A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision what could be minimum and the maximum value of u.

Answer

B

Explanation

Final K.E. =0.2J
Initial K.E. =`1/2mv_1^2+0`
`=1/2xx0.1xxu^2=0.05u^2`
`mv_1+mv_2=mu`

where `v_1 and v_2` are fiN/Al velocities of 1st and 2nd blcok respectivley.
`rarr v_1+v_2=u`………i
`(v_1-v_2)+e(u_1-u_2)=0`
`rarr euv_2-v_1`
`[u_2=0, u_1=u]`..........2
Adding equation 1 and 2
`2v_2=(1+e)u`
`rarr v_2=(u/2)(1+e)`
`:. v_1=u-u/2(1+e)`
`v_1=u/2(1-e)`
Given `1/2mv_1^2+1/2mv_2^2=0.2`
`v_1^2+v_2^2=4`
`rarr u^2/2(1+e^2)=4`
`rarr u^2=8/(1+e^2)`
For maximum value of u denomiN/Ator should be minimum
`rarr e=0`
`rarr u^2=8`
`rarr u=2m/s`