The figure shows a two-ended “rocket” that is initially stationary on a friction

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 = 7.40 kg) and blocks L and R (each of mass m = 2.70 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 2.60 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.70 s, block R is shot to the right with a speed of 2.60 m/s relative to the velocity that block C then has (after the second explosion). At t = 3.30 s, what are (a) the velocity of block C (including sign) and (b) the position of its center?

Explanation / Answer

Ans:-

initial momentum = 0
final momentum = 0 = 2.7kg•u + 10.1kg•v
Also, v - u = 2.6m/s
so v = u + 2.6m/s
so 0 = 2.7kg•u + 10.1kg•(2.6m/s + u) = 2.7kg•u + 26.26kg·m/s + 10.1kg•u
-12.8kg•u = 26.26kg·m/s
u = -2.05 m/s block L to the left
v = 2.6m/s + u = 0.55 m/s blocks C and R to the right

Repeat procedure for second explosion
initial momentum = 10.1kg•0.55m/s = 5.56 kg·m/s
final momentum = 5.56 kg·m/s = 7.4kg•u + 2.7kg•v
v = u + 2.6m/s
5.56 kg·m/s = 7.4kg•u + 2.7kg•(u + 2.6m/s) = 7.4kg•u + 2.7kg•u + 7.02kg·m/s
-1.46kg·m/s = 10.1kg•u
u = -0.145 m/s block C to the left answer (a)
v = 2.455 m/s block R to the right

(b) the position of the center of C is
x = 0.55m/s•0.7s - 0.145m/s•3s = 0.34m right of its initial location

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