(i) Explain why
A horse cannot pull a cart and run-in empty space
(ii) Explain why
Passengers are thrown forward from their seats when a speeding bus stops suddenly,
(iii) Explain why
It is easier to pull a lawn mower than to push it,
(iv) Explain why
A cricketer moves his hands backwards while holding a catch.
A horse cannot pull a cart and run-in empty space
(ii) Explain why
Passengers are thrown forward from their seats when a speeding bus stops suddenly,
(iii) Explain why
It is easier to pull a lawn mower than to push it,
(iv) Explain why
A cricketer moves his hands backwards while holding a catch.
(i) Step 1: Newton’s 3rd law of motion.
While trying to pull a cart, a horse pushes the ground backwards with some force.
The ground, in turn, exerts an equal and opposite reaction force upon the feet of the horse.
This reaction force causes the horse to move forward.
An empty space is devoid of any such reaction force.
Therefore, a horse cannot pull a cart and run-in empty space.
(ii)Step 1: Apply law of Inertia
Whenever a fast-moving bus stop suddenly, the lower portion of a passenger’s body, which is in contact with
the seat, suddenly comes to rest but the upper portion of the passenger’s body tends to remain in motion due
to inertia (first law).
As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.
(iii) Step 1: Force equilibrium
Let the force is applied at an angle θ, in pulling and pushing both conditions.
While pulling a lawn mower, the vertical component of applied force acts upwards. This reduces the effective weight of the mower.
Effective weight = mg−F sinθ
While pushing a lawn mower, the vertical component of applied force acts downward. This increases the effective weight of the mower.
Effective weight =mg+F sinθ
Since the effective weight of the lawn mower is lesser in first case,
Therefore, pulling the lawn mower is easier than pushing it.
(iv)Step 1: Newton’s second law of motion
The force applied by ball on the hand is given by, F=(mv−mu)Δt
Where, F= Stopping force experienced by the cricketer as he catches the ball.
m= Mass of the ball
u= Velocity of the ball before hitting the hand
v= Velocity of the ball after catch is taken =0
Δt= Time of impact of the ball with the hand.
We can see,
F∝1Δt
Therefore, if time of impact is increased, the impact force experienced by the cricketer’s hand will be lesser. While taking a catch, a cricketer moves his hand backwards to increase the time of impact (Δt) . This is turn results in the decrease in the stopping force, thereby preventing the hands from getting hurt.
While trying to pull a cart, a horse pushes the ground backwards with some force.
The ground, in turn, exerts an equal and opposite reaction force upon the feet of the horse.
This reaction force causes the horse to move forward.
An empty space is devoid of any such reaction force.
Therefore, a horse cannot pull a cart and run-in empty space.
(ii)Step 1: Apply law of Inertia
Whenever a fast-moving bus stop suddenly, the lower portion of a passenger’s body, which is in contact with
the seat, suddenly comes to rest but the upper portion of the passenger’s body tends to remain in motion due
to inertia (first law).
As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.
(iii) Step 1: Force equilibrium
Let the force is applied at an angle θ, in pulling and pushing both conditions.
While pulling a lawn mower, the vertical component of applied force acts upwards. This reduces the effective weight of the mower.
Effective weight = mg−F sinθ
While pushing a lawn mower, the vertical component of applied force acts downward. This increases the effective weight of the mower.
Effective weight =mg+F sinθ
Since the effective weight of the lawn mower is lesser in first case,
Therefore, pulling the lawn mower is easier than pushing it.
(iv)Step 1: Newton’s second law of motion
The force applied by ball on the hand is given by, F=(mv−mu)Δt
Where, F= Stopping force experienced by the cricketer as he catches the ball.
m= Mass of the ball
u= Velocity of the ball before hitting the hand
v= Velocity of the ball after catch is taken =0
Δt= Time of impact of the ball with the hand.
We can see,
F∝1Δt
Therefore, if time of impact is increased, the impact force experienced by the cricketer’s hand will be lesser. While taking a catch, a cricketer moves his hand backwards to increase the time of impact (Δt) . This is turn results in the decrease in the stopping force, thereby preventing the hands from getting hurt.
