Question

28. (a) An object of mass 1 kg travelling in a straight line with a velocity of 10
m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then
they both move off together in the same straight line. Calculate the total
momentum just before the impact and just after the impact. Also, calculate the
velocity of the combined object.​

Answer 1

$given$

$mass \: of \: the \: object = {m}^{1} \: = kg$

$mass \: of \: the \: wooden \: block = {m}^{2} \: = 5kg$

$velocity \: of \: the \: object = v {}^{1} = 10 \: m \: and \: s$

$since \: the \: wooden \: block \: is \: stationary$

$velocity \: of \: wooden \: blocks = {v}^{2} = 0$

$after \: collision \: they \: move \: together$

$mass \: of \: the \: system \: = m1 \: + m2$

$= 1 + 5 \\ = 6kg$

$let \: the \: velocity \: after \: collosion = v$

$finding \: momentum \: and \: after \: collosion$

$momentum \: before \\ = momentum \: of \: object \\ + momentum \: of \: block$

$= 1 \times 10 + 5 \times 0 \\ = 10 + 0 \\ = 10kg \: ms$

$momentum \: after \\ momentum \\ = total \: mass \times velocity \: after \\ = (m1 \: + m2) \: \times v \\ = 6v$

$since \: momentum \: is \: conserved \\ total \: momentum \: before \: collision = total \: momentum \: after \: collosion \\ 10 = 6v \\ \: v = \frac{10}{6} \\ v = \frac{5}{3} \: ms$

$therefore \\ momentum \: before \: collosion = 10 \: kg \: ms \\ since \: momentum \: before \: collision = momentum \: after \: collision$

$velocity \: of \: combined \: object = \frac{5}{3} = ms$