Chapter 09, Problem 041 The figure shows a two-ended rocket that is initialy sta

Question

Chapter 09, Problem 041 The figure shows a two-ended rocket that is initialy 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 = 6.60 kg) and blocks L and R (each of mass m 2.50 kg) on the left and right sides. Smal explosions can shoot either of the side hlocks away from block C and along thex axis. Here is the sequence: (1) At time t-D, block is shot to the left with a speed of 3.20 ms 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 3.20 elative to the velocity that block C then has (after the secod explosion). At t 3.20 s what are (a) the velooty of block C (including sign) and (b) the position of its center? (a) Number (b) Number Click if you would like to Show Work for this question: Units Units Open show Work

Explanation / Answer

(a) This is an exercise in momentum conservation (there are no external forces so the system's momentum is conserved.

The initial momentum is zero. After the first explosion blocks C and R move to the right with speed V1 and block L moves to the left with speed V1-3.20m/s. So, taking all speeds in m/s,

m (V1 - 3.20) + (m+M) V1 = 0

(M+2m) V1 = 3.20 m

V1 = m/(M+2m) * 3.20 = 0.6897 m/s

The momentum of CR just before the second explosion is (M+m) V1 .

After the second explosion M moves with speed V2 and m moves with speed (V2 + 3.20). Again, momentum conservation gives

M V2 + m ( V2 + 3.20) = (M+m) V1

(M+m) V2 = (M+m) V1 - 3.20 m

V2 = V1 - m/(M+m) * 3.20

= 0.6897 m/s – 0.8791 m/s = -0.1894 m/s

So the velocity of C after the second explosion will be -0.1894 m/s.

(b) As for its position:

it moves from rest to the right for 0.70s at a speed of 0.6897m/s. Therefore until the second explosion it will have travelled a distance:

s = v t = 0.6897/s * 0.70s = 0.4828 m

Then, for a period of (3.2 – 0.7) = 2.5 seconds, it travels to the left with a speed of 0.1894 m/s.

Its final position therefore is

x = 0.4828m - 0.1894 m/s * 2.5s

= 9.3 x 10^-3 m

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