Question

Question 28
A body of mass 300 g kept at rest breaks into two parts due to internal forces. One
part of mass 200 g is found to move at a speed of 12 m/s towards the east. What will
be the velocity of the other part?​

Answer 1

Answer:

$\bf \: Mass \: \: of \: \: one \: \: part \: \: m_1 \: \: = \: \: 200 gm$

$\bf \: Its \: \: velocity \: \: v_1 \: \: = \: \: 12 m/s \: \: towards \: \: east$

so,

Mass of other part

$m_2 = 300 - 200 = 100gm$

$\bf \: let \: \: its \: \: velocity \: \: be \: \: v_2 \\ \\ \bf \: applying \: \: \: conservation \: \: of \: \: linear \: \: momentum \: \\ \\ \bf \red{P_f -P_i}$

$\bf \:m_1v_1+m_2v_2 = 0 \\ \\</p><p>\bf \:200 × 12 +100 × v_1 = 0 \\ \\</p><p>v_2 = -24 m/s$

Velocity of other parts is 24m/s....

Answer 2

Step-by-step explanation:

Answer:

\bf \: Mass \: \: of \: \: one \: \: part \: \: m_1 \: \: = \: \: 200 gmMassofonepartm

1

=200gm

\bf \: Its \: \: velocity \: \: v_1 \: \: = \: \: 12 m/s \: \: towards \: \: eastItsvelocityv

1

=12m/stowardseast

so,

Mass of other part

m_2 = 300 - 200 = 100gmm

2

=300−200=100gm

\begin{gathered}\bf \: let \: \: its \: \: velocity \: \: be \: \: v_2 \\ \\ \bf \: applying \: \: \: conservation \: \: of \: \: linear \: \: momentum \: \\ \\ \bf \red{P_f -P_i}\end{gathered}

letitsvelocitybev

2

applyingconservationoflinearmomentum

P

f

−P

i

\begin{gathered}\bf \:m_1v_1+m_2v_2 = 0 \\ \\ \bf \:200 × 12 +100 × v_1 = 0 \\ \\ v_2 = -24 m/s\end{gathered}

m

1

v

1

+m

2

v

2

=0

200×12+100×v

1

=0

v

2

=−24m/s

Velocity of other parts is 24m/s....