The figure shows a two-ended The figure shows a two-ended ??rocket?? that is ini

Question

The figure shows a two-ended

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 = 7.80 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 timet = 0, block L is shot to the left with a speed of 2.70 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 2.70 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

On the first explosion, we treat L as a separate body from C and R.

Let

vL = velocity of A after the explosion
vCR = velocity of combined C and R after the explosion

As they are at rest initially, using conservation of momentum,

0 = mLvL + (mC + mR)vCR

Also, the relative velocity of the two objects is 2.70 m/s.

Thus,

vCR - vL = 2.70 m/s

Solving for vCR and VL,

vL = -2.1477 m/s
vCR = 0.55227 m/s

Now, for the second explosion,

Let

vC = velocity of C after explosion
vR = velocity of R after the explosion

Thus, by conservation of momentum,

(mC + mR)vCR = mCvC + mRvR

Also, using the relative velocity,

vR - vC = 2.70 m/s

Then, solving for vR and vC,

vC = -0.1419 m/s
vR = 2.5581 m/s

THUS:
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At t = 3.30 s,

vC = -0.142 m/s [ANSWER, PART A]

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It travelled at vCR = 0.55227 m/s for 0.80 s, then at vC from 0.80 s to 3.30 s (2.50 s).

Thus,

x = 0.55227 m/s * 0.8 s - 0.1419 m/s * 2.50 s

= 0.0871 m   [ANSWER, PART B]

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