Question

An object of mass 1 kg 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

Hey Dear,

◆ Answer -

Total momentum after impact = 10 kgm/s

Velocity of combined object = 1.67 m/s

● Explaination -

# Given -

m1 = 1 kg

m2 = 5 kg

u1 = 10 m/s

u2 = 0 m/s

# Solution -

Total momentum just before impact is -

pi = m1u1 + m2u2

pi = 1×10 + 5×0

pi = 10 kgm/s

By law of conservation of momentum,

pi = pf

pf = 10 kgm/s

Let v be the combined velocity of the object + block -

pf = (m1+m2) v

10 = (1+5) v

v = 10 / 6

v = 1.67 m/s

thanks dear...

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,

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

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

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