After several large firecrackers have been inserted into its holes, a bowling ba
ID: 1962013 • Letter: A
Question
After several large firecrackers have been inserted into its holes, a bowling ball is projected into the air using a homemade launcher and explodes in midair. During the launch, the 8-kg ball is shot into the air with initial speed v0 = 10.8 m/s at angle = 39°; it explodes at the peak of its trajectory, breaking into three pieces of equal mass. One piece travels straight up with a speed of 3.0 m/s. Another piece travels straight back with a speed of 2.0 m/s. What is the velocity of the third piece (speed and direction)? (Let up be the +y-direction and to the right be the +x-direction.)
speed:
Explanation / Answer
horizondal velocity at launch u=10.8cos(39) =8.393 during explosion u =8.393 and v=0 momentum=m*8.393 applying conservation of momentum in x derection 8.393m =-2m/3 + x*m/3 solving x= 27.18 applying momentum conservation in y direction 0=3*m/3 + y*m/3 y = -3 so total velocity = 27.18i -3j or total velocity =sqrt(27.18^2 + 3^2) =27.34 m/s direction =tani(-3/27.18) =-6.29 degree or 6.29 degree downward from horizondal