Question

Derivation of Law of conservation of
momentum​

Answer 1

Answer:

Explanation:

Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t.

A=m1(v1−u1) (change in momentum of particle A)

B=m2(v2−u2) (change in momentum of particle B)

FBA=−FAB (from third law of motion)

FBA=m2∗a2=m2(v2−u2)/t

FAB=m1∗a1=m1(v1−u1)/t

m2(v2−u2)/t=−m1(v1−u1)/t

m1u1+m2u2=m1v1+m2v2

Therefore, above is the equation of law of conservation of momentum where, m1u1+m2u2 is the representation of total momentum of particles A and B before collision and m1v1+m2v2 is the representation of total momentum of particles A and B after collision.

Answer 2

ᴀɴsᴡᴇʀ࿐

ᴅᴇғɪɴɪᴛɪᴏɴ:-

ᴍᴏᴍᴇɴᴛᴜᴍ ɪs ᴛʜᴇ ᴛʜᴇ ᴀᴍᴏᴜɴᴛ ᴏғ ᴍᴏᴛɪᴏɴ ᴏғ ᴀ ᴍᴏᴠɪɴɢ ʙᴏᴅʏ, ᴍᴇᴀsᴜʀᴇᴅ ᴀs ᴀ ᴘʀᴏᴅᴜᴄᴛ ᴏғ ɪᴛs ᴍᴀss ᴀɴᴅ ᴠᴇʟᴏᴄɪᴛʏ.

=>ᴍᴏᴍᴇɴᴛᴜᴍ= ᴍᴀss × ᴠᴇʟᴏᴄɪᴛʏ

ᴅɪᴍᴇɴsɪᴏɴᴀʟ ғᴏʀᴍᴜʟᴀ ᴏғ ᴍᴏᴍᴇɴᴛᴜᴍ

=>ᴘ = [ᴍ]x [ʟᴛ^(-1)]

ᴄᴏɴsᴇʀᴠᴀᴛɪᴏɴ ᴏғ ᴍᴏᴍᴇɴᴛᴜᴍ:-

ʟᴀᴡ ᴏғ ᴄᴏɴsᴇʀᴠᴀᴛɪᴏɴ ᴏғ ᴍᴏᴍᴇɴᴛᴜᴍ ɪs ᴅᴇғɪɴᴇᴅ ᴀs ᴀ ᴘʀɪɴᴄɪᴘʟᴇ ɪɴ ᴘʜʏsɪᴄs ᴛʜᴀᴛ sᴛᴀᴛᴇs ᴛʜᴀᴛ ғᴇᴡ ᴘᴀʀᴛs ɪɴ ᴀɴ ɪsᴏʟᴀᴛᴇᴅ sʏsᴛᴇᴍ ʀᴇᴍᴀɪɴɪɴɢ sᴛᴇᴀᴅʏ ᴀɴᴅ ᴜɴᴄʜᴀɴɢɪɴɢ ᴏᴠᴇʀ ᴛɪᴍᴇ ᴇᴠᴇɴ ᴡʜᴇɴ ᴏᴛʜᴇʀs ᴀʀᴇ ᴍᴏᴠɪɴɢ.