Question

Ans with explanation

Answer 1

In first case-
mass–»m
distance before stopped–»s
velocity with moving–»u

using equation of motion to find acceleration.

${v}^{2} - {u}^{2} = 2as \\ - {u}^{2} = 2as \\ a = - \frac{ {u}^{2} }{2s} \\ \\ force = ma \\ force(f) = m( - \frac{u}{2s} ) \\ \\ now \: in \: second \: case \: velocity \: became \: double \: and \: mass \: too \\ {v}^{2} - {(2u)}^{2} = 2as \\ a = \frac{ - 4 {u}^{2} }{2s} \\ force(F') = 2m( \frac{ - 4 {u}^{2} }{2s}) \\ F' = 8m \frac{ - { u}^{2} }{2s} \\ F' = 8f$
So, option D is correct.