Question

✔PHYSICS 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.

• I want a Good and Correct Explanation.
• Don't Dare To Spam!​

Answer 1

$\large\mathcal{\underbrace{\red{SOLUTION:-}}}$

✴️ Let, take

• Mass of the object = $\rm{m_1}$

• Velocity of the object before collision = $\rm{v_1}$

• Mass of the stationary wooden block = $\rm{m_2}$

• Velocity of the wooden block before collision = $\rm{v_2}$

GIVEN :-

• $\rm{m_1}$ = 1 kg

• $\rm{m_2}$ = 5 kg

• $\rm{v_1}$ = 10 m/s

• $\rm{v_2}$ = 0 m/s

TO FIND :-

1. The total momentum just before and just after the impact .
2. The velocity of the combined object .

CALCULATION :-

✍️ Total momentum before collision

$\rm{=\:m_1\:v_1\:+\:m_2\:v_2\:}$

$\rm{=\:1\times{10}\:+\:5\times{0}\:}$

$\bigstar\:\rm{\purple{\boxed{\implies\:Total\: momentum\:before\:collision\:=\:10\:kg.m.s^{-1}\:}}}$

• It is given that, after collision the object and the wooden block stick together.

$\longrightarrow\:\rm{Total\:mass\:of\:the\:combined\:system\:=\:m_1\:+\:m_2\:}$

$\longrightarrow\:\rm{Velocity\:of\:the\:combined\:object\:=\:v\:}$

✍️ According to the law of conservation of momentum,

$\checkmark\:\rm{\boxed{Total\:momentum\:before\:collision\:=\:Total\:momentum\:after\:collision\:}}$

$\rm{\implies\:m_1\:v_1\:+\:m_2\:v_2\:=\:(m1\:+\:m2)\:v\:}$

$\rm{\implies\:1\times{10}\:+\:5\times{0}\:=\:(1\:+\:5)\:v\:}$

$\rm{\implies\:v\:=\:\dfrac{10}{6}\:}$

$\bigstar\:\rm{\blue{\boxed{\implies\:v\:=\:\dfrac{5}{3}\:or\:1.67\:m/s\:}}}$

$\checkmark\:\rm{Total\:momentum\:just\:after\:the\:impact\:=\:(m_1\:+\:m_2)\:v\:}$

$\rm{=\:(1\:+\:5)\:\dfrac{5}{3}\:=\:6\times{\dfrac{5}{3}}\:}$

$\bigstar\:\rm{\green{\boxed{Total\:momentum\:just\:after\:the\:impact\:=\:10\:kg.m.s^{-1}\:}}}$

✍️ Hence, velocity of the combined object after collision is 1.67 m/s .

Answer 2

$\Large{ \bf{ \purple {SO} \blue{LU} \purple{TION:-}}}$

$\red{ \tt{ For \: Object :-}}$

$\star{ \sf{ \: m_1 = 1kg}}$

$\star{ \sf{ \: v_1 = 10m/s}}$

$\red{ \tt{For \: wooden \: block:-}}$

$\star{ \sf{\: m_2 = 5kg}}$

$\star{ \sf{ \: v_2 = 0}}$

$\red{ \tt{Momentum \: just \: before \: the \: impact:-}}$

$\star{ \sf{ \: p_1 = m_1 \times v_1 + m_2 \times v_2}}$

$\sf{ = 1 \times 10 + 5 \times 0}$

$\sf{ = 10kgm/s}$

Since Momentum before and after impact is same.

$\tt{ \small{Momentum \: before \: impact = Momentum \: after \: impact}}$

$\tt {\therefore{Momentum \: after \: collision \: is:}}$

$\star{ \sf{ \: p_2 = p_1 = 10kgm/s}}$

$\sf{Also, \: \: \: { \star{ \: p_2 = (m_1 + m_2)}}}$

$\star{ \tt{ \: v = 10}}$

$\star{ \tt{ \: v = \frac{10} {1 + 5}}}$

$\sf{ \: = \frac{ 10}{6}}$

$\purple{ \boxed{ \therefore v = 1.67m/s}}$

Have a nice day!!