Question

Aɴ ᴏʙᴊᴇᴄᴛ ᴏғ ᴍᴀss 1 ᴋɢ ᴛʀᴀᴠᴇʟʟɪɴɢ ɪɴ ᴀ sᴛʀᴀɪɢʜᴛ ʟɪɴᴇ ᴡɪᴛʜ ᴀ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ 10 ᴍs ᴄᴏʟʟɪᴅᴇs ᴡɪᴛʜ, ᴀɴᴅ sᴛɪᴄᴋs ᴛᴏ, ᴀ sᴛᴀᴛɪᴏɴᴀʀʏ ᴡᴏᴏᴅᴇɴ ʙʟᴏᴄᴋ ᴏғ ᴍᴀss 5 ᴋɢ. Tʜᴇɴ ᴛʜᴇʏ ʙᴏᴛʜ ᴍᴏᴠᴇ ᴏғғ ᴛᴏɢᴇᴛʜᴇʀ ɪɴ ᴛʜᴇ sᴀᴍᴇ sᴛʀᴀɪɢʜᴛ ʟɪɴᴇ. Cᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ᴛᴏᴛᴀʟ ᴍᴏᴍᴇɴᴛᴜᴍ ᴊᴜsᴛ ʙᴇғᴏʀᴇ ᴛʜᴇ ɪᴍᴘᴀᴄᴛ ᴀɴᴅ ᴊᴜsᴛ ᴀғᴛᴇʀ ᴛʜᴇ ɪᴍᴘᴀᴄᴛ. Aʟsᴏ, ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴛʜᴇ ᴄᴏᴍʙɪɴᴇᴅ ᴏʙᴊᴇᴄᴛ.
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Answer 1

Explanation:

hope it explains you.

REMEMBER:

1. INITIAL MOMEMTUM = FINAL MOMEMTUM.

Answer 2

$\huge\mathcal{\boxed{\fcolorbox{lime}{orange}{SOLUTION}}}$

☃️ Mass of the object $\bf{(m_1)}$ = 1 kg

☃️ Velocity of the object before collision $\bf{(v_1)}$ = 10 m/s

☃️ Mass of the stationary wooden block $\bf{(m_2)}$ = 5 kg

☃️ Velocity of the wooden block before collision $\bf{(v_2)}$ = 0 m/s

✴️ Total momentum before collision,

$\sf\red{m_1\:v_1\:+\:m_2\:v_2\:}$

$\sf{\implies\:1\times{10}\:+\:5\times{0}\:}$

$\sf{\implies\:10\:+\:0\:}$

$\sf{\implies\:10\:kg.m.s^{-1}}$

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

• Total mass of the combined system = $\bf{m_1\:+\:m_2}$

• Velocity of the combined object = v

✨ According to the law of conservation of momentum,

✔️ Total momentum before collision = Total momentum after collision

$\sf{\implies\:{\pink{m_1\:v_1\:+\:m_2\:v_2\:=\:(m_1\:+\:m_2)\:v\:}}}$

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

$\sf{\implies\:10\:=\:6\times{v}\:}$

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

$\bf{\color{lime}{\implies\:v\:=\:\dfrac{5}{3}\:m.s^{-1}}}$

__________________________

⭐ The total momentum after collision is 10 kg.m/s .

⭐ Total momentum just before the impact is 10 kg.m/s .

⭐ Total momentum just after the impact = $\sf{(m_1\:+\:m_2)\:v\:}$ = 6 × $\sf{\dfrac{5}{3}}$ = 10 kg.m/s .

⭐ Therefore, velocity of the combined object after collision is $\bf\blue{5/3\:m.s^{-1}}$ .