Question

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

Answer 1

Velocity of the ball at impact,

$v=\sqrt{2gh}$

Here h is the height and its value given 10 m.

$v=\sqrt{2\times9.8\times10} m/s$

$v=14 m/s$

If a lighter object moving with velocity v collide with heavy object which moving in the same direction  with velocity V, then after collision lighter object recoil in the opposite direction with velocity $v-2V$.

Given velocity of lift, $V=1 m/s$

Therefore, recoil velocity of the ball $=14-2\times1$

                                                                   $=12 m/s$                                                                

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