Kinetic and Potential Energy Questions in English

(B) Kinetic energy is the energy possessed by a body by virtue of its motion.
Every moving object possesses kinetic energy.
$A$ body uses kinetic energy to perform work.
For example,the kinetic energy of a hammer is used to drive a nail into a log of wood,and the kinetic energy of air is used to run windmills.

(N/A) If a body of mass $m$ is moving with a velocity $v$,then its kinetic energy $E_k$ is given by the expression:
$E_k = \frac{1}{2}mv^2$
Its $SI$ unit is Joule $(J)$.

(D) The expression for kinetic energy is $E_k = 1/2 mv^2$.
Given that the initial kinetic energy $E_k = 25\, J$ at a velocity $v = 5\, m\, s^{-1}$.
$(i)$ When the velocity is doubled,the new velocity becomes $v' = 2v = 10\, m\, s^{-1}$.
Since $E_k \propto v^2$,the new kinetic energy $E_k' = 1/2 m(2v)^2 = 4 \times (1/2 mv^2) = 4 \times 25\, J = 100\, J$.
$(ii)$ When the velocity is increased three times,the new velocity becomes $v'' = 3v = 15\, m\, s^{-1}$.
Similarly,the new kinetic energy $E_k'' = 1/2 m(3v)^2 = 9 \times (1/2 mv^2) = 9 \times 25\, J = 225\, J$.

(D) Given:
Mass $(m)$ $= 20 \, kg$
Initial velocity $(u)$ $= 5 \, m \, s^{-1}$
Final velocity $(v)$ $= 2 \, m \, s^{-1}$
According to the Work-Energy Theorem, the work done by a force is equal to the change in kinetic energy of the object.
Work done $(W)$ $= \Delta KE = KE_{final} - KE_{initial}$
$W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
$W = \frac{1}{2} m (v^2 - u^2)$
Substituting the values:
$W = \frac{1}{2} \times 20 \times (2^2 - 5^2)$
$W = 10 \times (4 - 25)$
$W = 10 \times (-21)$
$W = -210 \, J$
Since the question asks for the magnitude of work done by the force (which is a retarding force), the absolute value is $210 \, J$.

(B) We know that,potential energy $(PE) = mgh$.
Given:
Mass $(m) = 40\, kg$
Acceleration due to gravity $(g) = 9.8\, m/s^2$
Height $(h) = 5\, m$
Potential energy $= 40 \times 9.8 \times 5 = 1960\, J$.
According to the law of conservation of energy,the total mechanical energy remains constant.
When the object is half-way down,it has fallen a distance of $2.5\, m$. At this point,half of the initial potential energy is converted into kinetic energy.
Therefore,kinetic energy $= 1/2 \times 1960\, J = 980\, J$.

(C) The object is in motion,so its initial kinetic energy is $K_i = 1/2 mv^2$.
When the object comes to rest,its final velocity becomes $0$,so its final kinetic energy is $K_f = 0$.
According to the work-energy theorem,the work done on an object is equal to the change in its kinetic energy.
Work done $W = K_f - K_i$.
$W = 0 - 1/2 mv^2 = -1/2 mv^2$.
The negative sign indicates that the work is done against the direction of motion to bring the object to rest. The magnitude of the work required is $1/2 mv^2$.

(A) Mass of the car $m = 1500 \, kg$.
Velocity of the car $v = 60 \, km/h$.
Convert velocity to $m/s$: $v = \frac{60 \times 1000}{3600} = \frac{50}{3} \, m/s$.
The kinetic energy $(KE)$ of the car is given by the formula $KE = \frac{1}{2} mv^2$.
$KE = \frac{1}{2} \times 1500 \times (\frac{50}{3})^2 \, J$.
$KE = 750 \times \frac{2500}{9} \, J$.
$KE = 750 \times 277.777... \, J = 208333.3 \, J$.
According to the work-energy theorem, the work required to stop the car is equal to the change in its kinetic energy.
Work done $= KE_{initial} - KE_{final} = 208333.3 \, J - 0 \, J = 208333.3 \, J$.
Thus, the work required to stop the car is $208333.3 \, J$.

(D) The energy stored in an object due to its position or configuration is known as potential energy.
Since the water in a dam is stored at a certain height,it possesses gravitational potential energy due to its position relative to the ground.

