Question

A and B are two particles having the same mass m. A is moving along X-axis with a speed of `10ms^(-1)` and B is at rest. After undergoing a perfectly elastic collision with B, particle A gets scattered through an angle of `30^(@)`. What is th edirection of motion of B, and the speeds of A and B, after the collision?

Answer 1

Figure shows the particles A and B before collision , and figure shows the two particles after perfectly elastic collision in two dimenstions.
Here, `u=10ms^(-1), theta =30^(@), phi=?`
`upsilon_(1)=? V_(2)=?`
As is known from Art. special case
`theta+phi=90^(@)`
`:. phi=90^(@)-theta=90^(@)-30^(@)=60^(@)`
Using law of conservation of linear momentum along X-axis,
`u=upsilon_(1)costheta+upsilon_(2)cosphi`
`10=upsilon_(1)cos30^(@)+upsilon_(2)cos60^(@)-(upsilon_(1)sqrt(3))/(2)+(upsilon_(2))/(2)`
or `20=sqrt(3)upsilon_(1)+upsilon_(2)` ...(i)
Again, using law of conservation of linear momentum along Y-axis.
`0=upsilon_(1)sintheta-upsilon_(2)sinphi`
`=upsilon_(1)sin30^(@) -upsilon^(2)sin60^(@)=(upsilon_(1))/(2)-(upsilon_(2)sqrt(3))/(2)`
or `upsilon_(2)sqrt(3)=upsilon_(1)` ...(ii)
From (i), `20=sqrt(3)upsilon_(2)sqrt(3)+upsilon_(2)=4upsilon_(2)`
`upsilon_(2)=(20)/(4)=5m//s`
From (ii), `v_(1)=sqrt(3)upsilon_(2)=sqrt(3)xx5=1.732xx5m//s`
`=8.66 m//s`