Q41P Figure shows a two-ended 鈥渞ock... [FREE SOLUTION]

91影视

Chapter 9: Q41P (page 250)

Figure shows a two-ended 鈥渞ocket鈥� 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 ofmas m = 2.00 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/srelative 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 m/srelative to the velocity that block C then has. At t = 2.80 s, (a) What are the velocity of block C and (b) What are the positions of its center?

Short Answer

Expert verified

(a) The velocity of block C is $v_{2} = - (0.15) \hat{i} m / s$ .

(b) The position of its center is $v_{1} 鈭� t + v_{2} 鈭� t = 0.18 m$ .

Step by step solution

Step 1: Given data:

The mass of the block C is, $M_{c} = 6.00 kg$ .

The mass of the block L is, $m_{L} = 2.00 kg$ .

The mass of the block R is, $m_{R} = 2.00 kg$ .

At time t = 0 , speed of the block L is, $v_{L} = 3.00 m / s$ .

At time t = 0.80 s, speed of the block R is, $v_{R} = 3.00 m / s$ .

Determining the concept:

By using the conservation of momentum, find the velocity $v_{2}$ of block C. Also, by using the values $v_{1} and v_{2}$ , find the position of its center. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

The conservation of momentum is,

$m_{L} (v_{1} - v_{L}) + (m_{c} + m_{R} v_{1} = 0)$

Displacement is,

$x = v 鈭� t$

Where, $m_{L}, m_{c}, m_{R}$ are masses of block L, C, and R respectively. And v, $v_{1}$ are velocities,x is displacement, and $鈭� t$ is time period.

(a) Determining the velocity of block C:

The velocity of block L is $(v_{1} - 3) \hat{i}$ . Thus, momentum conservation gives,

$m_{L} (v_{1} - v_{L}) + (M_{c} + m_{R}) v_{1} = 0 m_{L} v_{1} + M_{c} v_{1} + m_{R} v_{1} = m_{L} v_{L} (m_{L} + M_{c} + m_{R}) v_{1} = m_{L} v_{L} v_{1} = \frac{m_{L} v_{L}}{m_{L} + M_{c} + m_{R}}$

Substitute known values in the above equation.

$v_{1} = \frac{(2.00 kg) (3.00 m / s)}{2.00 kg + 6.00 kg + 2.00 kg} = \frac{6.00 kg . m / s}{10.0 kg} = 0.60 m / s$

Thus, the velocity of block C and R together is $0.60 m / s$ .

At t = 0.80 s, momentum conservation gives,

$M_{c} v_{2} + m_{R} (v_{2} + 3 m / s) + (M_{c} + m_{R}) v_{1} M_{c} v_{2} + m_{R} v_{1} (3 m / s) = (M_{c} + m_{R}) v_{1} (m_{c} + m_{R}) v_{2} = (M_{c} + m_{R}) v_{1} - m_{R} (3 m / s) v_{1} = \frac{(M_{c} + m_{R}) v_{1} - m_{R} (3 m / s)}{(M_{c} + m_{R})}$

Substitute known values in the above equation.

$v_{2} = \frac{(6.00 kg + 2.00 kg) (0.60 m / s) - (2.00 kg) (3 m / s)}{(6.00 kg + 2.00 kg)} = \frac{(4.8 kg . m / s) - (6 kg . m / s)}{(8.00 kg)} = \frac{1.2 kg . m / s}{8.00 kg} = 0.15 m / s$