(A) According to the Law of Conservation of Energy,the total mechanical energy of a freely falling body remains constant.
At height $h$,the potential energy $(PE)$ is $mgh$ and kinetic energy $(KE)$ is $0$.
After falling a distance of $\frac{h}{2}$,the remaining height is $\frac{h}{2}$.
The potential energy at this point is $PE = mg(\frac{h}{2}) = \frac{1}{2}mgh$.
Since the total energy is $mgh$,the kinetic energy must be $KE = mgh - \frac{1}{2}mgh = \frac{1}{2}mgh$.
Therefore,at half the height,the body possesses half potential energy and half kinetic energy.

(B) The formula for kinetic energy $(K.E)$ is given by $K.E = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity.
Let the initial velocity be $v_1 = v$. The initial kinetic energy is $K_1 = \frac{1}{2}mv^2$.
When the velocity is tripled,the new velocity becomes $v_2 = 3v$.
The final kinetic energy is $K_2 = \frac{1}{2}m(3v)^2 = \frac{1}{2}m(9v^2) = 9 \times (\frac{1}{2}mv^2)$.
Therefore,the ratio of the initial kinetic energy to the final kinetic energy is $K_1 : K_2 = \frac{1}{2}mv^2 : 9(\frac{1}{2}mv^2) = 1:9$.

(A) Yes,an object can have mechanical energy even if its momentum is zero.
Mechanical energy is the sum of kinetic energy $(KE)$ and potential energy $(PE)$.
Momentum $(p)$ is defined as the product of mass $(m)$ and velocity $(v)$,i.e.,$p = mv$.
If momentum is zero $(p = 0)$,then the velocity of the object must be zero $(v = 0)$.
Since kinetic energy is given by $KE = \frac{1}{2}mv^2$,if $v = 0$,then $KE = 0$.
However,the object can still possess potential energy due to its position or configuration (e.g.,a ball held at a height above the ground).
Therefore,the total mechanical energy $(ME = KE + PE)$ can be non-zero because $PE$ is non-zero.

(C) The initial potential energy of the ball at height $h = 10\, m$ is given by $PE = mgh$.
Substituting the values,$PE = m \times 10\, m/s^2 \times 10\, m = 100m\, J$.
After striking the ground,the energy reduces by $40\%$,which means the remaining energy is $100\% - 40\% = 60\%$ of the initial energy.
Remaining energy = $0.60 \times 100m = 60m\, J$.
Let the new height reached by the ball be $h'$. The potential energy at this height is $PE' = mgh'$.
Equating the remaining energy to the new potential energy: $mgh' = 60m$.
$m \times 10 \times h' = 60m$.
$h' = 60 / 10 = 6\, m$.
Thus,the ball can bounce back to a height of $6\, m$.

(N/A) The relationship between kinetic energy $(K.E.)$ and momentum $(p)$ is given by $K.E. = \frac{p^2}{2m}$.
Given that both objects have the same momentum $(p_1 = p_2 = p)$,
Let $m_1$ be the mass of the light object and $m_2$ be the mass of the heavy object,where $m_1 < m_2$.
The kinetic energy of the light object is $(K.E.)_1 = \frac{p^2}{2m_1}$.
The kinetic energy of the heavy object is $(K.E.)_2 = \frac{p^2}{2m_2}$.
The ratio of their kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{\frac{p^2}{2m_1}}{\frac{p^2}{2m_2}} = \frac{m_2}{m_1}$.
Since $m_1 < m_2$,it follows that $\frac{m_2}{m_1} > 1$,which implies $(K.E.)_1 > (K.E.)_2$.
Therefore,the light object has a larger kinetic energy.

(B) stretched bow possesses potential energy due to a change in its shape (elastic potential energy).
When the string is released,this stored potential energy is transferred to the arrow.
As the bow returns to its original shape,the potential energy is converted into the kinetic energy of the arrow,causing it to move.

(A) The kinetic energy $(KE)$ of a body is given by the formula $KE = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity.
Given that the kinetic energy of both bodies is equal,we have $\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2$.
This implies $m_1v_1^2 = m_2v_2^2$,or $\frac{v_1^2}{v_2^2} = \frac{m_2}{m_1}$.
If $m_1 < m_2$ (light body has less mass than the heavy body),then $\frac{m_2}{m_1} > 1$.
Therefore,$\frac{v_1^2}{v_2^2} > 1$,which means $v_1^2 > v_2^2$,or $v_1 > v_2$.
Thus,the lighter body must be moving faster to possess the same kinetic energy as the heavier body.

