Question

Gayle runs at a speed of 2.75 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 8.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill.

What is their speed at the bottom of the hill if the total vertical drop is 15.0 m? Gayle's mass is 33.0 kg, the sled has a mass of 5.70 kg and her brother has a mass of 42.0 kg.

Answer 1

Given,

Vi = 2.75 m/s

Msled = 5.7kg

MGayle = 33kg

Mbro = 42 kg

Initially when Gayle jumps on sled, by momentum conservation,

MGayle * Vi = ( Msled + MGayle )*V

=> V =33 * 2.75 /(5.7+33) = 2.35 m/s

After descending a vertical height of 8m, energy will increase by ( Msled + MGayle )*g*8

So, new velocity Vn :

(1/2) ( Msled + MGayle )*Vn² = ( Msled + MGayle )*g*8 +(1/2)* ( Msled + MGayle )* V²

=> Vn = 12.85 m/s

Her brother jumps on at this point. Using momentum conservation

( Msled + MGayle ) *Vn = ( Msled + MGayle +Mbro)*V'

=> V' = 6.16 m/s

The slide goes down a further (15-8) = 7m

Using the same energy conversion as above

( Msled + MGayle +Mbro)*(1/2)*V'² + ( Msled + MGayle +Mbro)*g*7 =( Msled + MGayle +Mbro)*(1/2)*(V'')²

=> V'' = 13.24 m/s final velocity