Question

A block of mass 1 kg moving with a speed of
4 ms -, collides with another block of mass 2
kg which is at rest. The lighter block comes
to rest after collision. The loss in KE of the
system is​

Answer 1

Answer:

$\frac{1}{2}m {v}^{2} = \frac{1}{2} \times 1 \times {4}^{2} = 8 \: j$