Figure 9-55 shows a two-ended \'\'rocket\'\' that is initially stationary on a f

Question

Figure 9-55 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 = 7.50 kg) and blocks L and R (each of mass m = 1.40 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 3.10 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 3.10 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?

Explanation / Answer

1) initially at t = 0 s, velocity of block L relative to velocity of rest of the rocket is 3.10 m/s because velocity of the rest of the rocket is v1 then, velocity of L block is              v = (v1 -3.10 )i ^m/s As there are no external forces, the momentum of the system is conserved.    m(v1-3.10) + (M+m)v1 = 0                   ...... (1)
Here M = 7.50 kg         m = 1.40 kg plugthe values in eq 1 and solve for v1,       v1 = (0.421359 m/s) i^ 2) At t = 0.8 s velocity of the Cblock be v2 then apply conservation of momentum, velocity of right block  relative to the velocity that block C is 3.10 m/s then the velocity of right block is v2+3.10 then,        Mv2 + m(v2+3.10 m/s) = (M+m)v1       (7.50 kg)v2 + (1.4 kg)(v2+3.10 m/s) = (8.9 kg)(0.421359 m/s) By solving, we get
            v2 = (-0.06628 m/s) i^ -----------------------------------------------------------------------------------------------------
b) At t = 2.90 s, the velocity of block C is (-0.06628 m/s) i^ because it is constant. Then between t = 0 sec to t = 0.8 sec C has moved a distance of                x1 = v1t                  = (0.421359 m/s)(0.8 s)                    = 0.3370872 m                    = 0.34 m Then between t = 0.8 sec to t = 2.90 sec C has moved a distance of                x2 = v2t                  = (-0.06628 m/s)(2.1 s)                    = -0.139188 m                    = -0.14 m (nearly) The total displacement is             d = 0.3370872 m - -0.139188 m                = 0.1978992 m                = 0.2 m (nearly) Then between t = 0.8 sec to t = 2.90 sec C has moved a distance of                x2 = v2t                  = (-0.06628 m/s)(2.1 s)                    = -0.139188 m                    = -0.14 m (nearly) The total displacement is             d = 0.3370872 m - -0.139188 m                = 0.1978992 m                = 0.2 m (nearly)

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