Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

practice problems additional practice 21. a 4.00-kg model rocket is lau…

Question

practice problems
additional practice

  1. a 4.00-kg model rocket is launched, expelling burned fuel with a mass of 50.0 g at a speed of 625 m/s. what is the velocity of the rocket after the fuel has burned? hint: ignore the external forces of gravity and air resistance.
  2. a thread connects a 1.5-kg cart and a 4.5-kg cart. after the thread is burned, a compressed spring pushes the carts apart, giving the 1.5-kg cart a velocity of 27 m/s to the left. what is the velocity of the 4.5-kg cart?
  3. challenge carmen and judi row their canoe alongside a dock. they stop the canoe, but they do not secure it. the canoe can still move freely. carmen, who has a mass of 80.0 kg, then steps out of the canoe onto the dock. as she leaves the canoe, carmen moves forward at a speed of 4.0 m/s, causing the canoe, with judi still in it, to move also. at what speed and in what direction do the canoe and judi move if their combined mass is 115 kg?

Explanation:

Response
Problem 21

Step1: Define system and momentum

Initial total momentum $p_{initial}=0$ (system at rest). Let $m_r=4.00\ \text{kg}$, $m_f=0.0500\ \text{kg}$, $v_f=-625\ \text{m/s}$ (negative = fuel direction), $v_r$ = rocket velocity.

Step2: Apply conservation of momentum

$$p_{initial}=p_{final}$$
$$0 = m_r v_r + m_f v_f$$

Step3: Solve for $v_r$

$$v_r = -\frac{m_f v_f}{m_r}$$
$$v_r = -\frac{(0.0500\ \text{kg})(-625\ \text{m/s})}{4.00\ \text{kg}}$$
$$v_r = \frac{31.25}{4.00}\ \text{m/s} = 7.8125\ \text{m/s}$$

Step1: Define system and momentum

Initial total momentum $p_{initial}=0$ (system at rest). Let $m_1=1.5\ \text{kg}$, $v_1=-27\ \text{m/s}$ (left = negative), $m_2=4.5\ \text{kg}$, $v_2$ = 4.5-kg cart velocity.

Step2: Apply conservation of momentum

$$p_{initial}=p_{final}$$
$$0 = m_1 v_1 + m_2 v_2$$

Step3: Solve for $v_2$

$$v_2 = -\frac{m_1 v_1}{m_2}$$
$$v_2 = -\frac{(1.5\ \text{kg})(-27\ \text{m/s})}{4.5\ \text{kg}}$$
$$v_2 = \frac{40.5}{4.5}\ \text{m/s} = 9\ \text{m/s}$$

Step1: Define system and momentum

Initial total momentum $p_{initial}=0$ (system at rest). Let $m_c=80.0\ \text{kg}$, $v_c=4.0\ \text{m/s}$, $m_j=115-80=35\ \text{kg}$, $v_j$ = Judi/canoe velocity.

Step2: Apply conservation of momentum

$$p_{initial}=p_{final}$$
$$0 = m_c v_c + m_j v_j$$

Step3: Solve for $v_j$

$$v_j = -\frac{m_c v_c}{m_j}$$
$$v_j = -\frac{(80.0\ \text{kg})(4.0\ \text{m/s})}{35\ \text{kg}}$$
$$v_j = -\frac{320}{35}\ \text{m/s} \approx -9.14\ \text{m/s}$$

Answer:

The velocity of the rocket is $7.81\ \text{m/s}$ (opposite to the fuel direction)

---

Problem 22