Question

an object of mass 1 kg travelling in 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 the velocity of the combined object.​

Answer 1

Answer:

let object mass be m

wooden block mass be M

momentum before collision =momentum after collision

mu+M*0=(m+M)v (v=combined velocity )

10=6*v

v=10/6

v=1.66m/s

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$The \:velocity\:of\: both\: object\: after\: collision\: = 1.67 m\: {s}^{-1}.$

$Total\: momentum\: before\: collision \:=10kg\:{s}^{-1}.$

$Total \:momentum\: after\: collision\: = 10 kg\:m\:{s}^{-1} .$

$\bf{\underline {Given:}}$

$Mass\: of\: moving \:object, m_1=1kg$

$Mass\: of\: wooden \:block, m_2\:=\:5kg$

$Initial \:velocity\: of\: object, u_1=10m\:s^-1$

$Initial\: velocity \:of \:wooden\: block, \: u_2\:=\:0$

$\bf{\underline {To\:Find:}}$

$Final \:velocity \:of \:moving \:object\: and \:wooden\: block, \: v =\:?$

$Total\: momentum \:before\: collision and \:after \:collision/:=\:?$

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

We know that,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Substituting the values, we get

$1 × 10 + 5 × 0 = 1 × v + 5 × v$

$10 = v(1 +5)$

$10 = v × 6$

${v = \dfrac{10}{6} = 1.67m\:{s}^{-1}}$ .......... (1)

Total momentum of object and wooden block just before collision

${m_1u_1+m_2u_2 = 1 × 10+5×0 = 10kg\:m\:s^-1}$

Total momentum after collision

$m_1u_1+m_2u_2=m_1v_1+m_2v_2=v(m_1+m_2)$

(Since both the objects move with the same velocity $'v'$ after collision)

$=(1+5)×\dfrac{10}{6}$. ............. [from(1)]

$=6×\dfrac{10}{6}=10kg\:m\:s^-1$

Thus,

$\bf{The \:velocity\:of\: both\: object\: after\: collision\: = 1.67 m\: s^{-1} .}$

$\bf{Total\: momentum\: before\: collision \:=10kg\:s^{-1} .}$

$\bf{Total \:momentum\: after\: collision\: = 10 kg\:m\:s^{-1} .}$