English
Karnataka Board PUCPUC Science Class 11

A Force → F = → V × → a → V is the Velocity of the Particle and → a → a is a Constant Vector in the Horizontal Direction. with What Minimum Speed, a Particle of Mass M Be Projected

Advertisements
Advertisements

Question

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? 

Sum
Advertisements

Solution

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.
\[\vec{F} + m \vec{g} = 0\]
\[\Rightarrow \left( \vec{v} \times \vec{A} \right) + \vec{mg} = 0\]
\[ \Rightarrow \left( \vec{v} \times \vec{A} \right) = - \vec{mg}\]
\[\left| vA\sin\theta \right| = \left| mg \right|\] 
\[\therefore v = \frac{mg}{A\sin\theta}\]
v will be minimum when sinθ = 1.
⇒ θ = 90°
\[\therefore v_{\text{min}} = \frac{mg}{A}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Newton's Laws of Motion - Exercise [Page 80]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 5 Newton's Laws of Motion
Exercise | Q 21 | Page 80

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.


A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.


A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.


A spy jumps from an airplane with his parachute. The spy accelerates downward for some time when the parachute opens. The acceleration is suddenly checked and the spy slowly falls to the ground. Explain the action of the parachute in checking the acceleration.


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?


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.


Define linear momentum and state its S.I. unit.


State Newton's second law of motion. Under what condition does it take the form F = ma?


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)


A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.


Multiple Choice Question. Select the correct option.

The impulse of a body is equal to:


What causes motion in a body?


What do you mean by an impulsive force?


What do you mean by linear momentum of a body? A force causes an acceleration of 10 ms-2 in a body of mass 1 kg. What acceleration will be caused by the same force in a body of mass 4 kg?


A stone is dropped from a cliff 98 m high.
How long will it take to fall to the foot of the cliff?


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 ______.


The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?

  1. The force at t = (1/8) s on the particle is – 16π2 Am.
  2. The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.
  3. The particle is not acted upon by any force.
  4. The particle is not acted upon by a constant force.
  5. There is no impulse acting on the particle.

The position time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 0 s and t = 4 s.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×