Figure 9-56 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 = 2.90 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 m/s relative to the velocity that the explosion gives to block C?
I can not paste the graph here....sorry.
they three box are placed on the ground together in this order,L,C,R
the questions are:
(a) What is the velocity of block C at
______ m/s i
(b) What is the position of block C's center at that time?
x =_______ m
WEll, the sum of all momentums will be zero.So work out the velocities for c, LR at time zero, then at time-.8, you have the initial velocity (momentum) of LR, then the momentum of R with respect to C.
a) The velocity of block C at time t = 0.80 s is 1.50 m/s.
b) The position of block C's center at that time is 0.60 m.
To solve this problem, we can use the principle of conservation of momentum. Since the rocket is initially stationary, the total momentum before and after the explosions must be equal.
Step 1: Calculate the velocities of block C and the two side blocks (L and R) at time t = 0.
Given:
Mass of central block C (M) = 6.00 kg
Mass of side blocks (L and R) (m) = 2.90 kg
Velocity of block L relative to the velocity that the explosion gives the rest of the rocket = 3.00 m/s
Since the total momentum before and after the explosion must be equal, we can write:
(0) = (M + 2m)V_c + mV_L + mV_R
where V_c is the velocity of block C, V_L is the velocity of block L, and V_R is the velocity of block R.
Since block C is initially stationary and block L is shot to the left, we have the following initial velocities:
V_c = 0 m/s
V_L = -3.00 m/s (negative because it is shot to the left)
V_R = 0 m/s (since it is initially stationary)
Substituting these values into equation (0), we have:
(0) = (6.00 kg + 2(2.90 kg))(0 m/s) + (2.90 kg)(-3.00 m/s) + (2.90 kg)(0 m/s)
Simplifying,
(0) = -8.70 kg.m/s
Step 2: Calculate the velocities of block C and the two side blocks at time t = 0.80 s.
Given:
Velocity of block R relative to the velocity that the explosion gives to block C = 3.00 m/s
Time interval = 0.80 s
Since the explosions are happening separately, we can consider them independently.
(a) For block C:
Using conservation of momentum, we can write:
(0) = (M + 2m)V_c_final + mV_L + mV_R_initial
where V_c_final is the final velocity of block C after the explosions and V_R_initial is the initial velocity of block R before the explosion.
Substituting the given values:
(0) = (6.00 kg + 2(2.90 kg))(V_c_final) + (2.90 kg)(-3.00 m/s) + (2.90 kg)(0 m/s)
Simplifying,
0 = 17.80 kg.V_c_final - 8.70 kg.m/s
Solving for V_c_final:
V_c_final = (8.70 kg.m/s) / 17.80 kg
V_c_final = 0.4888 m/s
Therefore, the velocity of block C at that time is approximately 0.4888 m/s to the right (in the positive x direction).
(b) For the position of block C's center:
To determine the position, we need to find the displacement of block C from the origin.
Using the formula:
x = x_0 + V_0t + (1/2)at^2
Since the rocket is initially stationary, the initial position, x_0, is 0 m.
The initial velocity, V_0, is also 0 m/s.
The acceleration, a, can be found using Newton's second law:
F_net = ma
Considering the forces on block C:
F_net = mC.aC
F_net = mR.(V_R_final - V_R_initial)
Since block C and block R are the only ones experiencing a force, we have:
F_net = m.(V_R_final - V_R_initial)
Substituting the given values:
F_net = (2.90 kg)(3.00 m/s) = 8.70 kg.m/s
Now, using Newton's second law, we can write:
8.70 kg.m/s = (6.00 kg + 2(2.90 kg)).aC
Simplifying,
8.70 kg.m/s = 13.80 kg.aC
Solving for aC:
aC = (8.70 kg.m/s) / 13.80 kg
aC = 0.6304 m/s^2
Now, substituting the values into the position equation:
x = (0 m) + (0 m/s)(0.80 s) + (1/2)(0.6304 m/s^2)(0.80 s)^2
x = 0.2021 m
Therefore, the position of block C's center at that time is approximately 0.2021 m in the positive x direction.
To calculate the velocity of block C at a specific time, we need to consider the conservation of momentum for the system. The initial momentum of the system is zero since the entire system starts at rest.
(a) At time t = 0, block L is shot to the left with a velocity of 3.00 m/s. The momentum of block L is given by:
PL = mL * vL
Next, at time t = 0.80 s, block R is shot to the right with a velocity of 3.00 m/s relative to the velocity that the explosion gives to block C. Therefore, the velocity of block R relative to block C is 3.00 m/s in the opposite direction. The momentum of block R relative to block C is given by:
PRC = mR * (vR - vC)
Since momentum is conserved, the sum of the momenta before and after these explosions should be equal. Thus, we can equate the initial momentum (PL) before the explosions with the final momentum (PL + PRC) after the explosions:
PL = PL + PRC
Substituting the momenta expressions, we get:
mL * vL = mL * vL + mR * (vR - vC)
Since vL = 3.00 m/s, vR = 3.00 m/s, and solving for vC, we have:
6.00 kg * 3.00 m/s - 2.90 kg * (3.00 m/s - vC) = 0
Now, we can solve this equation to find the velocity of block C at that time.
(b) To find the position of block C's center at that time, we need to integrate the velocity of block C with respect to time. By using the information provided in the question, we can evaluate the integral to obtain the position.
I'll calculate the velocity of block C first:
mL * vL = mL * vL + mR * (vR - vC)
6.00 kg * 3.00 m/s = 6.00 kg * 3.00 m/s + 2.90 kg * (3.00 m/s - vC)
18.00 kg m/s = 18.00 kg m/s + 2.90 kg * 3.00 m/s - 2.90 kg * vC
0 = 8.70 kg m/s - 2.90 kg * vC
2.90 kg * vC = 8.70 kg m/s
vC = 8.70 kg m/s / 2.90 kg
vC = 3.00 m/s
Therefore, the velocity of block C is 3.00 m/s in the positive x-direction.
To find the position of block C's center at that time, we need to integrate the velocity of block C with respect to time.
Let's assume that at t = 0, the position of the center of block C is x = 0. Then, integrating the velocity of block C over the given time interval, we can find the position of block C's center.
∫(vC) dt = x
∫(3.00 m/s) dt = x
3.00 m/s * t = x
Since t = 0.80 s, we can substitute this value back into the equation to find the position of block C's center:
3.00 m/s * 0.80 s = x
x = 2.40 m
Therefore, the position of block C's center at that time is x = 2.40 m.