Home
Class 11
PHYSICS
A particle is acted upon by a force for ...

A particle is acted upon by a force for 1 second whose X component remains constant at `F_(x) = 30 N` but y and z components vary with time as shown in the graph. Calculate the magnitude of change in momentum of the particle in 1 s. What angle does the change in momentum `(Delta vec(P))` make with X axis ?

Text Solution

Verified by Experts

The correct Answer is:
`Deltap=10sqrt(29)Ns; theta = cos^(-1)((3)/(sqrt(29)))`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • KINETIC THEORY OF GASES

    ARIHANT|Exercise Level 3|4 Videos

  • NEWTONS LAWS OF MOTION

    ARIHANT|Exercise All Questions|111 Videos

Similar Questions

Explore conceptually related problems

The force acting on a particle of mass 20 kg is indicated by force - time graph as shown in figure. The magnitude of change in linear momentum of particle during time interval 0 to 6 second is

A particle is acted upon by a force which varies with position x as shown. F is in N and x is in m F 10 5 X 2 -10

Knowledge Check

  • A particle which is initially at rest at the origin, is subjection to an acceleration with x- and y- components as shown. After time t=5 , the particle has no acceleration. What is the magnitude of velocity of the particle at t=2 seconds ?

    A
    `10sqrt(5)m//s`
    B
    `5sqrt(10)m//s`
    C
    `5sqrt(5)m//s`
    D
    None of these

  • A particle which is initially at rest at the origin, is subjection to an acceleration with x- and y- components as shown. After time t=5 , the particle has no acceleration. What is the magnitude of average velocity of the particle between t=0 and t=4 seconds ?

    A
    `(5)/(2)sqrt(13)m//s`
    B
    `(5)/(2)sqrt(17)m//s`
    C
    `30m//s`
    D
    None of these

  • X- component of vec(a) is twice of its Y- component. If the magnitude of the vector is 5sqrt(2) and it makes an angle of 135^(@) with z-axis then the components of vector is:

    A
    `2sqrt(3), sqrt(3), -3`
    B
    `2sqrt(6), sqrt(6), -6`
    C
    `2sqrt(5), sqrt(5), -5`
    D
    None of these

    • Similar Questions

    Explore conceptually related problems

    A particle mass 1 kg is moving along a straight line y=x+4 . Both x and y are in metres. Velocity of the particle is 2m//s . Find the magnitude of angular momentum of the particle about origin.

    A particle of mass m is released from rest from point P at x = x_(0) on X-axis from origin O and falls vertically along y-axis as shown in Fig. What is the magnitude of the torque acting on the particle at time t , when it is at the point Q w.r.t.O ?

    The force F acting on a partical of mass m is indicated by the force-time graph shown below. The change in momentum of the particle over time interval from zero to 8 s is.

    A particle of mass 1 kg is moving along the line y = x + 2 (here x and y are in metres) with speed 2 m/s. The magnitude of angular momentum of particle about its origin is

    A particle moves along the parabolic path x = y^2 + 2y + 2 in such a way that Y-component of velocity vector remains 5ms^(-1) during the motion. The magnitude of the acceleration of the particle is

    Home
    Profile