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 Multiple Choice QuestionsMultiple Choice Questions


A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

  • 0.5

  • 0.25

  • 0.4

  • 0.8



According to law of conservation of linear momentum,

mv + 4m x 0 = 4mv' + 0

v' = v4e = relative velocity of separation Relative velocity of approach =v4ve = 14 = 0.25


A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is

  • a = gcosec θ

  • a= gsin θ

  • a = g tan θ

  • a = g cos θ


a = g tan θ

In non-inertial frame,

N sin θ = ma ...(i)
N cos θ = mg... (ii)

tan θ = a/g
a = g tan θ