Momentum of Skiers in a Completely Inelastic Collision

[HarleyM]

Homework Statement

In a movie stunt, a 65 kg skier starts from a rest position at the top of a hill 30m high. She slides down the hill to the bottom, where she collides with a 45 kg stationary skier. The collision is completely inelastic. Find the final velocity of the skiers.

Homework Equations

M1V1 +M2V2= (M1+M2)V
ET'=ET'
ET'=(M1+m2)gh ( ithink this is relevant?)

The Attempt at a Solution

ET'= (65+45)(9.8)(30)
ET'= 32,340

32,340= 1/2mv²



Sq RT[(32,340)(2)/(65)] = velocity of skier going down hill?

=31.5 m/s (100 km/h + seems kinda fast for a skier)

 

[technician]

I think you need to first find the speed of the 65kg skier from their starting height. You can then calculate momentum and go from there.
If the collision is completely inelastic then ONLY momentum is conserved in the collision.

 

[HarleyM]

How would I calculate the velocity of the skier? This is where I am confused.. Thats why I used the ET' calculation..

I know you don't want to give me the answer.. but any hints would be immensely appreciated.. I mean I know the mass and height of the skier, no idea how to calculate their velocity though..

thanks for your time and response!

 

[technician]

the skier gains KE as a result of 'falling' a distance of 30m.
You need to use PE = KE to calculate the speed of the 65kg skier.
Then it is a 'straight forward' momentum before = momentum after calculation.
Have a go and if you get stuck let me know where it goes wrong.

 

[HarleyM]

the skier gains KE as a result of 'falling' a distance of 30m.
You need to use PE = KE to calculate the speed of the 65kg skier.
Then it is a 'straight forward' momentum before = momentum after calculation.
Have a go and if you get stuck let me know where it goes wrong.

THANKS again!

So PE=KE
mgh=1/2mv²
√(2)(9.8)(30)= V
V= 24.25 m/s


Ek = 1/2mv²
=1/2(65)(24.3)²
=19,190.9 J (initial energy) ( don't think I have to do this)

PT=PT'

M1V1+M2V2=(M1+M2)V'
(65)(24.3)+(45)(0)=(110)V'
V'=(65*24.3)/(110)
V'=14.36 m/s

Ek= 1/2(M1+M2)V'²
=1/2(110)(14.36)²
=11 341.53


how does the final velocity of both look? As that's what I am most concerned about.

Again thank you so much for your help!

 

[technician]

That's it... you did not need to calculate KE in the question but well done.

 

[HarleyM]

Thank you so much ! Is there a rating system at PF?

 

[technician]

Don't know !... not bothered...