Question

An object of mass 1kg travelling in a straight line with a velocity of 10 m s-1 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

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,


Full Question: An object of mass 1 kg is found to be travelling with a velocity if 10 m/s along a straight line in turn colliding with a stationery wooden block having a mass of 5 kg. Calculate the total momentum of it just before and after the colliding is done. Calculate the total velocity of a combined system.

ANSWER :

Refer and consult to the following attachments for a detailed explanation and step by step solution with vital elaborations for certain formulas used into this query to arrive at a final conclusion that is,,,,


$\boxed{\bf{Total \: Momentum = 10 \: Kg \: m/s}}$


$\boxed{\bf{Total \: Velocity = \frac{5}{3} \: m/s}}$


HOPE IT HELPS YOU AND CLEARS THE DOUBTS TO FIND TOTAL MOMENTUM AND TOTAL VELOCITY OF THE GIVEN SYSTEM!!!!!!!

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,

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

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

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