Concepts Of Physics MCQ Edition [Volume 1]PhysicsCentre of Mass, Linear momentum, Collision
A person of mass M holding a bag of mass m slips from the roof of a tall building of height H and begins falling vertically, as shown in the figure. At a height h from the ground, the person observes that the ground below is quite hard, whereas a pond is located at a horizontal distance x from the line of fall. To save himself, he throws the bag horizontally (relative to himself) in the direction opposite to the pond
Options
- AVelocity: (M+m)x g m [ 2H - 2(H - h) ] , Landing position: (M+m)x m left to the line of fall
- BVelocity: mx g M [ 2H - 2(H - h) ] , Landing position: mx M left to the line of fall
- CVelocity: Mx g m [ 2H + 2(H - h) ] , Landing position: Mx m left to the line of fall
- DVelocity: Mx g m [ 2H - 2(H - h) ] , Landing position: Mx m left to the line of fall
Correct answer
D. Velocity: Mx g m [ 2H - 2(H - h) ] , Landing position: Mx m left to the line of fall
Step-by-step solution
Let the total time taken to fall from the roof to the ground be t . t = 2H g The time taken to fall from the roof to a height h from the ground is t₁ . t₁ = 2(H-h) g The remaining time for the person to reach the ground is t₂ . t₂ = t - t₁ = 2H - 2(H-h) g To land in the pond, the person must cover a horizontal distance x in time t₂ . The required horizontal velocity of the person v_M is: v_M = x t₂ = x g 2H - 2(H-h) Initially, the horizontal momentum of the system (person + bag) is zero. By conservation of linear m