Question

A 500 g cart is released from rest 1.00 m from the bottom of a frictionless, 30.0? ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure (Figure 1) shows the force during the collision.

After the cart bounces, how far does it roll back up the ramp?

Express your answer with the appropriate units.

Answer 1

h = height of ramp from where the cart was released = 1 Sin30 = 0.5 m

Using conservation of energy

Kinetic enegy at bottom = potential energy at top

(0.5) m V² = mgh

V = sqrt (2 gh) = sqrt (2 x 9.8 x 0.5) = 3.13 m/s

impulse from the graph is given as

I = F t = area of graph = (0.5) x height x base = (0.5) (200) (26.7 x 10^-3) = 2.67

let the final velocity after collision = V_f

change in momentum of cart = impulse

m ( V_f - (-3.13)) = 2.67

(0.5) ( V_f - (-3.13)) = 2.67

V_f = 2.21 m/s

let the height gained while travelling up = H

Using conservation of energy

Kinetic enegy at bottom = potential energy at top

(0.5) m V_f² = mgH

H = (0.5) (2.21)² /(9.8) = 0.25 m

distance along the ramp = d = H/Sin30 = 0.25 / 0.5 = 0.5 m