Question

You must provide the general ( algebraic ) solution and only substitute numerical values in the last line for full credit. Answers have to be on math/quad-ruled paper and clearly marked and readable. Problem 1: In a Freight Train Classification Yard, a freight cart of 6 tonnes ( 6,000 kg) is pushed over the hump ( Position 0 ), reaching a speed of Vi = 8 m/s at the bottom ( Positioni). 2 1.1: How high was the hill (assurning no friction)? At the bottom, the cart hits another cart (static) of mass me, the two carts interlock and proceed down the track. What is the end-speed v, if the second cart has the following mass: 1.2: m = 18 tonnes ? 1.3: m2 = 6 tonnes ? 1.4: m = 2 tonnes ? 1.5: Calculate the "ost energy in this collision by forming: Es E -Eow using only mi, ma and the speed at position "1"(.) on the right hand side: no numerical values nor the term Vz. Problem 2: A runner m = 75 [kg]) at the Red Bull "Toughest 400 meters" has to run up a Ski-Jump facility at an angle of 30° As=400m). B=30°(0.52 rad) 0 2.1: What is the power the runner has to expend to reach the top in 8 minutes ? Watch units !! 2.2: What is the power the runner has to expend to reach the top in 9 minutes ?? D 23: What is the finish time if the power output is maintained at 350 Watts ?

Answer 1

Given
1. Train fright cart mass m = 6*10^3 kg

the speed at poisition 1 is , v1 = 8 m/s
1.1

here by conservation of energy the gravitational potential energyat position 0 = kinetic energy at position1

   mgh = 0.5*m*v^2

   h = v^2/(2*g)

   h = 8^2/(2*9.8) m

   h = 3.265 m

the height of the hill is h = 3.265 m

1.2

   if it collides with m2 = 18 tonnes , and interlocked each other and moving with speedv2 = ?

from conservation of momentum

   P1 = P2

P1 is momentum of the system before collision and
P2 is momentum of the system afterthe collision

   P1 = m1*u1+m2*u2

   P2 = (m1+m2)v2

m1*u1+m2*u2 = (m1+m2)v2

v2 = m1*u1+m2*u2 / (m1+m2)

substituting the values

   v2 = (6000*8+180000*0)/(6000+18000) m/s = 2 m/s

1.3
   m2 = 6 tonnes

P1 = m1*u1+m2*u2

   P2 = (m1+m2)v2

m1*u1+m2*u2 = (m1+m2)v2

v2 = m1*u1+m2*u2 / (m1+m2)

substituting the values

   v2 = (6000*8+6000*0)/(6000+6000) m/s = 4 m/s

1.4
m2 = 2 tonnes

P1 = m1*u1+m2*u2

   P2 = (m1+m2)v2

m1*u1+m2*u2 = (m1+m2)v2

v2 = m1*u1+m2*u2 / (m1+m2)

substituting the values

   v2 = (6000*8+2000*0)/(6000+2000) m/s = 6 m/s

1.5

lost energy = E_begin- E_end

   E_lost = (k.e1 - k.e2)
k.e1 is kinetic energy of m1 at position1 and k.2 is the kientic energy of both (m1,m2) after postion 2

   E_lost = 0.5*m1*v1^2 - 0.5(m1+m2)v2^2

-------------

Given

runner mass m = 75 kg
Ds = 400 m
inclination is theta = 30 degrees

2.1
power of the runner P = ? if the runner reach the top in 8 min

We know that the power P = E/t = mgh/t
here h = dS*sin theta

   h = 400 sin30 = 200 m

   P = 75*9.8*200/(8*60) W

   P = 306.25 W

2.2
   t = 9 min

power P = E/t = mgh/t
here h = dS*sin theta

   h = 400 sin30 = 200 m

   P = 75*9.8*200/(9*60) W

   P = 272.22 W

2.3

t = ? if P = 350 W

   P = E/t = mgh/t

substituting the values

   350 = 75*9.8*200/t

   t = 420 s

   t = 7 min