for #8 i dont know what this question is asking me. first how do i tell if this

Question

for #8 i dont know what this question is asking me. first how do i tell if this question is asking me for impulse? it doesnt say anything about to solve for impulse...

i get the part where finding areas, but how do i know this is asking for impulse and initial velocity?

what where that 6*10^3 came from,, so damn confused on #8
explain in baby step because i dont understand this problem

#9 i got it.. .

In the hanging steel balls apparatus, when one ball at one end is dropped, why does one ball at the other end pop out at the same speed instead of two balls with half the speed, which would satisfy COM? A hooking mechanism on an aircraft carrier's deck can arrest a landing 6,000-Kg jetfighter with a force configuration as shown in the graph below. Determine the plane's maximum allowable touch-down speed. A 5Kg golf ball moving at -7m/s collides elastically head-on with a 15Kg bowling ball moving at +3 m/s. Find the vf of each ball after the collision. A moving train wagon collided with a smaller stationary one. The two wagons got tangled together and moved along the track at 6m/s after the collision. If the mass of the small wagon is one third that of the bigger wagon, what's the initial speed of the big wagon?

Explanation / Answer

#7 The answer says that if : at the other end two ball pop with half the speed, kinetic energy would not be conserved.

say taking two balls of mass m each as system,

change in KE = 0.5*m*v^2 - 0.5*m * (0.5v)^2 - 0.5*m * (0.5v)^2 = 0.25 *m *v^2

but if we take the case of other end ball popping with the same speed,

change in KE = 0.5*m*v^2 +0 - 0+ 0.5*m*v^2 =0

#8 Actually the question is asking for initial velocity (i,e touch down speed) and the final vel =0 from the relation: dp/dt = F (change in momentum / time )

=> deltaP= F*t (F*t is the area of given graph as u can see the unit of area is N*s)

since Vf=0 , so deltaP= m*Vi = F*t= 6*10^5 Ns

   => 6*10^3kg *Vi =   6*10^5 Ns

=> Vi= 100 m/s

touch down speed of plane is the speed of plane when its main gear wheel touches the surface, that implies its initial velocity.

6*10^3 kg is the mass of the aircraft.( mass* Vi = initial momentum of aircraft)

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