Laws of Motion

  • Question 485
    CBSEENPH11020334

    Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?

    • 4.9 ms-2 in horizontal direction

    • 9.8 ms-2 in vertical direction

    • zero

    • 4.9 ms-2 in vertical direction

    Solution

    D.

    4.9 ms-2 in vertical direction

    For the motion of block along inclined plane
    mg sin θ =ma
    a = g sin θ
    where a is along the inclined plane.
    The vertical component of acceleration is g sin2θ
    Therefore, the relative vertical acceleration of  A with respect to B is 
    g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)

    Question 486
    CBSEENPH11020335

    For a particle in uniform circular motion, the acceleration an at straight a with rightwards arrow on top point P (R, θ) on the circle of radius R is (Here θ is measured from the x–axis)

    • negative straight v over straight R space cos space straight theta space bold i with bold hat on top space plus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top
    • negative straight v over straight R space sin space straight theta space bold i with bold hat on top space plus space straight v squared over straight R space cos space straight theta space bold j with bold hat on top
    • negative straight v over straight R space cos space straight theta space bold i with bold hat on top space minus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top
    • straight v squared over straight R space straight i with hat on top space plus fraction numerator begin display style straight v squared end style over denominator begin display style straight R end style end fraction space straight j with hat on top

    Solution

    C.

    negative straight v over straight R space cos space straight theta space bold i with bold hat on top space minus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top
    straight a with rightwards arrow on top space equals space minus space straight V squared over straight R space cos space straight theta space straight i with hat on top space minus space straight V squared over straight R space sin space straight theta space straight j with hat on top
    Question 487
    CBSEENPH11020350

    A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

    • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2 over 3
    • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 half
    • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 third
    • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2

    Solution

    D.

    straight lambda subscript straight A over straight lambda subscript straight B space equals space 2
    By conservation of linear momentum
    mv space equals space mv subscript 1 space plus straight m over 2 straight v subscript 2
2 straight v space equals space 2 straight v subscript 1 space plus straight v subscript 2 space... space left parenthesis 1 right parenthesis
By space law space of space collision
straight e space equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
straight u space equals space straight v subscript 2 minus straight v subscript 1 space space... space left parenthesis 2 right parenthesis
by space equ. space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis
straight v subscript 1 space equals space straight v over 3 semicolon space straight v subscript 2 space equals space fraction numerator 4 straight v over denominator 3 end fraction
straight lambda subscript 1 space equals space straight h over straight p subscript 1 semicolon space straight lambda subscript 2 space equals space straight h over straight o subscript 2
straight lambda subscript 1 over straight lambda subscript 2 space equals space 2 over 1
    Question 488
    CBSEENPH11020376

    A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as

    • 17.6 kg ms−1 

    • 17.565 kg ms−1

    • 17.56 kg ms−1

    • 17.57 kg ms−1

    Solution

    A.

    17.6 kg ms−1 

    P = mv = 3.513 × 5.00 ≈ 17.6

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