Advertisements
Advertisements
प्रश्न
A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is ______.
विकल्प
frictional force along westward.
muscle force along southward.
frictional force along south-west.
muscle force along south-west.
Advertisements
उत्तर
A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is frictional force along south-west.
Explanation:
According to Newton’s second law of motion, only external forces can change the linear momentum of the system. The internal forces cannot change the linear momentum of system under consideration. If we take hockey players as a system, the external force which can change the direction of motion of the player is the force that must be friction between the ground and the shoes of player. The muscle force is the internal force, this cannot change the linear momentum of the player.
According to Newton’s Second Law, The rate of change of linear momentum of a body is equal to the external force applied on the body or F = dp/dt. So, the external force must be in the direction of change in momentum.
As shown in the diagram,
Let OA = p1
= Initial momentum of player northward
AB = p2
= Final momentum of the player towards west
Clearly OB = OA + AB
According to the problem, mass = 2 kg
The position of the particle is given here as a function of time, x(t) = pt + qt2 + rt3 By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
v = dx/dt = p + 2 qt + 3 rt2
If we again differentiate this equation w.r.t. time, we will get an acceleration of the particle as a function of time.
a = dv/dt = 0 + 2q + 6rt
At t = 2 s; a = 2q + 6 × 2 × r
= 2 q + 12 r
= 2 × 4 + 12 × 5
= 8 + 60
= 68 m/s
Force = F = ma
= 2 × 68
= 136 N
Change in momentum = p2 – p1
= AB – OA
= AB + (– OA)
= Clearly the change in momentum is OR will be along southwest, so the direction of force will also be along southwest.
APPEARS IN
संबंधित प्रश्न
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
An object is placed far away from all the objects that can exert force on it. A frame of reference is constructed by taking the origin and axes fixed in this object. Will the frame be necessarily inertial?
A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in the following figure. Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth h.
A force \[\vec{F} = \vec{v} \times \vec{A}\] is exerted on a particle in addition to the force of gravity, where \[\vec{v}\] is the velocity of the particle and \[\vec{A}\] is a constant vector in the horizontal direction. With what minimum speed, a particle of mass m be projected so that it continues to move without being defelected and with a constant velocity?
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.
In the following figure, m1 = 5 kg, m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.
Consider the situation shown in the following figure. Both the pulleys and the string are light and all the surfaces are frictionless.
- Find the acceleration of the mass M.
- Find the tension in the string.
- Calculate the force exerted by the clamp on the pulley A in the figure.
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 acceleration of the blocks A and B in the three situations shown in the following figure.
State Newton's second law of motion. Under what condition does it take the form F = ma?
A body of mass 200 g is moving with a velocity of 5 ms−1. If the velocity of the body changes to 17 ms−1, calculate the change in linear momentum of the body.
Why is it advantageous to turn before taking a long jump?
State Newton's second law of motion. Is Newton's first law of motion contained in Newton's second law of motion?
A stone is dropped from a cliff 98 m high.
How long will it take to fall to the foot of the cliff?
A stone is thrown vertically upward with a velocity of 9.8 m/s. When will it reach the ground?
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)
A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 ms−1, q = 4 ms−2 and r = 5 ms−3. The force acting on the body at t = 2 seconds is ______.
