Question

1. The person in Fig. 1 weighs 90 kg and the mass of the wagon (excluding the rocks) is 100 kg. Initially there are 50 one-kg rocks in the wagon. At t 0 the person starts to throw rocks out the back of the wagon at the rate of 1 rock per second. If the rocks leave the person's hand at a speed of 30 m/s, plot the position, X, and velocity U, of the wagon for t-1 to 50 seconds. Neglect wind drag on the person and the wagon. The initial conditions are: X 0 m and U 0 m/s at t-0 s (10 points) Case (a) Assume no friction loss.

Answer 1

As there are no external forces acting (Assumed that friction losses are zero) so we can apply Principle of Conservation of Linear Momentum.

Mass of remaining rocks at time t= [Mass_{(at t=0sec} - t] kg= (50-t) kg

M₁= Mass of Body 1= (Mass of Wagon+Mass of person+ Mass of remaining Rocks at that time)

= (100+90+50-t) kg

=(240-t) kg

_V1=velocity of Body 1

M2=Mass of body 2= mass of rock thrown away= 1kg

As per convention given in question, Velocity towards right is negative and that towards leftward is positive

V2=velocity of Body 2= (-30)i

M₁*V₁+ M₂*V₂=0

(240-t)*V₁ + 1*(-30i)=0

(240-t)*V₁ -30i=0

(240-t)*V₁ = 30i

V₁ = [30/ (240-t)] i m/s

for first second (t=1)

V_1(t=1sec)= [30/ (240-1)] i m/s

=0.1255 m/s [towards left as thats positive i direction]

Similarly we can calculate for any second by substituting value of t for that moment,

V_1(t=2sec)= [30/ (240-2)] i m/s

=0.1260 m/s

V_1(t=3sec)= [30/ (240-3)] i m/s

=0.1265 m/s

.........

V_1(t=47sec)= [30/ (240-47)] i m/s

=0.1554 m/s

V_1(t=48sec)= [30/ (240-48)] i m/s

=0.1562 m/s

V_1(t=49sec)= [30/ (240-49)] i m/s

=0.1570 m/s

V_1(t=50sec)= [30/ (240-50)] i m/s

=0.1579 m/s

For Calculation of X:

We know   

$int_{s1}^{s2}dS=int_{t}^{t+1}V*dt$

As we know V is constant for any particular second

S2-S1= V_t *(t+1-t)= V_t*(1)= V_t

At start (t=0) S₀=0

S₁-S₀=V_t=1

S₁-0=0.1255 meters

S₁=0.1255 meters

S2-S₁=V_t=2

S₂-0.1255=0.1260

S₂=0.2515

similarly we can calculate all the S_t