Question

Challenge !!

A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s (figure 9-E12) towards another block of equal mass kept at rest. The spring constant of the spring fixed 100 N/ m find the maximum compression of the spring.

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Answer 1

Answer:

ᴍᴀss ᴏғ ᴇᴀᴄʜ ʙʟᴏᴄᴋ ᴍᴀ ᴀɴᴅ ᴍʙ = 2kg.

ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴛʜᴇ sᴛ ʙʟᴏᴄᴋ, (ᴠ) = 1m/s

• VA = 1 m/s,
• VB = 0 m/s

sᴘʀɪɴɢ ᴄᴏɴsᴛᴀɴᴛ ᴏғ ᴛʜᴇ sᴘʀɪɴɢ = 100 N/m.

ᴛʜᴇ ʙʟᴏᴄᴋ ᴀ sᴛʀɪᴋᴇs ᴛʜᴇ sᴘʀɪɴɢ ᴡɪᴛʜ ᴀ ᴠᴇʟᴏᴄɪᴛʏ 1m/s

• ᴀғᴛᴇʀ ᴛʜᴇ ᴄᴏʟʟɪsɪᴏɴ, ɪᴛ’s ᴠᴇʟᴏᴄɪᴛʏ ᴅᴇᴄʀᴇᴀsᴇs ᴄᴏɴᴛɪɴᴜᴏᴜsʟʏ ᴀɴᴅ ᴀᴛ ᴀ ɪɴsᴛᴀɴᴛ ᴛʜᴇ ᴡʜᴏʟᴇ sʏsᴛᴇᴍ (ʙʟᴏᴄᴋ ᴀ + ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ sᴘʀɪɴɢ + ʙʟᴏᴄᴋ ʙ) ᴍᴏᴠᴇ ᴛᴏɢᴇᴛʜᴇʀ ᴡɪᴛʜ ᴀ ᴄᴏᴍᴍᴏɴ ᴠᴇʟᴏᴄɪᴛʏ.

Let that velocity be V.

• Using conservation of energy,

$\sf \: \frac{1}{ 2 } M_AV_A ^2 +\frac{1}{ 2 }M_BV_B^ 2 =\frac{1}{ 2 } M_Av^2 + \frac{1}{ 2 }M_Bv^2 +\frac{1}{ 2 } kx^2$

$\sf \: \frac{1}{2} × 2(1)2 + 0 = \frac{1}{2} × 2× {v}^{2} + \frac{1}{2}× 2 × v2 + \frac{1}{2} {x}^{2} × 100$

(ᴡʜᴇʀᴇ x = ᴍᴀx. ᴄᴏᴍᴘʀᴇssɪᴏɴ ᴏғ sᴘʀɪɴɢ)

$\sf \: ⇒1 = {2v}^{2} + 50 {x}^{2} \: \: \: \: \: - - - - - - (1)$

ᴀs ᴛʜᴇʀᴇ ɪs ɴᴏ ᴇxᴛᴇʀɴᴀʟ ғᴏʀᴄᴇ ɪɴ ᴛʜᴇ ʜᴏʀɪᴢᴏɴᴛᴀʟ ᴅɪʀᴇᴄᴛɪᴏɴ, ᴛʜᴇ ᴍᴏᴍᴇɴᴛᴜᴍ sʜᴏᴜʟᴅ ʙᴇ ᴄᴏɴsᴇʀᴠᴇᴅ.

$\sf \: M_AV_A+M_BV_B=(M_AM_B)V$

$\sf \implies2×1=4×v$

$\implies\sf \: v= \frac{1}{2} m/s \: \: \: \: \: \: - - - - - - (2)$

• Putting in equation (1)

$\sf \: 1=2× \frac{1}{4} +50x^2$

$\sf \implies \frac{1}{2} =50x^2$

$\sf \implies \: x^2= \frac{1}{100} m^2$

$\sf \implies \: x = \frac{1}{10} m$

$\sf \implies \: x = 0.1 \: m$

$\sf \implies \: x = 10 \: cm$

Answer 2

Answer:

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Step-by-step explanation:

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