Two skaters A and B of masses 20 kg and 40 kg respectively stand facing each other 10 m apart. They pull each other by a inextensible rope. How far each has moved when they meet?
Answers
Answer:
c
Explanation:
Correct option is
C
A moves 3.5 metres and B moves 2.5 metres.
Here no any external force is applied taking two stakers A and B as a system.
then center of mass of both stakers will not change therefore they meat at center of mass, taking axis as shown in figure 1.
X
com
=
50+70
50×0+70×6
=
120
70×6
=3.5m from 50kg sketers.
X
com
= X-coordinate of center of mass.
So A mover 3.5cm
B mover (6 - 3.5) = 2.5m
Option C-correct
Given: Two skaters A and B of masses 20 kg and 40 kg respectively stand facing each other 10 m apart. They pull each other by a inextensible rope.
To find: Distance each skater has moved when they meet
Explanation: Let the mass of skater A be m1 and mass of skater B be m2.
Let skater A be at the origin (0,0). Since they are 10 m apart facing each other, therefore skater B is at (10,0).
As there is no external force other acting on the system of skaters A and B, so the skaters would meet at the centre of mass at the end, and the centre of mass will not get displaced from its place.
The coordinates of the center of mass is given by the formula:
m1*x1 + m2*x2 / m1+ m2, m1* y1 + m2*y2 / m1+ m2
m1 = 20 kg
m2 = 40 kg
x1 =0, y1 =0
x2 = 10, y2= 0
Using the values in the formula, the coordinates are:
20*0+40*10/ 20+40, 20*0+40*0/ 20+40
= 400/60, 0
= 6.67,0
Since the centre of mass is at 6.67 m with respect to skater A, therefore skater A travels 6.67 m.
Distance travelled by skater B
= Distance between skaters- Distance travelled by skater A
= 10-6.67
= 3.33 m
Therefore, skater A travels 6.67 m and skater B travels 3.33 m when they meet.