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 = 2.00 kg) on the left and rigid sides. Small explosions can shoot either of the side blocks away from block C and along the x-axis. Here is the sequence: At time t = 0, block L is shot to the left with a speed of 3.0 m/s relative to the velocity that the explosion gives the rest of the rocket. 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 block C then has. At t = 2.80s, What are; the velocity of block C and the position of its center?

Explanation / Answer

given that ::

mass of block C, m1 = 6 kg

mass of block L, m2 = 2 kg

mass of block R, m3 = 2 kg

here is the sequence :

1. at time, t = 0, block L is shoot to the left with a speed of 3 m/s relative to the velocity that the explosion gives the rest of the rocket is given as -

for the first explosion :

Initial momentum, p = 0

Final momentum, 0 = m2 u + (m1 + m2)v                                    { eq. 1 }

where, M = m1 + m2

inserting the values in above eq.

0 = (2 kg) u + (6 kg + 2 kg) v

0 = 2 kg.u + 8 kg.v                                   { eq. 2 }

then    v - u = 3 m/s

v = u + 3 m/s                                        { eq. 3 }

inserting the value of 'v' in eq.2,

0 = 2 kg.u + 8 kg.(u + 3 m/s)

0 = 2 kg.u + 8 kg.u + 24 kg.m/s    

- 10 kg.u = 24 kg.m/s

u = - 2.4 m/s

now, inserting the value of 'u' in eq. 3,

v = (- 2.4 m/s) + 3 m/s

v = 0.6 m/s

velocity of block L to the left side is, u = -2.4 m/s

velocity of block R & C to the right side is, v = 0.6 m/s

(2) Next, at time t = 0.8 sec, block R is shoot to the right with a speed of 3 m/s relative to the velocity that block C. then, the velocity of the block C is given as ::

for the second explosion -

Initial momentum, p = M v = (8 kg) (0.6 m/s)

p = 4.8 kg.m/s

Final momentum, p = m1 u + m2 v       { eq. 4 }

inserting the values in eq.4,

(4.8 kg.m/s) = (6 kg) u + (2 kg) v { eq. 5 }

again using eq.3 :    v = u + 3 m/s

inserting the value of 'v' in eq.5,

(4.8 kg.m/s) = (6 kg) u + (2 kg) (u + 3 m/s)

(4.8 kg.m/s) = (6 kg.u) + (2 kg.u) + (6 kg.m/s)

- 1.2 kg.m/s = 8 kg.u

u = - 0.15 m/s

inserting the value of 'u' in eq. 3,

v = u + 3 m/s

v = (- 0.15 m/s + 3 m/s)

v = 2.85 m/s

the velocity of block C to the left is u = - 0.15 m/s.

velocity of the block R to the right is, v = 2.85 m/s

b. position of its center is given as ::

x = vt1 + ut2                            { eq. 6 }

where, t1 = 0.8 sec         &      t2 = 2.8 sec

inserting the values in eq.6,

x = (0.6 m/s) (0.8 sec) + (- 0.15 m/s) (2.8 sec)

x = 0.48 m/s - 0.42 m/s

x = 0.06 m/s

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