(B) When a watch spring is wound,work is done on it,which causes a change in its shape. This work is stored in the spring in the form of elastic potential energy. As the spring gradually unwinds,this stored energy is converted into kinetic energy to move the gears of the watch.

(B) The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2}mv^2$.
Given: Mass $(m)$ = $1\, kg$,Kinetic Energy $(KE)$ = $1\, J$.
Substituting the values into the formula: $1 = \frac{1}{2} \times 1 \times v^2$.
$v^2 = 2$.
$v = \sqrt{2} \approx 1.414\, m/s$.
Therefore,the body will have a kinetic energy of $1\, J$ at a speed of approximately $1.414\, m/s$.

(B) rolling stone is in motion. The energy possessed by an object due to its motion is known as kinetic energy. Therefore,a rolling stone possesses kinetic energy.

(B) running athlete is in motion. The energy possessed by an object due to its motion is known as kinetic energy. Therefore,a running athlete possesses kinetic energy.

(B) The kinetic energy $(KE)$ of an object is given by the formula $KE = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity.
Since both the car and the truck are moving with the same velocity $(v)$,the kinetic energy is directly proportional to the mass $(m)$ of the object $(KE \propto m)$.
Given that the mass of the truck is greater than the mass of the car,the truck will have more kinetic energy.

(N/A) The kinetic energy $(KE)$ of a body is given by the formula $KE = \frac{1}{2}mv^2$.
As the body is thrown vertically upwards,its velocity $(v)$ decreases due to the gravitational force acting downwards.
At the highest point,the velocity of the body becomes zero $(v = 0)$.
Consequently,the kinetic energy at the highest point becomes $KE = \frac{1}{2}m(0)^2 = 0$.
Therefore,the change in kinetic energy is equal to the initial kinetic energy $(KE_{initial} - 0 = KE_{initial})$,which is entirely converted into potential energy at that point.

(B) As the ball moves vertically upwards,its kinetic energy is converted into potential energy due to the work done against gravity.
At the highest point,the velocity of the ball becomes zero,meaning the kinetic energy is zero.
According to the law of conservation of energy,the total mechanical energy remains constant.
Therefore,at the point where velocity is zero,the potential energy reaches its maximum value.

(A) The expression for kinetic energy is $KE = \frac{1}{2} mv^{2}$.
Since the kinetic energy is directly proportional to the mass $(KE \propto m)$ when the speed $(v)$ is constant,the object with the greater mass will have more kinetic energy.
Given that the mass of the horse $(210\, kg)$ is greater than the mass of the dog $(25\, kg)$,the horse possesses more kinetic energy.

(N/A) Potential energy is the energy possessed by an object by virtue of its position or configuration.
The expression for gravitational potential energy is $PE = mgh$,where $m$ is the mass of the object,$g$ is the acceleration due to gravity,and $h$ is the height above the reference level.
The $SI$ unit of potential energy is the joule $(J)$.

(A) Kinetic energy $(K)$ is defined by the relation $K = \frac{1}{2} m v^{2}$.
To relate this to momentum $(p = mv)$,we multiply and divide the right-hand side by $m$:
$K = \frac{1}{2} \frac{m^2 v^2}{m}$
Since $p = mv$,we can substitute $p^2$ for $m^2 v^2$:
$K = \frac{p^2}{2m}$
Rearranging this to solve for momentum,we get $p = \sqrt{2mK}$.

(N/A) When we wind our watch,we perform work on the spring inside the watch. As a result,this energy is stored in the spring in the form of elastic potential energy. This stored elastic potential energy is gradually released to power the mechanism of the watch throughout the day.

(B) The relationship between momentum $(p)$,mass $(m)$,and kinetic energy $(K)$ is given by the formula $p = \sqrt{2mK}$.
Since the kinetic energy $(K)$ is the same for both bodies,the momentum is directly proportional to the square root of the mass $(p \propto \sqrt{m})$.
Therefore,the body with the larger mass will have a greater momentum.
Thus,the heavy body will have more momentum than the light body.

(A) The relationship between kinetic energy $(K)$,momentum $(p)$,and mass $(m)$ is given by the formula $K = \frac{p^2}{2m}$.
Since the momentum $(p)$ is the same for both bodies,the kinetic energy is inversely proportional to the mass $(K \propto \frac{1}{m})$.
Therefore,the body with the smaller mass (the lighter body) will have a larger denominator,resulting in a greater value for kinetic energy.
Thus,the lighter body has more kinetic energy than the heavier body.

