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 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

Explanation / Answer

PART A

PE = mgh

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

0.5*mv^2 = mgh

v = sqrt(2*gh)

Conserving momentum before and after collision:

m*v = (m+2m)*v'

v' = speed of the combined mass

v' = mv/3m = v/3 = sqrt(2gh)/3

PART B

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

v1 = final speed of mass m

v2 = speed of mass 2m

v = v1 + 2*v2 (eq1)

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

v^2 = v1^2 + 2*v2^2 (eq2)

h = 1/(2*9.8) m

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

Solving both equations we get :

v1 = -1/3 = -v/3

v2 = 2/3 = 2v/3

Speed of the 1st mass = v/3

Height reached = h'

sqrt(2*g*h') = v/3 = 2*g*h' = v^2/9 = 2*g*h/9

h' = h/9

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects