A particle of mass 2kg is projected vertically upward in a lift with a velocity 10m/s relative to the lift from a height 10m above the floor of lift. When particle strikes the floor of lift what will

Question

A particle of mass 2kg is projected vertically upward in a lift with a velocity 10m/s relative to the lift from a height 10m above the floor of lift. When particle strikes the floor of lift what will be the velocity of the particle just before the strike and the impulse imparted on it by lift respectively. [At $t = 0, V =$ $10 m / s & a = 5 m / s^{2}$ (Both upwards) for the lift]. Particle does not rebound after it strikes the floor of lift :-

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In the lift frame g eff ​ = g + a = 15 m / s 2 . Strike: − 10 = 10 t − 2 1 ​ ( 15 ) t 2 ⇒ t = 2 s . Relative strike speed: v rel 2 ​ = 1 0 2 + 2 ( 15 ) ( 10 ) = 400 ⇒ v rel ​ = 20 m / s downward. Lift's speed at t = 2 s : V = 10 + 5 ( 2 ) = 20 m / s upward; so the particle's ground-frame velocity = 20 − 20 = 0 m / s . Impulse by the floor = m Δ v = 2 ( 20 − 0 ) = 40 N − s upward. Answer: (A)

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