InterviewSolution
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InterviewSolution
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Two sphere `A and B` of masses `m_(1) and m_(2)` respectivelly colides. A is at rest initally and `B` is moving with velocity `v` along x-axis. After collision `B` has a velocity `(v)/(2)` in a direction perpendicular to the original direction. The mass `A` moves after collision in the direction.A. same as that of `B`B. opposite to that of `B`C. `theta=tan^(-1)(1//2)`to the `x-`axisD. `theta=tan^(-1)(-1//2)`to the `x-`axis |
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Answer» Correct Answer - D Here, mass of `A=m_(1), ` mass of `B=m_(2)` Initial velocity of `A,U_(1)=0` Initial velocity of `B,u_(2)=upsilon, `along `x-` axis. ltbr. After collision, let final velocity of `A` along `x-` axis be `v_(x)` and final velocity `A` along `y-` axis be `upsilon_(y)`. Final velocity of `B=(upsilon)/( 2)` along `Y-` axis. Applying principle of conservation of linear momentum along `x-` axis. `m_(1)xx0+ m_(2) upsilon=m_(1) upsilon_(x)+ m_(2)xx0` `upsilon_(x)=( m_(2)upsilon)/(m_(1)) ....(i)` Applying principle of conservation of linear momentum along `Y-axis`, `m_(1)xx0 + m_(2)xx0=m_(1)upsilon_(y)+m_(2)((upsilon)/(2))` or `upsilon^(y)=-(m_(2)upsilon)/(2m_(1)) ...(ii)` If mass `A` moves at angle `theta` with the `X-`axis, then `tan theta=(upsilon_(y))/( upsilon_(x))=-(m_(2)upsilon)/(2m_(1))xx(m_(1))/(m_(2)upsilon)=-(1)/(2)` `:. theta= tan^(-1)((1)/(2)) ` to the `x-`axis |
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