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A body of mass 5kg starts from origin wi...

A body of mass 5kg starts from origin with a velocity `baru = (30hati +40hatj)ms^(-1)` . If a constant force `F=-(hati+5hatj)N` acts on the body, then pick up the correct statements

A

y - component of the velocity becomes zero at t = 40 second

B

X- component of the velocity becomes zero at t = 180 second

C

body will never attain permanent rest

D

the direction of motion changes only once along x and y- directions

Text Solution

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The correct Answer is:
A, B, C, D
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A body of mass 5kg starts from the origin with an initial velocity of barU= (30i+40j)m//s . A constant force of F= (-hati - 5 hatj)N acts on the body . Find the time in which they- component of the velocity becomes zero .

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Knowledge Check

  • A body of mass 5kg starts from the origin with an initial velocity vecu = 30hati +40hatj ms^(-1) . If a constant force vecF=-(hati+5hatj)N acts on the body, the time in which the y-component of the velocity becomes zero is :

    A
    5 s
    B
    20 s
    C
    40 s
    D
    80 s

  • A body of mass 5kg starts from the origin with an initial velocity vecu = 30 hati + 40 hatj ms^(-1) . If a constant force vecF = -(hati + 5hatj) acts on the body, the time in which the y-component of the velocity becomes zero is :

    A
    5s
    B
    20s
    C
    40s
    D
    80s

  • A body is projected from the top of a tower with a velocity baru = 3hati + 4hatj + 5hatk ms^-1, where hati, hatj and hatk are unit vectors along east, north and vertically upwards respectively. If the height of the tower is 30m, horizontal range of the body on the ground is (g = 10ms^-2)

    A
    12m
    B
    9m
    C
    25m
    D
    15m

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