Question

Indicate the minimum number of tosses by each skater required to avoid collision.
Number of tosses by A =
Number of tosses by B =

Answer 1

First time when A tosses the ball linear momentum conserbe.

$70\times 1=10\times 6+70v$

$\frac{10}{70}$ right

time taken by ball to reach B $t_1=\frac{10}{7}sec$

delay in toss again by B is 1 sec.

total time $=(\frac{10}{7}+1)sec$

momentum conservation when B catch..

$70-60=80u=\frac{10}{80}=\frac{1}{8}$

again B tossed momentum conserbe when B through with speed 5 ms ball.

$\frac{70}{8}=10(5+\frac{1}{8})+70v^{\prime }\frac{70}{8}=50+\frac{10}{8}+{70}^{\prime }$

$v<0$ it means $B$ goes right.Similarly when the ball reach $A.$ $A$ also goes left without tosses the ball then number of tosses required to avoid collision .

So $A$ need - one toss, $B$ need - one toss.