Two friends on roller-skates are standing
5 m apart facing each other. One of them throws a ball of 2 kg towards the other,
who catches it. How will this activity
affect the position of the two? Explain
your answer.
Answers
Explanation:
Let us say mass of the friends is m kg each. For example...
We apply the conservation of momentum principle here. Let us say the ball of 2 kg is thrown with a velocity u by one friend on the left, towards his friend on the right. We take u as the horizontal component of velocity. Initially the friends were stationary. Let the friend on the left move further left with a velocity v.
m v + 2 kg * u = 0 v = - 2 u / m
Let us say the second friend move further to right with a velocity V after catching the ball.
2 * u + m * 0 = (m+2) * V
V = 2 u / (m+2)
The friends separate with a relative speed =
V - v = 2u / (m+2) + 2 u / m
= 2 u [ 2m + 2 ] / m (m+2)
= 2 u * 2m / m * m = 4 u / m if m >>2 kg
❤️❤️❤️❤️❤️❤️❤️
Hey mate here is your answer.
Explanation:
Separation between them will increase. Initially the momentum of both of them are zero as they are at rest. In order to conserve the momentum the one who throws the ball would move backward. The second will experience a net force after catching the ball and therefore will move backwards that is in the direction of the force.