A skateboarder of mass m rolls starting from rest down a ram

A skateboarder of mass m rolls, starting from rest, down a ramp of height h at the local skate park. Unfortunately another skater 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 cannon 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.

Skaterboarder with mass m Ramp height h Skaterboarder with mass 2m

Solution

a)

The potential energy stored by the 1st skater at the top of the ramp:

PE = mgh

At the bottom, the PE is converted to Kinetic energy (KE)

So, 0.5*mv^2 = mgh

So, v = sqrt(2*gh)

Now, conserving momentum before and after collision:

m*v = (m+2m)*v\' <-------- v\' = speed of the combined mass

So, v\' = mv/3m = v/3 = sqrt(2gh)/3 <-----answer

b)

In this case, conserving momentum:

m*v = m*v1 + 2m*v2 <------ v1 = final speed of mass m, v2 = speed of mass 2m

So, v = v1 + 2*v2 ------- (1)

Conserving energy:

0.5*m*v^2 = 0.5*m*v1^2 + 0.5*2m*v2^2

So, v^2 = v1^2 + 2*v2^2 ---- (2)

Let h = 1/(2*9.8) m <---- we can safely assume this value for simplifying our calculation

So, v = sqrt(2*gh) = sqrt(2*9.8*1/(2*9.8)) = 1 m/s

So, Solving both equations we get :

v1 = -1/3 = -v/3

v2 = 2/3 = 2v/3

So, speed of the 1st mass = v/3

So, let the height reached be h\'

So, sqrt(2*g*h\') = v/3

So, 2*g*h\' = v^2/9 = 2*g*h/9

So, h\' = h/9 <-------answer