocity of $1.875 \ m \ s^{-1}$.

(N/A) Given: Initial velocity $u = 0 \ m/s$,Distance $S = 800 \ m$,Time $t = 10 \ s$,Mass $m = 800 \ kg$.
Using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
Substituting the values: $800 = (0 \times 10) + \frac{1}{2} \times a \times (10)^2$.
$800 = 0 + \frac{1}{2} \times a \times 100$.
$800 = 50a$.
$a = \frac{800}{50} = 16 \ m/s^2$.
Now,calculating the force using Newton's second law: $F = ma$.
$F = 800 \ kg \times 16 \ m/s^2 = 12800 \ N$.

(N/A) Given: Force $F = -50 \ N$ (retarding force),mass $m = 200 \ kg$,initial velocity $u = 15 \ m \ s^{-1}$,final velocity $v = 0 \ m \ s^{-1}$.
$1$. Calculate acceleration $(a)$:
Using Newton's second law,$F = ma$,so $a = F / m = -50 / 200 = -0.25 \ m \ s^{-2}$.
$2$. Calculate time $(t)$:
Using the first equation of motion,$v = u + at$,we get $0 = 15 + (-0.25)t$.
$0.25t = 15$,therefore $t = 15 / 0.25 = 60 \ s$.
$3$. Calculate distance $(S)$:
Using the second equation of motion,$S = ut + 1/2 at^2$.
$S = (15 \times 60) + 1/2 \times (-0.25) \times (60)^2$.
$S = 900 - 0.125 \times 3600 = 900 - 450 = 450 \ m$.

(A) Given: Mass $m = 80 \, kg$,initial velocity $u = 30 \, m s^{-1}$,final velocity $v = 0 \, m s^{-1}$,time $t = 5 \, s$.
Using the first equation of motion: $v = u + at$
$0 = 30 + a \times 5$
$5a = -30$
$a = -6 \, m s^{-2}$.
The negative acceleration (retardation) is $6 \, m s^{-2}$.
Using Newton's second law of motion,the frictional force $F = m \times a$ (magnitude):
$F = 80 \, kg \times 6 \, m s^{-2} = 480 \, N$.

(FALSE) The statement is False.
Friction is the resistance to motion between two surfaces in contact. In general,the force of friction (specifically fluid friction or drag) is significantly lower in liquids compared to the friction between two solid surfaces. Solids have higher intermolecular forces and surface irregularities that create greater resistance to motion compared to the flow characteristics of liquids.