Laws of Motion

Laws of Motion

Easy Questions
Moderate Questions
Difficult Questions

Mechanics

Physical World, Units and Measurements
Motion in a Straight Line
Motion in a Plane
Laws of Motion
Work, Energy and Power
Gravitation
Mechanical Properties of Solids
Mechanical Properties of Fluids
System of Particles and Rotational Motion
  1. A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be​








  1. View Hint View Answer Discuss in Forum


    N = m a sin θ + mg cos θ .....(1)
    Also, m g sin θ = m a cos θ ....(2)
    From (1) & (2), a = g tan θ .
    or,

    ∴ N = mg
    sin² θ
    		 + mg cos θ
    cos θ

    =
    mg
    		(sin² θ + cos² θ) =
    mg
    cos θcos θ

    or, N =
    mg
    cos θ

    Correct Option: A


    N = m a sin θ + mg cos θ .....(1)
    Also, m g sin θ = m a cos θ ....(2)
    From (1) & (2), a = g tan θ .
    or,

    ∴ N = mg
    sin² θ
    		 + mg cos θ
    cos θ

    =
    mg
    		(sin² θ + cos² θ) =
    mg
    cos θcos θ

    or, N =
    mg
    cos θ

  1. A particle of mass m is moving with a uniform velocity v1. It is given an impulse such that its velocity becomes v2. The impulse is equal to








  1. View Hint View Answer Discuss in Forum

    Impulse = final momentum – initial momentum = m (v2 – v1)

    Correct Option: D

    Impulse = final momentum – initial momentum = m (v2 – v1)

  1. Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively :-








  1. View Hint View Answer Discuss in Forum


    Before cutting the string
    kx = T + 3 mg ...(i)
    T = mg ...(ii)
    ⇒ kx = 4mg
    After cutting the string T = 0

    aA =
    4mg - 3mg
    3m


    aA =
    g
    		↑
    3

    and aB =
    mg
    		 = g ↓
    m

    Correct Option: A


    Before cutting the string
    kx = T + 3 mg ...(i)
    T = mg ...(ii)
    ⇒ kx = 4mg
    After cutting the string T = 0

    aA =
    4mg - 3mg
    3m


    aA =
    g
    		↑
    3

    and aB =
    mg
    		 = g ↓
    m

  1. A balloon with mass ‘m’ is descending down with an acceleration ‘a’ (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration ‘a’?​








  1. View Hint View Answer Discuss in Forum

    Let upthrust of air be Fa then
    For downward motion of balloon
    Fa = mg – ma
    mg – Fa = ma
    For upward motion
    Fa – (m – ∆m)g = (m – ∆m)a

    Therefore ∆m =
    2ma
    g + a

    Correct Option: A

    Let upthrust of air be Fa then
    For downward motion of balloon
    Fa = mg – ma
    mg – Fa = ma
    For upward motion
    Fa – (m – ∆m)g = (m – ∆m)a

    Therefore ∆m =
    2ma
    g + a

  1. Three blocks with masses m, 2 m and 3 m are connected by strings as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)








  1. View Hint View Answer Discuss in Forum


    From figure
    F = 6 mg,
    As speed is constant, acceleration a = 0
    ∴ 6 mg = 6ma = 0, F = 6 mg
    ∴ T = 5 mg , T′ = 3 mg
    T″ = 0
    Fnet on block of mass 2 m
    = T – T' – 2 mg = 0
    Alternate :
    ∵ v  = constant
    so, a = 0, Hence, Fnet = ma = 0

    Correct Option: D


    From figure
    F = 6 mg,
    As speed is constant, acceleration a = 0
    ∴ 6 mg = 6ma = 0, F = 6 mg
    ∴ T = 5 mg , T′ = 3 mg
    T″ = 0
    Fnet on block of mass 2 m
    = T – T' – 2 mg = 0
    Alternate :
    ∵ v  = constant
    so, a = 0, Hence, Fnet = ma = 0