Two skaters A and B are moving in opposite directions such as they collide head-on and
immediately become entangled. Skater A has a mass of 40kg and is moving with a
velocity 4m/s eastwards while the other skater B has a mass of 60kg and is moving with
a velocity 3 m/s westwards. Assume that the frictional force acting between the skaters
and the ground is negligible.
i) Calculate the total momentum of the two skaters before collision
ii) With what velocity the skaters will move after the collision?
iii) In which direction, the two skaters will move after the collision?
Answers
Given: Mass of A, M₁ = 40kg
Mass of B, M₂ = 60kg
Velocity of A, V₁ = 4 m/s eastward
Velocity of B. V₂ = 6m/s westward
To find: (a) Total momentum of the two skaters before the collision
(b) Velocity that skaters will move after the collision
(c) Direction the two slaters will move after the collision.
Solution:
(a) Momentum of the skater A = M₁V₁
= 40×4 = 160kgm/s
Momentum of skater B will be = M₂V₂
= 60×(-3) = -180 m/s
Negative in the momentum of skater B is because we are considering the motion in the positive x-axis to be positive, and in the negative x-axis to be negative.
The total momentum of the two skaters before collision will be M₁V₁ +M₂V₂
= 160 + ( -180)
= -20 m/s
(b) The two skaters are moving on two different sides, so when they collide they both will move with a common velocity.
Since the momentum is being conserved, therefore
M₁V₁ + M₂V₂ = (M₁+M₂)V'
Where V' is the common velocity
40×4 + (60×-3) = (40+60) V'
-20/100 = V'
-0.2m/s = V'
Therefore, the velocity with which they move after the collision is 0.2 m/s in a westward direction.
(c) since the velocity after the collision is negative, and we have assumed that the direction at the negative x-axis is westward.
Therefore, the direction with which the two skaters will move after the collision will be westward.