Question

A ball after falling from a height of 10 m strikes the roof of a lift which is descending down with a velocity of 1m/s. what is the recoil velocity of the ball?

Answer 1

Let the mass of the lift be m₂ and that of the ball be m₁.

Then u be the recoil velocity.

Then :

U = ((m₂ - m₁) u₁ / m₂ + m₁) + (2m₂u₂) / m₁ + m₂

This simplifies to :

U = u₁ + 2u₂

Where u₁ = velocity of the ball

u₂ = velocity of the lift.

From the kinematics equation below, we can get the final velocity of the ball.

v² - u² = 2gs

U = 0

v² = 2 × 10 × 10 = 200

V = √200 = 14.14 m/s

Given that the lift is moving downwards, our equation becomes :

U = u₁ - u₂

U = 14.14 - 1 = 13.14

= 13.14 m/s

Answer 2

Let the mass of ball be  m₁ and initial velocity u₁ and mass of  lift  be  m₂ and  initial velocity be  m₂. Let v₁ and  v₂ are  the velocity of  ball and  lift  after collision  respectively.

u₁ = 10 m/sec

u₂ = 1 m /sec

Now,

According  to  law of conservation of linear momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁(u₁ - v₁) = m₂(v₂ - u₁).........................i)

For Elastic Collision,

1/2m₁u₁² + 1/2m₂u₂² = 1/2m₁v₁² + 1/2m₂v₂²

m₁(u₁² - v₁²) = m₂(v₂² - u₁²) ........................ii)

dividing  eq i) and  ii)

$\frac{u_1^{2}-v_1^{2}}{u_1-v_1}=\frac{v_2^{2}-u_2^{2}}{v_2-u_1}\\\\\frac{(u_1+v_1)(u_1-v_1)}{u_1-v_1}=\frac{(v_2+u_2)(v_2-u_2)}{v_2-u_2}\\\\u_1+v_1=v_2+u_2\\\\v_2=u_1-u_2+v_1......................iii)\\\\\text{putting this value in equ i)}\\\\m_1(u_1-v_1)=m_2(u_1+v_1-2u_2)\\\\-v_1(m_1+m_2)=u_1(m_2-m_1)-2m_2u_2\\\\v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+(\frac{2m_2}{m_1+m_2})u_2\\\\\text{Since,lift is very heavier than ball}\\\\m_2>>m_1\\\\\text{mass of lift is negligible in comparison of ball}\\\;\\v_1=u_1+2u_2\\\\v_1=10+2\times1\\\\v_1=12\;\;m/s$

Note:- A block diagram is attached