A block of mass 0.50 kg is moving with speed of `2.00 ms^(-1)` on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collisions is

Options

A. 0.16 J
B. 1.00 J
C. 0.67 J
D. 0.34 J

Answer

c

Explanation

Initial kinetic energy of the system
`KE_i=1/2"mu"^2+1/2M(0)^2=1/2xx0.5xx2xx2+0=1J`
For collision applying conservation of linear momentum `mxxu=(m+M)xxv`
`:. 0.5xx2=(0.5+1)xxvimplies v=2/3 m//s`
Final kinetic energy of the system is
`KE_f=1/2(m+M)v^2=1/2(0.5+1)xx2/3xx2/3=1/3J`
`:.` Energy loss during collision=`(1-1/3)J=0.67J`