c
Let percentage decrease in momentum \(= X \%\)

Decrease in momentum \(mv _1- mv _2= m \left( v _1- v _2\right)=(0.0 X ) mv 1\)

we can write it like this - \(v _2=(1-0.0 X ) v _1 \quad\)....(1)

Change in \(K E-\frac{1}{2} m \left( v _1^2- v _2^2\right)=0.19 \frac{1}{2} mv _1^2\)

\(\Rightarrow v _2^2=0.81 v _1^2\)

By equation \((1)\) we get \((1-0.0 X )^2=0.81\)

\(\Rightarrow X =10\)