Advertisements
Advertisements
प्रश्न
Explain why a cricketer moves his hands backwards while holding a catch.
Advertisements
उत्तर
According to Newton’s second law of motion, we have the equation of motion:
F = ma = `m (trianglev)/(trianglet)` ....(i)
Where
F = Stopping force experienced by the cricketer as he catches the ball
m = Mass of the ball
Δt = Time of impact of the ball with the hand
From equation (i), it is deducible that there is an inverse proportionality between the impact force and the impact time, i.e.
`F prop 1/(trianglet)` ......(ii)
Equation (ii) demonstrates that as the time of impact increases, the force experienced by the cricketer decreases and vice versa.
During a catch, a cricketer pulls his hands backward to lengthen the time of impact (Δt). This leads to a reduction in the stopping force, which helps protect the cricketer's hands from harm.
APPEARS IN
संबंधित प्रश्न
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.
Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.
Find the mass M of the hanging block in the following figure that will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
In the following figure shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling by a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight recorded on the machine? What force should he exert on the rope to record his correct weight on the machine?
Show that the rate of change of momentum = mass × acceleration. Under what condition does this relation hold?
A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : The velocity acquired by the body
A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate : The magnitude of the force
A bullet of mass 50 g moving with an initial velocity 100 m s-1 strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate: (i) Initial momentum of the bullet, (ii) Final momentum of the bullet, (iii) Retardation caused by the wooden block and (iv) Resistive force exerted by the wooden block.
A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. (Take g = 10 m s−2)
Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.
A motorcycle of mass 100 kg is running at 10 ms−1. If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of a motorcycle.
State two factors which determine the momentum of a body.
What causes motion in a body?
Why is it advantageous to turn before taking a long jump?
A stone is dropped from a cliff 98 m high.
What will be its speed when it strikes the ground?
A stone is thrown vertically upward with a velocity of 9.8 m/s. When will it reach the ground?
A ball is thrown vertically downward with an initial velocity of 10 m/s. What is its speed 1 s later and 2 s later?
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
According to Newton's Second Law of Motion, what quantity is directly proportional to the applied force?