(N/A) Let the initial velocity of the body be $u$ and the final velocity be $v$ after traveling a distance $S$ under the influence of a constant force $F$.
According to the third equation of motion, $v^{2} - u^{2} = 2aS$, where $a$ is the acceleration.
From this, the distance $S$ can be expressed as $S = \frac{v^{2} - u^{2}}{2a}$.
According to Newton's second law of motion, the force applied is $F = ma$.
The work done $W$ by the force is defined as $W = F \times S$.
Substituting the values of $F$ and $S$ into the work equation:
$W = (ma) \times \left( \frac{v^{2} - u^{2}}{2a} \right)$.
Simplifying the expression by canceling $a$:
$W = \frac{1}{2}m(v^{2} - u^{2}) = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$.
Since kinetic energy $K = \frac{1}{2}mv^{2}$, the expression becomes $W = K_{final} - K_{initial} = \Delta K$.
Thus, the work done by the force is equal to the change in the kinetic energy of the body.

(N/A) Potential energy is the energy possessed by a body by virtue of its position or configuration. It is a scalar quantity because it has only magnitude and no direction.
$(b)$ $A$ stretched string of a bow is an example of a body having potential energy due to its change in configuration.

(N/A) In a loudspeaker,electrical energy is transformed into sound energy.
$(b)$ In a solar battery,light energy is transformed into electrical energy.

(N/A) Yes,a body at rest has no momentum,i.e.,$p = 0$. However,it can possess potential energy $(U)$ due to its position or configuration. Therefore,its total energy $E = K + U = U$ is not zero. For example,a stone lying on a roof or a wound-up spring of a clock.
$(b)$ No,a body cannot have momentum without having energy. Momentum $(p = mv)$ implies that the body is in motion $(v \neq 0)$. If a body is in motion,it must possess kinetic energy $(K = \frac{1}{2}mv^2)$. Since kinetic energy is a component of total energy,a body with momentum must possess energy.

(D) $(i)$ The energy possessed by an object due to its position or configuration is called potential energy. The $SI$ unit of potential energy is joule $(J)$.
$(ii)$ Given: mass of the body $(m) = 50 \, kg$,height $(h) = 10 \, m$,and acceleration due to gravity $(g) = 10 \, m s^{-2}$.
The formula for potential energy $(PE)$ is $PE = m \times g \times h$.
Substituting the values: $PE = 50 \, kg \times 10 \, m s^{-2} \times 10 \, m = 5000 \, J$.
Therefore,the potential energy of the body is $5000 \, J$.

(N/A) The more the bow is bent, the greater is the elastic potential energy it will possess. Arjun might have told Nakul to bend the bow more so that it has more elastic potential energy. When the arrow is released, this energy is converted into the kinetic energy of the arrow. As a result, it would have moved faster and a longer distance.
$(b)$ The values imbibed by Arjun are knowledge, helpfulness, and patience.

(A) Let the two objects have masses $m_1$ and $m_2$ such that $m_2 > m_1$. Let their common momentum be $p$.
The kinetic energy $K$ is related to momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.
For the first object (lighter),$K_1 = \frac{p^2}{2m_1}$.
For the second object (heavier),$K_2 = \frac{p^2}{2m_2}$.
The ratio of their kinetic energies is $\frac{K_1}{K_2} = \frac{p^2 / 2m_1}{p^2 / 2m_2} = \frac{m_2}{m_1}$.
Since $m_2 > m_1$,it follows that $\frac{m_2}{m_1} > 1$,which implies $K_1 > K_2$.
Therefore,the lighter object has a larger kinetic energy.

(N/A) Potential energy is the energy stored in an object due to its position or configuration. Gravitational potential energy of an object at a height $h$ above the ground is defined as the work done in raising it from the ground to that point against gravity.
For an object of mass $m$ raised through a height $h$, the work done $W$ is given by:
$W = \text{force} \times \text{displacement} = (m \times g) \times h = mgh$
Thus, the potential energy $PE = mgh$.
$(b)$ Given:
Mass $m = 10 \, kg$
Height $h = 6 \, m$
Acceleration due to gravity $g = 9.8 \, m s^{-2}$
Potential energy $PE = mgh = 10 \times 9.8 \times 6 = 588 \, J$.

