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.
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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
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
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