A skateboarder of mass m rolls, starting from rest, down a ramp of height h at t

Question

A skateboarder of mass m rolls, starting from rest, down a ramp of height h at the local skate park. Unfortunately another skater standing at the bottom of the ramp, who has mass 2m, didn't get out of the way fast enough, causing the two skaters to collide. Assume that the ramp is frictionless. 2. A skateboarder of mass m rolls, starting from rest, down a ramp of height h at the local skate park. Unfortunately another skater standing at the bottom of the ramp, who has mass 2m, didn't get out of the way fast enough, causing the two skaters to collide. Assume that the ramp is frictionless. a. If the two skateboarders become entangled and cannot separate, what is their speed after the collision? Give you answer in terms of the experimental variables and g. b. Now consider the scenario if the collision between the skateboarders is perfectly elastic. What will be the maximum height that the first skateboarder (with mass m) reaches as she travels back up the ramp? Give you answer in terms of experimental variables and g.

Explanation / Answer

For this problem we will use the conservation of energy to find the speed of the skateboard in the bottom of the ramp before the crash

Emi = m g h

Emf = ½ m V2

Emi = Emf

mgh = ½ m V2

V2= 2gh

this is the speed just before the crash, we now use the conservation of momentum

Part a)

pi = m V

pf = (m+ 2m) Vf

pi = pf

m V = (m+2m) Vf

Vf = m/(3m) V

Vf = sqrt (2gh)

Part b)

If elastic shock

pf = m V1f + 2m V2f

pi =pf

m V1 = m V1f + 2m V2f

(V1-V1f) = 2V2f (1)

We write the conservation of kinetic energy

½ m V12 = ½ m V1f2 + ½ 2m V2f2

V12- V1f2= 2V2f2

Using (a+b)(a-b)= a2– b2

(V1+V1f) (V1 – V1f) = 2 V2f2 (2)

(V1+V1f) (V1 – V1f)  = 2 1/22 (V1 – V1f)2

calculate from 1 y 2

(V1+V1f) =1/2 (V1 – V1f)

3/2 V1f =-1/2 V1

V1f = - V1

V1f = - sqrt (2gh)

This is the speed of the shock body 1 goes up the ramp, in this case we use the conservation of energy to find the maximum height of the rise

Ei = ½ m V1f2

Ef= m g H

Ei = Ef

½ m V1f2= m g H

H = 1/2g V1f2

H = 1/2g ( sqrt (2gh))2

H = 1/2g 1/9 2gh

H = 1/9 h

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