An old firework shell is hanging at rest by a thread at a height H above level g
ID: 1902890 • Letter: A
Question
An old firework shell is hanging at rest by a thread at a height H above level ground. It explodes into three pieces, with masses mA, mB and mC, respectively. Initially, piece A moves horizontally toward the west at speed vA. Initially, piece B moves vertically upward at speed vB. (Ignore air resistance, assume constant free-fall acceleration g, and disregard any possible interference by the thread.) a. In what order do the pieces land on the level ground below? Explain how you know. b. How much time elapses between the first and last impacts? Explain/show your reasoning. c. What is the largest distance between any two of the three impact points? Explain/show your reasoning.Explanation / Answer
masses mA, mB and mC, VAi=-va i VBi=vb j inital momentum =final momentum 0=mAVA+mBVB+mcVC =>VC=+mA/mC VA i -mB/mC VB j a)H=.5g*tA^2 =>tA=sqrt(2H/g) b)for b it goes up and comes down with saem velocity so 2VB=g*t T=2VB/g T is time of flight t to hit ground is H=vBt+.5gt^2 a)For a part first particle C hits the ground then particle A hits the ground and B hits the ground C has some downward velocity so it reaches first downward A is in freefall in y direction B take some extra time to go up and come down so C goes down first followed by BothA&then B b)Time elapsed=2VB/g+T1-T2 where T1 is the best root satisfying H=vBt+.5gt^2 where T2 is root satisying H=VBmB/mCt+.5g*t^2 if mb and mc are equal then Time elapsed=2VB/g c)the largest impact distance is between the C and A particles as they land far away from current positions and in opposite directions d=sqrt(2H/g)*VA+VA*mA/mB *T2 where T2 is root satisying H=VBmB/mCt+.5g*t^2