Two ice skaters are practicing a lift. They skate directly towards each other. The woman has a mass of 60 kg and a velocity of 3 m/s towards the east. The man has a mass of 100 kg and is skating at a velocity of 4 m/s towards the west. They time it badly and collide and bounce off each other’s outstretched hands. The woman moves off at a velocity of 5 m/s towards the west. With what velocity does the man move off and in which direction?
Given:
M1 = 60 kg, V1 = 3 m/s.
M2 = 100kg, V2 = -4 m/s.
V3 = -5 m/s = velocity of M1 after the collision.
V4 = ? = velocity of M2 after the collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
60*3 + 100*(-4) = 60*(-5) + 100*V4.
-220 + 300 = 100V4,
V4 = 0.8 m/s, East.
Thanks Henry you helped me with my schoolwork!!!
Thanks u helped with my homework
Please can somebody confirm if 0.1m/s West is the correct answer?
Letting East be +ve, and West be -ve
60x(+3) + 100x(-4) = 60x(-3.5) + 100x(?)
180 – 400 = -210 + 100(?)
-220 + 210 = 100(?)
100(?) = -10
? = 0.1m/s West
momentum is conserved
you're off by a factor of ten
Henry2,
Thank you for this, now that I look again I see that I had the right method but when I came to do the calculation for some reason I wrote down V3 as 3.5m/s rather than 5m/s as stated in the quesiton and as correctly noted by you! Silly careless error by me!!
Thanks again.
You are welcome.
P.S. Sometimes I make strange mistakes too.
Why did you make the 4 and 5 negative?
To solve this problem, we need to use the law of conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision.
To calculate the velocity at which the man moves off after the collision, we can set up an equation using the conservation of momentum.
First, let's define the initial momentum (before the collision) and the final momentum (after the collision) for the system.
Initial Momentum = Momentum of Woman + Momentum of Man
Final Momentum = Momentum of Woman + Momentum of Man
We know that the woman has a mass of 60 kg, a velocity of 3 m/s towards the east, and moves off with a velocity of 5 m/s towards the west. Therefore, her momentum before the collision is given by:
Momentum of Woman (initial) = mass of Woman * velocity of Woman (initial)
= 60 kg * 3 m/s = 180 kg·m/s towards the east
Her momentum after the collision is given by:
Momentum of Woman (final) = mass of Woman * velocity of Woman (final)
= 60 kg * 5 m/s = 300 kg·m/s towards the west
Similarly, the man has a mass of 100 kg, a velocity of 4 m/s towards the west, and his velocity after the collision is what we need to calculate.
Assuming east to be positive and west to be negative, we can write the equation for conservation of momentum as:
Initial Momentum = Final Momentum
(60 kg * 3 m/s) + (100 kg * (-4 m/s)) = (60 kg * (-5 m/s)) + (100 kg * velocity of Man (final))
Simplifying the equation, we get:
180 kg·m/s - 400 kg·m/s = -300 kg·m/s + (100 kg * velocity of Man (final))
-220 kg·m/s = -300 kg·m/s + (100 kg * velocity of Man (final))
Now, we can solve for the velocity of the man:
-220 kg·m/s + 300 kg·m/s = 100 kg * velocity of Man (final)
80 kg·m/s = 100 kg * velocity of Man (final)
Dividing both sides of the equation by 100 kg, we find:
velocity of Man (final) = 80 kg·m/s / 100 kg
velocity of Man (final) = 0.8 m/s towards the east
Therefore, the man moves off with a velocity of 0.8 m/s towards the east after the collision.
In summary, the man moves off with a velocity of 0.8 m/s towards the east, while the woman moves off with a velocity of 5 m/s towards the west.