English
Karnataka Board PUCPUC Science Class 11

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 ______. - Physics

Advertisements
Advertisements

Question

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 ______.

Options

  • frictional force along westward.

  • muscle force along southward.

  • frictional force along south-west.

  • muscle force along south-west.

MCQ
Fill in the Blanks
Advertisements

Solution

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 + rtBy 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.

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Laws of Motion - Exercises [Page 30]

APPEARS IN

NCERT Exemplar Physics [English] Class 11
Chapter 5 Laws of Motion
Exercises | Q 5.6 | Page 30

RELATED QUESTIONS

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.


Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to

  1. A,
  2. B along the direction of string. What is the tension in the string in each case?

A car accelerates on a horizontal road due to the force exerted by.


A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block by another string. Find the tensions in the two strings. Take g = 10 m/s2.


Two blocks A and B of mass mA and mB , respectively, are kept in contact on a frictionless table. The experimenter pushes block A from behind, so that the blocks accelerate. If block A exerts force F on block B, what is the force exerted by the experimenter on block A?


A particle of mass 0.3 kg is subjected to a force F = −kx with k = 15 N/m. What will be its initial acceleration if it is released from a point x = 20 cm?


The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?


In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (In the following figure), m1 = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds; (b) find the tension in the string; (c) find the force exerted by the clamp on the pulley.


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 500 g block in the following figure.


Write the mathematical form of Newton's second law of motion. State the conditions if any.


State the magnitude and direction of the force of gravity acting on the body of mass 5 kg. Take g = 9.8 m s-2.


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)


A pebble is thrown vertically upwards with a speed of 20 m s-1. How high will it be after 2 s? (Take g = 10 m s-2)


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.


Which of the following has the largest inertia?


What do you mean by the conservation of momentum? Briefly, explain the collision between two bodies and the conservation of momentum.


In the previous problem (5.3), the magnitude of the momentum transferred during the hit is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×