Question

an object of mass 1 kg travelling in a straight line with a velocity of 10 MS -1 colour lights with and stick to a stationary wooden block of mass 5 kg then they both move of together in the same straight line calculate the total no just before the impact and just after the impact also calculate the velocity of the combined object

Answer 1

m 1= 1kg
m2= 5kg
u1=10ms-1
u2=0
Initial momentum =(1×5) + (5×0) kgms-1
=5kgms-1
Final momentum =5kgms-1
Let combined velocity be v.
→(m1+m2)v =5kgms-1

→v6 =5ms-1

→v = 5/6ms-1

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,

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

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

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