Question

Two particles $A$ and $B$ start simultaneously from the same point and move in a horizontal plane. $A$ has an initial velocity $u_1$ due east and acceleration $a_1$ due north. $B$ has an initial velocity $u_2$ due north and acceleration $a_2$ due east. Which of the following statements is(are) correct?

• A. Their paths must intersect at some point.
• B. They must collide at some point.
• C. They will collide only if $a_1{u}_1=a_2{u}_2$.
• D. If $u_1>u_2$ and $a_1<a_2$, the particles will have the same speed at some point of time.

Answer 1

Answer: (A), (C), (D)

The correct options are
A Their paths must intersect at some point.
C They will collide only if $a_1{u}_1=a_2{u}_2$.
D If $u_1>u_2$ and $a_1<a_2$, the particles will have the same speed at some point of time.

For A ,
$S_A=u_{1}t\hat{i}+\frac{1}{2}{a}_1{t}^2\hat{j}$
For B ,
$S_B=\frac{1}{2}{a}_2{t}^2\hat{i}+u_{2}t\hat{j}$
For these particles to meet , the displacement in east and north directions should be same .
$u_{1}t=\frac{1}{2}{a}_2{t}^2$
Hence , $t=\frac{2u_1}{a_2}$
and
$\frac{1}{2}{a}_1{t}^2=u_{2}t$
Hence , $t=\frac{2u_2}{a_1}$
These two times should be equal for the particles to collide .
$⟹\frac{2u_2}{a_1}=\frac{2u_1}{a_2}\to a_1{u}_1=a_2{u}_2$
The path of A is a parabola of the form $Y=\frac{a_1{X}^2}{2u_1}$
The path of B is of the form $X=\frac{a_2{Y}^2}{2u_2}$
These 2 parabolae meet at $(0,0)$ and $(\frac{2a_2}{u_2}\left({\frac{u_1{u}_2^2}{a_1{a}_2^2}}^{\frac{2}{3}}\right),2\left({\frac{u_1{u}_2^2}{a_1{a}_2^2}}^{\frac{1}{3}}\right))$
For the speeds to be equal ,
$u_1^2+a_1^2{t}^2=u_2^2+a_2^2{t}^2$
Hence , $t^2=\frac{u_1^2-u_2^2}{a_2-a_1}$
If $u_1>u_2$ and $a_2>a_1$ , t will have a real value .