Question

Two balls are 8.0 meters apart and moving directly towards each other. If the first ball is moving at a speed of 2.5 ms–1 with respect to the ground and the second ball 3.5 ms–1 with respect to the ground, where will they collide? 

Answer 1

$s = 8m$

$v_1 =2.5\frac{m}{s}\newline v_2 =3.5\frac{m}{s}\newline s_1 = v_1t\newline s_2 = v_2t\newline$

$s = s_1+s_2$

$v_1t +v_2t = s$

$t= \frac{s}{v_1+v_2}=\frac{8}{6}\approx 1.33s$

$s_1 =1.33*2.5\approx3.33m$

$s_2 =s - s_1= 4.67m$

$\text{Answer: distance collision point :}$

$3.33m \text{ from the beginning of the movement of 1 ball}$

$4.67 m \text{ from the beginning of the movement of 2 ball}$