(N/A) No,the total energy of the body is conserved. As a body falls down,its height from the ground decreases,and hence its potential energy decreases. Simultaneously,its velocity and kinetic energy increase. The decrease in potential energy is exactly equal to the gain in kinetic energy. Thus,the sum of potential energy and kinetic energy (mechanical energy) remains constant at any instant of time.
$(b)$
$(i)$ Potential energy is maximum at the initial height $h$,where the object is at rest.
$(ii)$ Kinetic energy is maximum at the ground level,just before the object hits the surface,as the velocity is highest at this point.

(N/A) flying bird possesses potential energy because of its height above the ground level,which is defined with respect to the Earth's surface.
It also possesses kinetic energy due to the velocity with which it is flying through the air.
The expression for potential energy $(PE)$ is: $PE = mgh$,where $m$ is the mass of the bird,$g$ is the acceleration due to gravity,and $h$ is the height above the ground.
The expression for kinetic energy $(KE)$ is: $KE = \frac{1}{2}mv^2$,where $m$ is the mass of the bird and $v$ is its velocity.

(2:3) Kinetic energy is the energy possessed by an object due to its motion.
$(b)$ Given the ratio of masses $\frac{m_1}{m_2} = \frac{2}{3}$ and the velocity $v_1 = v_2 = 108 \, km/h$.
The formula for kinetic energy is $KE = \frac{1}{2}mv^2$.
Since the velocity is the same for both,the ratio of kinetic energies is $\frac{KE_1}{KE_2} = \frac{\frac{1}{2}m_1v^2}{\frac{1}{2}m_2v^2} = \frac{m_1}{m_2}$.
Therefore,the ratio of their kinetic energy is $2:3$.

(N/A) Kinetic energy: The energy possessed by a body by virtue of its motion is called kinetic energy.
Potential energy: The energy possessed by a body by virtue of its position or configuration is called potential energy.
The expression for kinetic energy is $KE = \frac{1}{2} mv^2$.
Given:
Mass $(m)$ = $10 \, kg$
Velocity $(v)$ = $10 \, m s^{-1}$
Calculation:
$KE = \frac{1}{2} \times 10 \times (10)^2$
$KE = 5 \times 100$
$KE = 500 \, J$
The kinetic energy of the stone is $500 \, J$.

(D) $(i)$ Consider a body of mass $m$ initially at rest,i.e.,$u = 0$,on a frictionless surface. Let a constant force $F$ act on the body,producing an acceleration $a$. After a displacement $S$,the body attains a velocity $v$. Using the equation of motion $v^2 - u^2 = 2aS$,we get $a = v^2 / (2S)$.
The work done $W$ by the force is $W = F \times S$. Since $F = ma$,we have $W = (ma) \times S = m \times (v^2 / 2S) \times S = 1/2 mv^2$.
This work done is stored as kinetic energy $(KE)$. Thus,$KE = 1/2 mv^2$. The $SI$ unit of kinetic energy is Joule $(J)$.
$(ii)$ Given: Mass $m = 10 \, kg$,velocity $v = 5 \, m s^{-1}$.
$KE = 1/2 mv^2 = 1/2 \times 10 \times (5)^2 = 5 \times 25 = 125 \, J$.

(N/A) Given: Mass of car $m_1 = 750 \,kg$,Velocity $v = 54 \,km \cdot h^{-1} = 54 \times \frac{5}{18} \,m \cdot s^{-1} = 15 \,m \cdot s^{-1}$.
Initial Kinetic Energy $(KE_1)$:
$KE_1 = \frac{1}{2} m_1 v^2 = \frac{1}{2} \times 750 \times (15)^2 = 375 \times 225 = 84375 \,J$.
When a passenger of mass $50 \,kg$ sits in the car,the new total mass $m_2 = 750 \,kg + 50 \,kg = 800 \,kg$.
New Kinetic Energy $(KE_2)$:
$KE_2 = \frac{1}{2} m_2 v^2 = \frac{1}{2} \times 800 \times (15)^2 = 400 \times 225 = 90000 \,J$.

(D) Let the mass of each body be $m$.
The kinetic energy $(KE)$ of the first body is given by $K_{1} = \frac{1}{2}mv^{2} \quad \dots(i)$.
The kinetic energy $(KE)$ of the second body is given by $K_{2} = \frac{1}{2}m(3v)^{2} = \frac{1}{2}m(9v^{2}) = \frac{9}{2}mv^{2} \quad \dots(ii)$.
To find the ratio of their kinetic energies,we divide equation $(ii)$ by equation $(i)$:
$\frac{K_{2}}{K_{1}} = \frac{\frac{9}{2}mv^{2}}{\frac{1}{2}mv^{2}} = \frac{9}{1}$.
Thus,the ratio of their kinetic energies is $9:1$.

