When a body of mass `m_(1)` moving with uniform velocity `40 ms^(-1)` collides with another body of mass `m_(2)` at rest, then the two together begin

Question

When a body of mass `m_(1)` moving with uniform velocity `40 ms^(-1)` collides with another body of mass `m_(2)` at rest, then the two together begin

When a body of mass `m_(1)` moving with uniform velocity `40 ms^(-1)` collides with another body of mass `m_(2)` at rest, then the two together begin to move with uniform velocity of `30 ms^(-1)`. The ratio of the masses (i.e., `m_(1)//m_(2)`) of the two bodies will be
A. `1 : 3`
B. `3 : 1`
C. `1 : 1.33`
D. `1 : 0.75`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
Initial momentum of the system
`=m_(1)xx40+m_(2) xx0=40 m_(1)`
Final momentum of the system `=(m_(1)+m_(2))xx30`
By the law of conservation of momentum,
Initial momentum = Final momentum
`40 m_(1) =(m_(1)+m_(2))30`
`40 m_(1)-30 m_(1) =30 m_(2)`
`10 m_(1) = 30 m_(2) implies m_(1)/m_(2) =3/1`

4 views Answered 2023-02-05 15:29