Question

Two skaters have weight in the ratio $4:5$ and are $9m$ apart, on a smooth frictionless surface. They pull on a rope stretched between them. The ratio of the distance covered by them when they meet each other will be

• A. $5:4$
• B. $4:5$
• C. $25:16$
• D. $16:25$

Answer 1

Answer: (A)

$5:4$

Let mass of 2 skaters be 4m and 5m. Let 4m mass move distance x meters and 5m mass moves 9-x distance.

$s=ut+\frac{1}{2}a{t}^2$

$x=0+\frac{1}{2}\frac{F}{4m}{t}^{2}9-x=0+\frac{1}{2}\frac{F}{5m}{t}^{2}ratio=\frac{x}{9-x}=\frac{\frac{1}{2}\frac{F}{4m}{t}^2}{\frac{1}{2}\frac{F}{4m}{t}^2}=5:4$