The figure below shows a two-ended \"rocket\" that is initially stationary on a

Question

The figure below 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 = 6.00 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 3.00 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.80 s, block R is shot to the right with a speed of 3.00 m/s relative to the velocity that the explosion gives to block C? (a) What is the velocity of block C at t = 2.80 s? . m/s (b) What is the position of block C's center at that time?

Explanation / Answer

initial momentum = 0 final momentum = 0 = 2.3kg•u + 8.3kg•v Also, v - u = 2.2m/s so v = u + 2.2m/s so 0 = 2.3kg•u + 8.3kg•(2.2m/s + u) = 2.3kg•u + 18.26kg·m/s + 8.3kg•u -10.6kg•u = 18.26kg·m/s u = -1.72 m/s ? block L to the left v = 2.2m/s + u = 0.48 m/s ? blocks C and R to the right Repeat procedure for second explosion initial momentum = 8.3kg•0.48m/s = 3.96 kg·m/s final momentum = 3.96 kg·m/s = 6kg•u + 2.3kg•v v = u + 2.2m/s 3.96 kg·m/s = 6kg•u + 2.3kg•(u + 2.2m/s) = 6kg•u + 2.3kg•u + 5.06kg·m/s -1.1kg·m/s = 8.3kg•u u = -0.133 m/s ? block C to the left ? answer (a) v = 1.867 m/s ? block R to the right (b) the position of the center of C is x = 0.48m/s•0.8s - 0.133m/s•2s = 0.12m right of its initial locaion

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