Question

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

"Mass of object m_1 = 1 kg

Velocity of the object = 10 m/s

Mass of wooden block m_2 = 5 kg

Velocity of the wooden block = 0 m/s

After collision the velocity will be “v”

Before collision, total momentum will be

m_1 × u_1 + m_2 × u_2 = 1 × 10 + 5 × 0

Total momentum = 10 kg.m/s

After collision, total momentum will be

(m_1 + m_2) v = (1 + 5) v = 6v kg.m/s

According to the conservation of momentum law,

6v = 10

v = 10/6 = 5/3 m/s

Therefore, after impact, the momentum will be = 6 × 5/3 m/s = 10 m/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,

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

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

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