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Particle A makes a perfectly elastic col...

Particle A makes a perfectly elastic collision with anther particle B at rest. They fly apart in opposite direction with equal speeds. If the masses are `m_(A)&m_(B)` respectively, then

A

`1/2`

B

`1/3`

C

`1/4`

D

`1/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
B
By the principle of conservation of linear momentum
`m_(1)u_(1)+m_(2) xx0=m_(1)v_(1)+m_(2)(-v_(1))`
`:. M_(1) u_(1) = (m_(1) -m_(2) ) v_(1)" "…(i)`
`:. (u_(1))/(v_(1))=(m_(1)_m_(2))/(m_(1))" "...(2)`
and by the principle of conservation of K.E
`1/2 m_(1) u_(1)^(2)=1/2 (m_(1)+m_(2))v_(1)^(2)" "...(3)`
Dividing (3) by (1) , we get
`u_(1)=((m_(1)+m_(2))/(m_(1)-m_(2)))v_(1)`
`:.(u_(1))/(v_(1))=(m_(1)+m_(2))/(m_(1)-m_(2))" "...(4)`
`:.` From (2) and (4) , we get
`(m_(1)-m_(2))/(m_(1))=(m_(1)+m_(2))/(m_(1)-m_(2))`
`:. m_(1)^(2)+m_(2)^(2)-2m_(1)m_(2)=m_(1)^(2)+m_(2)m_(1)`
`:. m_(2)^(2)=3m_(1)m_(2) :. (m_(1))/(m_(2)) =1/3 " or, " (mA)/(mg) = 1/3`
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