A metal ball of mass `2 kg` moving with speed of `36 Km//h` has a collision with a stationary ball of mass `3 kg`. If after collision, both the ball move together, the loss in Kinetic energy due to collision is :

Options

A. 40 J
B. 60 J
C. 100 J
D. 140 J

Answer

B

Explanation

(b) `v = 36 km//h = 10 m//s`
By law of conservation of momentum
`2 xx 10 = (2 + 3) V rArr V = 4 m//s`
Loss in `K.E. = (1)/(2) xx 2 xx (10)^2 - (1)/(2) xx 5 xx (4)^2`
=`60 J`
Alternatively,
From law of conservation of momentum, we have
`m_1 u_1 = m_2 u_2 = (m_1 + m_2) v`
`rArr v = (m_1 u_1 + m_2 u_2)/((m_1 + m_2))`
Given,
`m = 2kg, u_1 = 36 xx (5)/(18) = 10 m//s`,
`m_2 = 3kg, u_2 = 0`
:. `v = (2 xx 10 + 3 xx 0)/(2 + 3) = 4 m//s`
Loss in kinetic energy is
`Delta K = Delta K' -Delta K''`
=`{(1)/(2) m_1 u_1^2 + (1)/(2) m_2 m_2^2}-{(1)/(2) (m_1 + m_2)v^2}`
=`(1)/(2) xx 2 xx (10)^2 - (1)/(2) xx 5 xx (4)^2`
=`100 - 40 = 60 J`.