Question

The figure shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6.70 kg) and blocks L and R (each of mass m = 1.50 kg) on the left and right sides. Small explosions can shoot either of the side blocks away from block C and along the x axis. Here is the sequence: 

(1) At time t = 0, block L is shot to the left with a speed of 2.70 m/s relative to the velocity that the explosion gives the rest of the rocket. 

(2) Next, at timet = 0.50 s, block R is shot to the right with a speed of 2.70 m/s relative to the velocity that block C then has (after the second explosion). At t = 2.90 s, what are (a) the velocity of block C(including sign) and (b) the position of its center?

Answer 1

answer) first on applying conservation of momentum on first explosion

1.5(-2.7+vo)+(6.7+1.5)vo=0

-4.05+1.5vo+8.2v0=0

9.7vo=4.05

vo=0.4175 m/s

again we have after the 2nd explosion

8.2vo=6.7vc+1.5(vc+2.7)

replacing the value of vo in above eqn we get

8.2*0.4175=8.2vc+4.05

vc=-0.6265/8.2=-0.0764 m/s

answer to part a) -0.0764 m/s or -0.076 m/s

b) we know distance is

d=v*t

d=0.4175*0.5-0.0764(2.9-0.5)

d=0.20875-0.18336

d=0.0254 m

so the answer is 0.0254 m or 0.025 m