(A) Given: Mass,$m = 2 \, kg$,Velocity,$v = 0.1 \, m s^{-1}$.
The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v^2$.
Substituting the given values into the formula:
$KE = \frac{1}{2} \times 2 \times (0.1)^2$
$KE = 1 \times 0.01$
$KE = 0.01 \, J$.
Therefore,the kinetic energy of the body is $0.01 \, J$.

(B) Given: Mass of the body,$m = 100 \, g = 0.1 \, kg$.
Kinetic energy,$KE = 20 \, J$.
We know that the formula for kinetic energy is $KE = \frac{1}{2} m v^2$.
Rearranging the formula to solve for velocity $v$:
$v^2 = \frac{2 \times KE}{m}$.
Substituting the values:
$v^2 = \frac{2 \times 20}{0.1} = \frac{40}{0.1} = 400$.
Taking the square root on both sides:
$v = \sqrt{400} = 20 \, m s^{-1}$.

(B) The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the object.
$1$. If we double the mass $(m' = 2m)$,the new kinetic energy becomes $KE' = \frac{1}{2}(2m)v^2 = 2 \times KE$. Thus,the kinetic energy doubles.
$2$. If we double the velocity $(v' = 2v)$,the new kinetic energy becomes $KE'' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times KE$. Thus,the kinetic energy increases four times.
Comparing the two,doubling the velocity has a greater effect on the kinetic energy than doubling the mass.

(A) Given: Mass $m = 120 \, kg$,initial velocity $u = 25 \, m s^{-1}$,final velocity $v = 40 \, m s^{-1}$.
According to the work-energy theorem,the work done $W$ on an object is equal to the change in its kinetic energy.
$W = \Delta KE = \frac{1}{2} m (v^2 - u^2)$
Substituting the values:
$W = \frac{1}{2} \times 120 \times (40^2 - 25^2)$
$W = 60 \times (1600 - 625)$
$W = 60 \times 975$
$W = 58500 \, J$
Thus,the work done is $58500 \, J$.

(B) Initial potential energy of the ball at height $h = 10 \, m$ is $E = mgh$.
After striking the ground,the energy reduces by $40 \%$,which means the ball retains $60 \%$ of its initial energy.
Energy retained $= 0.60 \times mgh$.
When the ball bounces back to a new height $h'$,its potential energy at that height will be $mgh'$.
Equating the retained energy to the new potential energy: $mgh' = 0.60 \times mgh$.
Canceling $mg$ from both sides,we get $h' = 0.60 \times h$.
Substituting $h = 10 \, m$,we get $h' = 0.60 \times 10 \, m = 6 \, m$.
Therefore,the ball bounces back to a height of $6 \, m$.

(N/A) Given: Mass $m = 10 \, kg$,Height $h = 20 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$.
$1$. Kinetic Energy $(KE)$ at the ground:
According to the law of conservation of energy,the potential energy $(PE)$ at the top is converted into kinetic energy $(KE)$ at the ground.
$PE = mgh = 10 \times 10 \times 20 = 2000 \, J$.
Therefore,$KE = 2000 \, J$.
$2$. Speed $(v)$ just before hitting the ground:
Using the formula $KE = \frac{1}{2} mv^2$:
$2000 = \frac{1}{2} \times 10 \times v^2$
$2000 = 5 \times v^2$
$v^2 = 400$
$v = \sqrt{400} = 20 \, m s^{-1}$.

(N/A) Given: Mass of car $m_{c} = 750\, kg$,Velocity $v = 54\, km h^{-1} = 54 \times \frac{5}{18} = 15\, m s^{-1}$.
Initial Kinetic Energy $(KE_{i})$ = $\frac{1}{2} m_{c} v^{2} = \frac{1}{2} \times 750 \times (15)^{2} = 375 \times 225 = 84375\, J$.
When a passenger of mass $50\, kg$ sits in the car,the new total mass $M = 750 + 50 = 800\, kg$.
New Kinetic Energy $(KE_{f})$ = $\frac{1}{2} M v^{2} = \frac{1}{2} \times 800 \times (15)^{2} = 400 \times 225 = 90000\, J$.