Chapter 09, Problem 041. The figure shows a two-ended rocket that is initially s

Question

Chapter 09, Problem 041. 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 = 5.30 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 rn/s relative to the velocity that block C then has (after the second explosion). At t = 3.10 s, what are (a) the velocity of block C (including sign) and (b) the position of its center? (a) Number Units (b) Number Units

Explanation / Answer

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 m/s. So, taking all speeds in m/s,

m (V1 - 3) + (m+M) V1 = 0

(M+2m) V1 = 3 m

V1 = 3m/(M+2m) = 0.75 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). Again, momentum conservation gives

M V2 + m ( V2 + 3) = (M+m) V1

(M+m) V2 = (M+m) V1 - 3 m

V2 = V1 - m/(M+m) * 3
= 0.75 m/s - 1.01 m/s = -0.26 m/s

So the velocity of C after the second explosion will be -0.26 m/s.

As for its position:

it moves from rest to the right for 0.80s at a speed of 0.75m/s. Therefore until the second explosion it will have travelled a distance:

s = v t = 0.75m/s * 0.80s = 0.6 m

Then, for a period of 3.1 seconds, it travels to the left with a speed of 0.26 m/s.
Its final position therefore is

x = 0.6m - 0.26 m/s * 3.1s
= -0.20 m

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