Laws of Motion

  • Question 517
    CBSEENPH11020501

    It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :

    • (0,1)

    • (.89,.28)

    • (.28,.89)

    • (0,0)

    Solution

    B.

    (.89,.28)

    For collision of a neutron with deuterium:

    Applying conservation of momentum:

    mv + 0 = mv1 + 2mv2 .....(i)
    v2 -v1 = v ...... (ii)

    Therefore, Collision is elastic, e = 1

    From equ (i) and equ (ii) v1 = -v/3

    Pd = 12mv2 -12mv1212mv2 = 89 = 0.89

    Now, for the collision of neutron with carbon nucleus

    Applying conservation of momentum

    mv + 0 = mv1 + 12mv2 ....; (iii)

    v = v2-v1  ....(iv)

    v1 = -1113 vPc = 12mv2 - 12m1113v212mv2 = 48169 0.28

    Question 518
    CBSEENPH11020513

    A car is negotiating a curved road of radius R. The road is banked at angle straight theta. The coefficient of friction between the tyres of the car and the road is straight mu subscript straight s. The maximum safe velocity on this road is,

    • square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root
    • square root of straight g over straight R open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tanθ end fraction close parentheses end root
    • square root of straight g over straight R squared open parentheses fraction numerator mu subscript s plus tan theta over denominator 1 minus mu subscript s tan theta end fraction close parentheses end root
    • square root of g R squared open parentheses fraction numerator straight mu subscript straight s plus tanθ over denominator 1 minus straight mu subscript straight s tanθ end fraction close parentheses end root

    Solution

    A.

    square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root

    A car is negotiating a curved road of radius R. The road is banked at angle straight theta and the coefficient of friction between the tyres of car and the road is straight mu subscript straight s.
    The given situation is illustrated as:

    In the case of vertical equilibrium,
    N cos straight theta = mg + f1 sin straight theta
    rightwards double arrowmg = N cos straight theta space minus space straight f subscript 1 space sin space straight theta    ... (i)
    In the case of horizontal equilibrium,
    straight N space sin space straight theta space plus space straight f subscript 1 space cosθ space equals space mv squared over straight R  ... (ii)
    Dividing Eqns. (i) and (ii), we get
    straight v squared over Rg equals fraction numerator sin space theta space plus space mu subscript s space cos theta over denominator cos space theta space minus space mu subscript s space sin space theta end fraction space open square brackets f subscript 1 proportional to mu subscript s close square brackets

rightwards double arrow v space equals square root of R g open parentheses fraction numerator tan space theta space plus mu subscript s over denominator cos space theta space minus space mu subscript s space sin space theta space end fraction close parentheses end root
rightwards double arrow space v space equals space square root of R g open parentheses fraction numerator tan space theta space plus space mu subscript s over denominator 1 minus mu subscript s space tan space theta end fraction close parentheses end root

    Question 519
    CBSEENPH11020529

    A particle of mass m is driven by a machine that delivers a constant power K watts. If the particle starts from rest, the force on the particle at time t is

    • square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent
    • square root of mk space t to the power of negative 1 divided by 2 end exponent
    • square root of 2 mk end root space t to the power of negative 1 divided by 2 end exponent
    • 1 half square root of m t end root to the power of negative 1 divided by 2 end exponent

    Solution

    A.

    square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent

    As the machine delivers a constant power
    So F, v =constant = k (watts)
    rightwards double arrow space straight m dv over dt. straight v space equals straight k
rightwards double arrow integral vdv space equals straight k over straight m integral dt
rightwards double arrow straight v squared over 2 equals straight k over straight m straight t space space
rightwards double arrow space straight v equals space square root of fraction numerator 2 straight k over denominator straight m end fraction end root straight t
Now comma space force space on space the space particles space is space given space by
straight F equals space straight m fraction numerator d straight v over denominator d straight t end fraction space equals straight m straight d over dt open parentheses fraction numerator 2 kt over denominator straight m end fraction close parentheses to the power of 1 divided by 2 end exponent
equals space square root of 2 km end root space open parentheses 1 half straight t to the power of negative 1 half end exponent close parentheses
equals space square root of mk over 2 end root. straight t to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent

    Question 520
    CBSEENPH11020538

    Three blocks A, B, and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a 14 N is applied to the 4 kg block, then the contact force between A and B is

    • 2 N

    • 6 N

    • 8 N 

    • 18 N 

    Solution

    B.

    6 N

    Given, mA = 4 kg
    mB = 2 kg 
    => mC =1 kg

    So total mass (M)  = 4+2+1 = 7 kg
    Now, F = Ma 
    14 = 7a
    a=2 m/s2

    F-F' = 4a
    F' = 14-4x2
    F' = 6N

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