Question

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?

Answer 1

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

Answer 2

momentum is conserved

you're off by a factor of ten

Answer 3

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.

Answer 4

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.

Answer 5

You are welcome.

P.S. Sometimes I make strange mistakes too.

Answer 6

Why did you make the 4 and 5 negative?

Answer 7

Thanks Henry you helped me with my schoolwork!!!

Answer 8

Thanks u helped with my homework

Answer 9

Please take this off ur site my students are cheating