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A particle of mass m is moving in yz-pla...

A particle of mass `m` is moving in yz-plane with a unifrom velocity `v` with its trajectory running parallel to +ve y-axis and intersecting z-axis at `z = a` Fig. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is :

A

`mv a hat(e)_(x)`

B

`2mv a hat(e)_(x)`

C

`ymv hat(e)_(x)`

D

`2y mv hat(e)_(x)`

Text Solution

Verified by Experts

The correct Answer is:
B
The initial velocity is `v_(1)= v hat(e)_(y)` and after reflection from the wall, the final velocity is `v_(f)= - v hat(e)_(y)`. The trajectory is described as position vector `vec(r)= y hat(e)_(y) + a hat(e)_(z)` Hence, the change in angular momentum is `vec(r) xx m (v_(f)- v_(j))= 2 mv a hat(e)_(x)`
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Knowledge Check

  • A particle of mass m is moving with constant velocity v parallel to the x-axis as shown in the figure. Its angular momentum about origin O is

    A
    mvb
    B
    mva
    C
    `mv sqrt(a^(2)+b^(2))`
    D
    `mv(a+b)`

  • A particle is moving in the xy-plane with a constant velocity along a line parallel to the X-axis away from the origin. The magnitude of its angular momentum about the origin.

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    is zero
    B
    remains constant
    C
    goes on increasing
    D
    goes on decreasing

  • A particle is moving along a straight line parallel to x-axis with constant velocity. Its angular momentum about the origin

    A
    decreases with time
    B
    increases with time
    C
    remains constant
    D
    is zero